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# Selina solutions for Concise Mathematics Class 8 ICSE chapter 8 - Profit, Loss and Discount [Latest edition]

## Chapter 8: Profit, Loss and Discount

Exercise 8 (A)Exercise 8 (B)Exercise 8 (C)Exercise 8 (D)Exercise 8 (E)Exercise 8 (F)
Exercise 8 (A) [Page 90]

### Selina solutions for Concise Mathematics Class 8 ICSE Chapter 8 Profit, Loss and Discount Exercise 8 (A) [Page 90]

Exercise 8 (A) | Q 1 | Page 90

Megha bought 10 note-books for Rs.40 and sold them at Rs.4.75 per note-book. Find, her gain percent.

Exercise 8 (A) | Q 2 | Page 90

A fruit-seller buys oranges at 4 for Rs.3 and sells them at 3 for Rs.4 Find his profit percent.

Exercise 8 (A) | Q 3 | Page 90

A man buys a certain number of articles at 15 for Rs. 112.50 and sells them at 12 for Rs.108. Find ;
(i) his gain as percent;
(ii) the number of articles sold to make a profit of Rs.75.

Exercise 8 (A) | Q 4 | Page 90

A boy buys an old bicycle for Rs. 162 and spends Rs. 18 on its repairs before selling the bicycles for Rs. 207. Find his gain or loss percent.

Exercise 8 (A) | Q 5 | Page 90

An article is bought from Jaipur for Rs. 4,800 and is sold in Delhi for Rs. 5,820. If Rs. 1,200 is spent on its transportations, etc. ; find he loss or the gain as a percent.

Exercise 8 (A) | Q 6 | Page 90

Mohit sold a T.V. for Rs. 3,600; gaining one-sixth of its selling price. Find :
(i) the gain
(ii) the cost price of the T.V.
(iii) the gain percent.

Exercise 8 (A) | Q 7 | Page 90

By selling a certain number of goods for Rs. 5,500; a shopkeeper loses equal to one-tenth of their selling price. Find :
(i) the loss incurred
(ii) the cost price of the goods
(iii) the loss as a percent.

Exercise 8 (A) | Q 8 | Page 90

The selling price of a sofa-set is 4/5 times of its cost price. Find the gain or the loss as a percent.

Exercise 8 (A) | Q 9 | Page 90

The cost price of an article is 4/5 times of its selling price. Find the loss or the gain as a percent.

Exercise 8 (A) | Q 10 | Page 90

A shopkeeper sells his goods at 80% of their cost price. Find the percent gain or losses?

Exercise 8 (A) | Q 11 | Page 90

The cost price of an article is 90% of its selling price. What is the profit or the loss as a percent ?

Exercise 8 (A) | Q 12 | Page 90

The cost price of an article is 30 percent less than its selling price. Find, the profit or loss as a percent.

Exercise 8 (A) | Q 13 | Page 90

A shop-keeper bought 300 eggs at 80 paise each. 30 eggs were broken in a transaction and then he sold the remaining eggs at one rupee each. Find, his gain or loss as a percent.

Exercise 8 (A) | Q 14 | Page 90

A man sold his bicycle for Rs.405 losing one-tenth of its cost price, find :
(i) its cc price;
(ii) the loss percent.

Exercise 8 (A) | Q 15 | Page 90

A man sold a radio-set for Rs.250 and gained one-ninth of its cost price. Find ;
(i) its cost price;
(ii) the profit percent.

Exercise 8 (B) [Page 93]

### Selina solutions for Concise Mathematics Class 8 ICSE Chapter 8 Profit, Loss and Discount Exercise 8 (B) [Page 93]

Exercise 8 (B) | Q 1.1 | Page 93

Find the selling price, if:
C.P. = Rs. 950 and profit = 8%

Exercise 8 (B) | Q 1.2 | Page 93

Find the selling price, if:
C.P. = Rs. 1,300 and loss = 13%

Exercise 8 (B) | Q 2.1 | Page 93

Find the cost price, if :
S.P. = Rs. 1,680 and profit = 12%

Exercise 8 (B) | Q 2.2 | Page 93

Find the cost price, if :
S.P. = Rs. 1,128 and loss = 6%

Exercise 8 (B) | Q 3 | Page 93

By selling an article for Rs.900; a man gains 20%. Find his cost price and the gain.

Exercise 8 (B) | Q 4 | Page 93

By selling an article for Rs.704; a person loses 12%. Find his cost price and the loss.

Exercise 8 (B) | Q 5.1 | Page 93

Find the selling price, if :
C.P. = Rs.352; overheads = Rs.28 and profit = 20

Exercise 8 (B) | Q 5.2 | Page 93

Find the selling price, if :
C.P. = Rs.576; overheads = Rs.44 and loss = 16%

Exercise 8 (B) | Q 6 | Page 93

If John sells his bicycle for Rs. 637, he will suffer a loss of 9%. For how much should it be sold, if he desires a profit of 5%?

Exercise 8 (B) | Q 7 | Page 93

A man sells a radio-set for Rs.605 and gains 10%. At what price should he sell another radio of the same kind, in order to gain 16%?

Exercise 8 (B) | Q 8 | Page 93

By selling a sofa-set for Rs.2,500; the shopkeeper loses 20%. Find his loss percent or profit percent; if he sells the same sofa-set for Rs.3150.

Exercise 8 (B) | Q 9 | Page 93

Mr. Sinha sold two tape-recorders for Rs.990 each; gaining 10% on one and losing 10% on the other. Find his total loss or gain as a percent on the whole transaction.

Exercise 8 (B) | Q 10 | Page 93

A tape-recorder is sold for Rs. 2,760 at a gain of 15% and a C.D. player is sold for Rs. 3,240 at a loss of 10% Find :
(i) the C.P. of the tape-recorder
(ii) the C.P. of the C.D. player.
(iii) the total C.P. of both.
(iv) the total S.P. of both
(v) the gain % or the loss % on the whole

Exercise 8 (B) | Q 11 | Page 93

Rajesh sold his scooter to Rahim at 8% loss and Rahim, in turn, sold the same scooter to Prem at 5% gain. If Prem paid Rs. 14,490 for the scooter; find :
(i) the S.P. and the C.P. of the scooter for Rahim
(ii) the S.P. and the C.P. of the scooter for Rajesh

Exercise 8 (B) | Q 12 | Page 93

John sold an article to Peter at 20% profit and Peter sold it to Mohan at 5% loss. If Mohan paid Rs.912 for the article; find how much did John pay for it?

Exercise 8 (C) [Page 95]

### Selina solutions for Concise Mathematics Class 8 ICSE Chapter 8 Profit, Loss and Discount Exercise 8 (C) [Page 95]

Exercise 8 (C) | Q 1 | Page 95

A stationery buys pens at 5 for Rs.28 and sells them at a profit of 25 %. How much should a customer pay; if he buys
(i) only one pen ;
(ii) three pens?

Exercise 8 (C) | Q 2 | Page 95

A fruit-seller sells 4 oranges for Rs. 3, gaining 50%. Find :
(i) C.P. of 4 oranges,
(ii) C.P. of one orange.
(iii) S.P. of one orange.
(iv) profit made by selling one orange.
(v) a number of oranges brought and sold in order to gain Rs. 24.

Exercise 8 (C) | Q 3 | Page 95

A man sells 12 articles for Rs. 80 gaining 33 1/3%. Find the number of articles bought by the man for Rs. 90.

Exercise 8 (C) | Q 4 | Page 95

The cost price of 20 articles is the same as the selling price of 16 articles. Find the gain percent.

Exercise 8 (C) | Q 5 | Page 95

The selling price of 15 articles is equal to the cost price of 12 articles. Find the gain or loss as a percent.

Exercise 8 (C) | Q 6 | Page 95

By selling 8 pens, Shyam loses equal to the cost price of 2 pens. Find his loss percent.

Exercise 8 (C) | Q 7 | Page 95

A shop-keeper bought rice worth Rs.4,500. He sold one-third of it at 10% profit.
If he desires a profit of 12% on the whole; find :
(i) the selling price of the rest of the rice ;
(ii) the percentage profit on the rest of the rice.

Exercise 8 (C) | Q 8 | Page 95

Mohan bought a certain number of note-books for Rs.600. He sold 1/4 of them at 5 percent loss. At what price should he sell the remaining note-books so as to gain 10% on the whole?

Exercise 8 (C) | Q 9 | Page 95

Raju sells a watch at a 5% profit. Had he sold it for Rs.24 more; he would have gained 11%. Find the cost price of the watch.

Exercise 8 (C) | Q 10 | Page 95

A man sold a bicycle at 5% profit. If the cost had been 30% less and the selling price Rs.63 less, he would have made a profit of 30%. What is the cost price of the bicycle?

Exercise 8 (C) | Q 11 | Page 95

Renu sold an article at a loss of 8 percent. Had she bought it at 10% less and sold for Rs.36 more; she would have gained 20%. Find the cost price of the article.

Exercise 8 (D) [Page 98]

### Selina solutions for Concise Mathematics Class 8 ICSE Chapter 8 Profit, Loss and Discount Exercise 8 (D) [Page 98]

Exercise 8 (D) | Q 1 | Page 98

An article is marked for Rs. 1,300 and is sold for Rs. 1,144; find the discount percent.

Exercise 8 (D) | Q 2 | Page 98

The marked price of a dinning table is Rs. 23,600 and is available at a discount of 8%. Find its selling price.

Exercise 8 (D) | Q 3 | Page 98

A wrist-watch is available at a discount of 9%. If the list-price of the watch is Rs. 1,400; find the discount is given and the selling price of the watch.

Exercise 8 (D) | Q 4 | Page 98

A shopkeeper sells an article for Rs. 248.50 after allowing a discount of 10%. Find the list price of the article.

Exercise 8 (D) | Q 5 | Page 98

A shop-keeper buys an article for Rs.450. He marks it at 20% above the cost price. Find :
(i) the marked price of the article.
(ii) the selling price, if he sells the articles at 10 percent discount.
(iii) the percentage discount given by him, if he sells the article for Rs.496.80

Exercise 8 (D) | Q 6 | Page 98

The list price of an article is Rs.800 and is available at a discount of 15 percent. Find :
(i) selling price of the article ;
(ii) cost price of the article, if a profit of 13 1/3 % is made on selling it.

Exercise 8 (D) | Q 7 | Page 98

An article is marked at Rs. 2,250. By selling it at a discount of 12%, the dealer makes a profit of 10%. Find :
(i) the selling price of the article.
(ii) the cost price of the article for the dealer.

Exercise 8 (D) | Q 8 | Page 98

By selling an article at a 20% discount, a shopkeeper gains 25%. If the selling price of the article is Rs. 1,440 ; find :
(i) the marked price of the article.
(ii) the cost price of the article.

Exercise 8 (D) | Q 9 | Page 98

A shopkeeper marks his goods at 30 percent above the cost price and then gives a discount of 10 percent. Find his gain percent.

Exercise 8 (D) | Q 10 | Page 98

A ready-made garments shop in Delhi, allows a 20 percent discount on its garments and still makes a profit of 20 percent. Find the marked price of a dress which is bought by the shop-keeper for Rs.400.

Exercise 8 (D) | Q 11 | Page 98

At 12% discount, the selling price of a pen is Rs. 13.20. Find its marked price. Also, find the new selling price of the pen, if it is sold at 5% discount.

Exercise 8 (D) | Q 12 | Page 98

The cost price of an article is Rs. 2,400 and it is marked at 25% above the cost price. Find the profit and the profit percent, if the article is sold at a 15% discount.

Exercise 8 (D) | Q 13 | Page 98

Thirty articles are bought at Rs. 450 each. If one-third of these articles be sold at a 6% loss; at what price must each of the remaining articles be sold in order to make a profit of 10% on the whole?

Exercise 8 (D) | Q 14 | Page 98

The cost price of an article is 25% below the marked price. If the article is available at a 15% discount and its cost price is Rs. 2,400; find:
(i) Its marked price
(ii) its selling price
(iii) the profit percent.

Exercise 8 (D) | Q 15.1 | Page 98

Find a single discount (as a percent) equivalent to following successive discounts:
20% and 12%

Exercise 8 (D) | Q 15.2 | Page 98

Find a single discount (as a percent) equivalent to following successive discounts:
10%, 20% and 20%

Exercise 8 (D) | Q 15.3 | Page 98

Find a single discount (as a percent) equivalent to following successive discounts:
20%, 10% and 5%

Exercise 8 (D) | Q 16.1 | Page 98

Find the single discount (as a percent) equivalent to successive discounts of:
80% and 80%

Exercise 8 (D) | Q 16.2 | Page 98

Find the single discount (as a percent) equivalent to successive discounts of:
60% and 60%

Exercise 8 (D) | Q 16.3 | Page 98

Find the single discount (as a percent) equivalent to successive discounts of:
60% and 80%

Exercise 8 (E) [Page 101]

### Selina solutions for Concise Mathematics Class 8 ICSE Chapter 8 Profit, Loss and Discount Exercise 8 (E) [Page 101]

Exercise 8 (E) | Q 1 | Page 101

Rajat purchases a wrist-watch costing Rs 540. The rate of sales tax is 8%. Find the total amount paid by Rajat for the watch.

Exercise 8 (E) | Q 2 | Page 101

Ramesh paid Rs 345.60 as sales tax on a purchase of Rs 3,840. Find the rate of sales tax.

Exercise 8 (E) | Q 3 | Page 101

The price of a washing machine, inclusive of sales tax is Rs 13,530/-. If the sales tax is 10%, find its basic cost price.

Exercise 8 (E) | Q 4 | Page 101

Sarita purchases biscuits costing Rs 158 on which the rate of sales tax is 6%. She also purchases some cosmetic goods costing Rs 354 on which the rate of sales tax is 9%. Find the total amount to be paid by Sarita.

Exercise 8 (E) | Q 5 | Page 101

The price of a T.V. set inclusive of sales tax of 9% is Rs 13,407. Find its marked price. If sales tax is increased to 13%, how much more does the customer has to pay for the T.V.?

Exercise 8 (E) | Q 6 | Page 101

The price of an article is Rs 8,250 which includes sales tax at 10%. Find how much more or less does a customer pay for the article, if the sales tax on the article:

(i) increases to 15%
(ii) decreases to 6%
(iii) increases by 2%
(iv) decreases by 3%

Exercise 8 (E) | Q 7 | Page 101

A bicycle is available for Rs 1,664 including sales tax. If the list price of the bicycle is Rs 1,600, find :

(1) the rate of sales tax.

(2) the price, a customer will pay for the bicycle if the sales tax is increased by 6%.

Exercise 8 (E) | Q 8 | Page 101

When the rate of sale-tax is decreased from 9% to 6% for a coloured T.V.; Mrs. Geeta will save Rs 780 in buying this T.V. Find the list price of the T.V.

Exercise 8 (E) | Q 9 | Page 101

A shopkeeper sells an article for Rs. 21384 including 10% sales tax. However, the actual rate of sales tax is 8%. Find the extra profit made by the dealer.

Exercise 8 (F) [Page 104]

### Selina solutions for Concise Mathematics Class 8 ICSE Chapter 8 Profit, Loss and Discount Exercise 8 (F) [Page 104]

Exercise 8 (F) | Q 1 | Page 104

Some goods/services cost Rs. 16,000 and the rate of GST on them is 12%. Find the amount of the bill, in the case of:

1. intra-state transaction.
2. inter-state transaction.
Exercise 8 (F) | Q 2 | Page 104

John belongs to Delhi. He buys goods, worth Rs. 25,000 from a shop in Delhi. If the rate of GST is 5%, find how much money in all, will Does John pay for these goods?

Exercise 8 (F) | Q 3 | Page 104

Find the amount of bill for the following inter-state transaction of goods/services:
Cost of transaction = 30,000; discount = 30% and GST=28%.

Exercise 8 (F) | Q 4 | Page 104

For both the following inter-state transaction of services, find the total amount of bill.
(i) Cost of services = 5,000, discount = 20% and GST = 12%
(ii) Cost of services = 12,500, discount = 40% and GST = 18%

Exercise 8 (F) | Q 5 | Page 104

A shopkeeper in Indore sells 20 identical articles for ₹450 each. Find the amount of bill if he gives a 20% discount and then charges GST = 28%

Exercise 8 (F) | Q 6 | Page 104

A dealer in Bihar supplied goods to a dealer in Mumbai. The dealer in Mumbai buys :
(i) 40 articles for 800 each at 30% discount
(ii) 75 articles for 1,000 each at 20% discount.

If the rate of GST, on the whole, is 12%, find how much will the dealer at Mumbai pay to a dealer in Bihar.

## Chapter 8: Profit, Loss and Discount

Exercise 8 (A)Exercise 8 (B)Exercise 8 (C)Exercise 8 (D)Exercise 8 (E)Exercise 8 (F)

## Selina solutions for Concise Mathematics Class 8 ICSE chapter 8 - Profit, Loss and Discount

Selina solutions for Concise Mathematics Class 8 ICSE chapter 8 (Profit, Loss and Discount) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CISCE Concise Mathematics Class 8 ICSE solutions in a manner that help students grasp basic concepts better and faster.

Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. Selina textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

Concepts covered in Concise Mathematics Class 8 ICSE chapter 8 Profit, Loss and Discount are Overhead Expenses, To Find S.P., When C.P. and Gain (Or Loss) Percent Are Given, To Find C.P., When S.P. and Gain (Or Loss) Percent Are Given, Concept of Discount, Computation of Tax, Goods and Service Tax (Gst), Gst Comprises of, Concept of Discount.

Using Selina Class 8 solutions Profit, Loss and Discount exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Selina Solutions are important questions that can be asked in the final exam. Maximum students of CISCE Class 8 prefer Selina Textbook Solutions to score more in exam.

Get the free view of chapter 8 Profit, Loss and Discount Class 8 extra questions for Concise Mathematics Class 8 ICSE and can use Shaalaa.com to keep it handy for your exam preparation


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# How to determine which gas is more soluble

I have a chemistry homework problem that asks whether $Br_2$ or $O_2$ is more soluble in water in room temperature. However, I can't seem to find a way to figure it out, short of looking it up.

Is there a guiding rule for such figures, such as relating to molar mass?

• The answer the question is seeking might be simpler than you think. The most obvious difference between oxygen and bromine is that bromine is a liquid not a gas a room temperature. Might this be significant? – matt_black Jan 18 '14 at 13:52

$\ce{O_2}$ and $\ce{Br_2}$ have no permanent dipole so the only interactions are Debye forces (interaction between water permanent dipole and a gas molecule induced dipole) and London forces (between instantaneous dipoles).
This is the reason why van der waals forces are stronger for bigger atoms. This would explain why $\ce{Br_2}$ has a better solubility than $\ce{O_2}$ in water.
London interactions are very important if you look at liquefaction temperature of gases. For instance, it explains why $\ce{Ar}$ becomes a liquid below 87K while $\ce{He}$ become a liquid only below 4K.

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## General Chemistry: Principles and Modern Applications (10th Edition)

$[H_3O^+] = 0.026 \space M$ $[HS{O_4}^-] = 0.014 \space M$ $[S{O_4}^{2-}] = 6.0 \times 10^{-3} \space M$
1. Draw the ICE table for the first and second reactions: $$\begin{vmatrix} Compound& [ H_2SO_4 ]& [ HS{O_4}^- ]& [ H_2O ]\\ Initial& 0.020 & 0 & 0 \\ Change& -0.020& +0.020& +0.020\\ Equilibrium& 0& 0.020& 0.020 \\ \end{vmatrix}$$ Since $H_2SO_4$ is a strong acid, it will react completely, so x = 0.020 $$\begin{vmatrix} Compound& [ HS{O_4}^- ]& [ S{O_4}^{2-} ]& [ H_3O^+ ]\\ Initial& 0.020 & 0 & 0.020 \\ Change& -x& +x& +x\\ Equilibrium& 0.020 -x& 0 +x& 0.020 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ S{O_4}^{2-} ][ H_3O^+ ]}{[ HS{O_4}^- ]}$$ $$K_a = \frac{(x)( 0.020 + x)}{0.020 - x}$$ 3. Assuming $0.020 \gt\gt x:$ $$K_a = \frac{(x)( 0.020 )}{0.020}$$ $$x = \frac{K_a \times [ HS{O_4}^- ]_{initial}}{ 0.020 } = \frac{ 1.1 \times 10^{-2} \times 0.020 }{ 0.020 }$$ $x = 0.011$ 4. Test if the assumption was correct: $$\frac{ 0.011 }{ 0.020 } \times 100\% = 55.0 \%$$ The percent is greater than 5%, therefore, the approximation is invalid. 5. Return for the original expression and solve for x: $$K_a = \frac{( 0.020 + x)(x)}{[ HS{O_4}^- ]_{initial} - x}$$ $$K_a [ HS{O_4}^- ] - K_a x = 0.020 x + x^2$$ $$x^2 + (K_a + 0.020 ) x - K_a [ HS{O_4}^- ] = 0$$ $$x_1 = \frac{- ( 1.1 \times 10^{-2} + 0.020 )+ \sqrt{( 1.1 \times 10^{-2} + 0.020 )^2 - 4 (1) (- 1.1 \times 10^{-2} ) ( 0.020 )} }{2 (1)}$$ $$x_1 = 6.0 \times 10^{-3}$$ $$x_2 = \frac{- ( 1.1 \times 10^{-2} + 0.020 )- \sqrt{( 1.1 \times 10^{-2} + 0.020 )^2 - 4 (1) (- 1.1 \times 10^{-2} ) ( 0.020 )} }{2 (1)}$$ $$x_2 = -0.037$$ - The concentration cannot be negative, so $x_2$ is invalid. $$x = 6.0 \times 10^{-3}$$ 6. Determine the hydronium concentration: $$[H_3O^+] = x + [H_3O^+]_{initial} = 6.0 \times 10^{-3} + 0.020 = 0.026$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.026 ) = 1.59$$ $[H_3O^+] = 0.026 \space M$ $[HS{O_4}^-] = 0.020 - x = 0.014 \space M$ $[S{O_4}^{2-}] = x = 6.0 \times 10^{-3} \space M$


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# 20. Given f(x) = and g(x) = x 3 find the following: a_ (f . g)(x) b the domain of

###### Question:

20. Given f(x) = and g(x) = x 3 find the following: a_ (f . g)(x) b the domain of (f g)(x) in interval notation C. (g o f)(x) d. the domain of (g o f)(x) e [ X

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# Could PPCG benefit from a new sorting mode for answers?

The default sorting for answers (by votes) seems somewhat inappropriate here on PPCG. In any case it tends to put early answers over late answers, because once a few are there, few people scroll to the bottom to see newer answers and even fewer people sort by activity.

If you consider that better answers will usually take longer to come up with and implement, that means the answers that deserve the votes are actually at a disadvantage.

This is problematic for different reasons on different challenge types:

### Popularity Contest

Sorting by votes is still useful here, because that actually shows you the ranking of the answers by the winning criterion. However, for that very reason (that votes determine the winner) a bias towards early answers can't be in the interest of the community.

### Other Challenges

Here, sorting by votes doesn't even make really make sense in the first place. The reason this is the default sorting on other SE sites is that it's a decent heuristic for which answer is the best. But we don't need that, because the quality of an answer can be determined objectively by the winning criterion. Of course, sorting by winning criterion is not an option, but sorting by votes doesn't seem useful either. Hence, some other sorting could at least help later and better answers get the votes they deserve.

So, I'd say we should think about different, more appropriate, sorting modes and also consider whether the default sorting should be changed. If we can come to a community consensus here, we might be able to bring this to the attention of SE staff.

The questions to discuss are: are the existing sorting modes sufficient or do we need new ones (which ones)? Should a different (new or existing) sorting mode be the default?

• – Ilmari Karonen Aug 21 '14 at 14:09
• @IlmariKaronen While useful this doesn't really solve the problem, because it's especially about new/one-time users who get here via HNQ or otherwise and are just here to vote on popular questions. – Martin Ender Aug 21 '14 at 14:12
• True. I was just pointing out that a similar issue has been discussed here before, and that, based on the feedback I've got for the script (and Nathan Osman's earlier similar script), a significant fraction of CGSE users do seem to feel that the existing sorting methods are inadequate. That said, the issue you raise seems to be somewhat different, as it mainly concerns popularity-contest questions, whereas my script is mainly useful for code-golf. Still, I felt it was potentially relevant enough to at least link to. – Ilmari Karonen Aug 21 '14 at 14:19

# Measuring the Strength of the FGITW Effect

Although this answer isn't a suggestion for a solution, it should provide a better understanding of the problem. A proper understanding is necessary to create a solution.

The FGITW effect ("fastest gun in the west" effect) is the phenomenon in which the earliest answers receive the most votes and always stay at the top of the answer list. The feedback loop is this:

early answer -> some votes -> top of the answer list -> more visibility -> even more votes

Although I think everyone recognizes the existence of FGITW, people might not have an idea as to its power. How can we measure the FGITW effect in an objective way? With the help of user Eric Tressler, some data was collected and measured.

We decided to look at the RPSLS tournament as an example of FGITW. This challenge was posted about 3.5 weeks ago. Here is the data (collected by Eric):

// Score, date posted rank, #votes for Rock Paper Scissors Lizard Spock
n = 65
64  28 18
62  55 1
62  65 2
61  64 1
60  29 5
58  27 3
57  32 2
53  61 1
51  42 1
51  16 20
49  31 2
49  49 1
47  34 3
46  56 1
45  24 4
44  48 1
43  40 4
42  20 4
42  53 1
41  12 2
41  13 3
40  23 3
40  46 2
39  62 1
38  38 4
38  33 2
38  43 2
37  50 1
35  15 8
35  58 1
35  54 1
34  3  4
33  10 3
32  4  8
32  44 1
32  14 5
32  17 2
31  19 8
31  45 2
31  51 1
30  1  3
30  60 1
29  22 4
28  25 3
27  57 1
25  30 2
24  2  7
24  39 3

The three variables are:

• Score (s), showing how well the post performed in the contest
• Date rank (d), with lower numbers being the earliest answers
• Votes (v), which is just the net number of votes (up - down)

We wanted to find the relationships between these variables. We used Spearman's Rank Correlation Coefficient (Rho) to measure the correlation between the variables.

Dates-Scores:

The above dates-scores graph shows how performance was related to submission time. You can see how the very best entries (high 50s and 60s) came later in the competition, while the earliest submissions generally performed below-average (the average score was 34.6615).

We calculated rho = 0.243443, so there was a moderately-weak correlation between newer posts and higher scores. This is as expected, since the later bots could be tailored to the competition. Also, the best posts would probably take longer to write.

Given a sample size of 65, the standard error for rho is approximately 0.0791, so these results are significant.

Dates-Votes:

This is the interesting part - the FGITW. Here is the graph, which speaks for itself.

This is the FGITW in full force, you can clearly see how the most up-voted answers are typically early answers. The most-voted answers were all early in the competition. After a certain point (the latest 18), no answer has more than 2 upvotes. Although we would expect older posts to have some more votes, simply by virtue of being older, it is clear that the FGITW Effect is strong.

We calculated rho = -0.766193, which is a moderately strong correlation between earlier posts and more votes. This is objective evidence as to the strength of the FGITW, and I believe that the age of the post is one of the most important factors in determining the number of votes an answer will receive.

Scores-Votes:

While the relationship between dates and votes should show the strength of the effect, the relationship between scores and votes should show why this is a problem.

The above graph shows the relationship between performance and the number of votes. With the exception of a few outliers, this third graph looks rather similar to the second graph. The submissions with lower scores actually had more votes than most of the submissions with higher scores.

We calculated rho = -0.250174, which is a moderately weak correlation between "higher score" and "fewer votes."

A negative correlation is the opposite of what we should want. Submissions with better performance should receive more votes. I suspect that this is due to the FGITW as well. As we saw above, earlier posts receive more votes, while later posts perform better. It turns out that this actually causes a negative correlation between score (performance) and votes.

• As a side note (since this is not the main issue here), if the winning answer (or the successive winning answers in the case of a koth that runs during a few weeks) has been accepted, it would have given it more visibility (and thus theoretically more votes), hence the importance of accepting answers even if that means unaccepting later when a better answer comes along. – plannapus Aug 20 '14 at 6:21
• @plannapus But that still only fixes the problem for one answer at a time. PhiNotPi, it should also be noted that the two outliers on the final plot are also outliers on the previous plot, so they aren't outliers because they go a lot of votes for being good, but they are outliers because they were good despite being early posts. – Martin Ender Aug 20 '14 at 13:24

## Random Sorting

I propose that a purely random default sorting would solve the above problems. It would simply remove the bias towards any kind of answer and would distribute the votes more fairly. Of course, the existing sorting by activity will also remove the bias towards early answers, but the fact that it's deterministic is still problematic.

Especially in cases where there are so many answers that few people read all of them, some answers are just almost never seen because they are too far down the list (depending on which sorting is active). If the sorting was random by default, people still wouldn't scroll through all of them, but at least each answer would get the same number of views on average.

I would even go so far to suggest that this default sorting is not remembered between visiting different questions (currently, if you switch to sorting by activity, all questions will be sorted by activity until you switch back).

Note: random sorting has been suggested before a few years ago back on SO but without much response. I don't even think it's that useful on other SE sites, but PPCG could really use it.

### Pros

• Trivial to implement.
• All answers are equally likely to be seen. So there is no bias towards any answers, hence it's as fair as it gets.

### Cons

• All answers are equally likely to be seen. Even the bad ones.

(However, this would lead to really bad answers getting more downvotes and hence being more likely to be deleted, resulting in less overall clutter. So maybe this is another pro after all.)

• I think we need a compromise that gives new posts an equal footing without being random. Ideally we want good posts at the top, with posts with few votes getting some time at the top but only until they have been judged, then not seeing the top again if they were not well received in the first few days. – trichoplax Aug 12 '14 at 14:24
• @githubphagocyte that sound's great, but might be somewhat hard to implement. Good idea though. – tomsmeding Aug 12 '14 at 20:37
• @tomsmeding I think it might be easier than you'd think. Even something as simple as votes/time has that basic behaviour. The algorithm described by Randall Munroe in my answer also has that behaviour, although on this site we'd need to include time too as we don't use downvotes in quite the same way. – trichoplax Aug 12 '14 at 21:51

# Randall Munroe - a statistical estimation ordering

When reddit had a similar problem with new comments getting lost at the bottom, Randall Munroe brought in a new sorting algorithm. Obviously don't just vote for this because you like Randall Munroe - it's an interesting algorithm well worth reading about before you decide.

It only requires the information that stack exchange already uses to sort by votes anyway (upvotes and downvotes), but deals elegantly with precisely the problem we're discussing. The example at the end demonstrates this nicely.

The algorithm has the best features of sorting by votes, sorting randomly, and sorting by time, without actually using any randomness or time, just upvotes and downvotes.

In short: The Randall Munroe algorithm works well on reddit because it uses both upvotes and downvotes to measure how statistically likely a post is to get to the top, and orders by the expected future position rather than just its current votes. So it doesn't need time because the downvotes give a percentage measure rather than just a plain count of upvotes The problem with applying that to stack exchange is that downvotes are a rare thing here - most answers have zero downvotes, so we might need to take a different approach that bases the statistical measure on upvotes, downvotes, and time. These are all already available to the sorting code of stack exchange so it is feasible to implement.

• Ah - I see Firefly beat me to it by 59s... :) – trichoplax Aug 12 '14 at 13:42
• Actually this isn't the same algorithm Firefly is describing - reddit uses "hot" and "best". This one is "best" and isn't about time but statistical confidence. – trichoplax Aug 12 '14 at 13:48
• This is a very interesting approach though as mention in several places, our downvotes don't work like the ones on reddit, so we would either need to use time, (unique) question views, total votes or a combination thereof. I'd also suggest using a more optimistic approach to the statistics, using the best score with 5% confidence for instance, thus favoring new answers during their first few votes, even if things don't look superb. – overactor Aug 12 '14 at 14:53
• It could also be interesting to use higher statistical confidence for lower rep and users and people that are not logged in, users with higfh rep would use a low statistical significance. This way trusted users get presented with more content that still needs a concensus and new users see the entries that can safely be assumed to be at least decent to good. – overactor Aug 12 '14 at 15:01
• I don't think that this method would have much impact, since downvotes are really quite rare. It is entirely feasible that the top N posts to a question all have 0 downvotes, in which case the ordering would be the same and we would still encounter the problem of the first posts getting all the visibility and votes. – PhiNotPi Aug 14 '14 at 15:00
• @PhiNotPi yes I agree. That's why I think it would need to include time too. – trichoplax Aug 14 '14 at 15:04

## Reddit-style sorting

I don't remember exactly how it works, but Reddit's sorting algorithm (or HN's, or ...) takes both time and votes into account to present "hot" content, i.e. content climbing faster than typical or with higher than typical votecount. In practice high-voted content tends to be the norm, with new or rising content spliced in occasionally.

Our situation is a bit different: we don't want sufficiently old answers to suddenly drop quickly the same way as Reddit does, and we have an additional "active" parameter separate from the time since initial answer. Still, Reddit's (or a similar service's) algorithm could serve as a base for a good sorting algorithm. Reddit-style sorting might also be of interest in other SE subsites as well (moreso than random sorting, at least), so there's more reason for SE to provide it.

I think it'd be unfair to high-voted entries with completely random sorting, and even though we don't need the vote-based sorting for questions with something other than popularity-contest as the winning criterion it still makes sense to promote high-voted answers since they presumably follow the winning criterion well (or otherwise entertain the voters).

• Well SE has this feature for questions already (the HNQ sidebar). Nevertheless, this would probably be a nice solution for most challenges, but for pop-cons it would still bias some answers which may or may not be fair (in this case, the bias might be reasonable, because it favours new answers which get lots of votes, which means they are likely actually better). – Martin Ender Aug 12 '14 at 13:45
• Since you got in just before me, feel free to edit in the link from my answer and combine it into one. The "hot" sorting algorithm is actually unrelated to the one I mention, and I agree wouldn't suit our purposes, but Randall Munroe's "best" sorting algorithm maintains an order based on statistical confidence in the comment which favours new answers but only until they have enough votes to order them better - so old answers still stay high if voted highly, and never move off the top spot for long. – trichoplax Aug 12 '14 at 13:45
• @githubphagocyte I think we shoud keep them separate as the posts are about different algorithms. Actually, my suggestion isn't so much about using an exact algorithm as coming up with a new one based on it... which is a bit vague I guess. The biggest issue with applying Reddit's Hot as is would be how old posts fall off the frontpage. – FireFly Aug 12 '14 at 13:48

## Not enough information given

I can't agree to change the default sort at all unless more evidence is given that the basic premise is true.

PhiNotPi gives some data showing that the FGITW effect happens, and I don't doubt that. However, the only correlation it really shows is between early answers and votes. For one question. With tons of answers.

To be clear, 80 answers is just not typical. The site as a whole averages a bit under 10 answers per question. Certain challenges might garner 30+, but there aren't very many, and they just drag the average up. I don't think the default sort for all questions should be changed for the sake of those few.

...few people scroll to the bottom to see newer answers and even fewer people sort by activity.

This could be true. It also might not be. A simpler explanation to explain FGITW could be that early answers get more page views, so have a better chance of getting an upvote. Even if every user scrolls to see every answer on every question, the early ones still get seen more.

Changing the sort method won't change that, except (most likely) on questions that go 30+ answers and paginate. I think it likely that on most questions, people read at least the majority of answers, because there aren't very many anyway. In addition, since the vast majority of challenges are , there are typically <10 short answers.

• Note: This doesn't apply to the mobile app. There are no sort options there (votes only), and you can't even get to page 2 of answers from what I can tell. In addition, you can't jump to the newest answer/edit from the main page like you can on the actual site. It's a damn mess. – Geobits Aug 20 '14 at 13:25

# Vote Order + Semi-Random Permutation

A combination of votes and a randomizer might be best.

One method of accomplishing this is to take the placement of each answer (1, 2, 3, etc.) and add a random real number 0-5 to that number. Then, sorting by these new numbers gives the new order. The goal is to create a semi-random permutation of the order such that no answer is too far from its original location.

## Pros:

• The most upvoted answers are still near the top, with just enough randomization that the not-quite-as-upvoted answers also get the change to be near the top.
• Answers with no votes are still lower down, but aren't always dead last.
• The "catchy" answers are still visible to random visitors, which helps create interest in this site.
• Simpler than other methods.
• Changing the 0-5 range to something bigger (like 0-10) is an easy way to increase randomness. As the range becomes larger, it is closer to the random sorting. If is is smaller, it is closer to vote order.

## Cons:

• Newer answers still have an advantage if the choice of range is small, as highly-voted answers will remain visible.
• If there are a ton of 0-score answers, most of those answers will end up being swapped around with each other.
• The con of newer answers/highly voted answers still having the advantage outweighs the pros in this one, and thus won't solve the problem. – Isiah Meadows Aug 14 '14 at 23:30

## Why sorting by votes is a good idea

PPCG benefits a lot from people stumbeling upon popular questions, when they open thsi question, we definitely want to showcase the best and most interesting answers. In many categories, there is a very clear difference between good and interesting answers. This answer explains this better than I ever could..

## Why sorting by votes is a terrible idea

Well, basically, for all the reasons you mentioned, late answers don't get the attention they deserve and the gap between early answers and late answers only gets bigger over time (especially when answers exceed one page).

## Problems with random sorting

While it is more fair and we could still let new users (or people who aren't logged in) keep their ranking by votes. It's still valuable to have answers which are obviously good near the top and terrible answers (or very unremarkable ones) buried. Random sorting won't do this and whenever I visit a question with a lot of answers (the sort of question where this ranking would be desirable) I'll be presented with mostly 'meh' answers.

## An alternative Ranking system

We could opt for a ranking systems that still largely bases on votes but favors newer answers, similar to how hot network questions work A possible formula for this could look something like: votes on this anwer/votes on all answers + a*votes on this anwer/votes on all answer since this answer was posted where a is a carefully selected weight. This would give late answers a fighting chance when they are first posted. If they gather some upvotes then, they'll stay up and even float up, if being at the top doesn't gather them upvotes though, they'll drop down again.

A random addend could also be added to this score to bring some life into things.

• Two things about the specifics of your ranking: you don't need to divide the first term by all votes because it's just a constant that's the same for all answers (and hence can be absorbed into a). Secondly, it would need an additional tweak to place completely new answer without any votes at the top (maybe just count the post itself as one upvote for the purposes of this ranking). Regardless, the same comment applies as posted on FireFly's proposal. – Martin Ender Aug 12 '14 at 13:48
• @MartinBüttner Good point, the formula was more to illustrate the idea rather than a serious suggestion. The formula in githubphagocyte's answer is better by a wide margin. Though it's not time based, it still operates under teh same assumption, that new answers that gathered relatrively many upvotes are good enough to be at the top without having the same absolute vote count of older asnwers. – overactor Aug 12 '14 at 13:56
• My suggestion will probably also need to include time since we don't have the same level of downvotes that reddit does (as ours mean something different). – trichoplax Aug 12 '14 at 14:27
• The only problem with this approach is that although it gives new answers the advantage they need early on, it gives a permanent advantage over all older answers, even after the passage of time when they should be being treated near to equally. This difference will be most pronounced when there are no more votes coming in, but the existing vote distribution can never sort the answers fairly even after a long period of time. – trichoplax Aug 12 '14 at 14:31
• @githubphagocyte you're right, when time approaches infinite (or a few weeks) answers should be treated almost equally. – overactor Aug 12 '14 at 14:35
• Example: 1 answer posted, 50% of votes ever come in. 2nd answer comes in, the remaining 50% of votes ever are shared equally by the 2 equally good answers. 1st answer has 0.75 + 0.75a. 2nd answer has 0.25 + 0.5a. This may be the other way around from my intuition... It doesn't settle down to a fair ratio though. – trichoplax Aug 12 '14 at 14:35
• I think something as simple as Martin's votes/time gives plenty of advantage early on but still evens out with increasing time. – trichoplax Aug 12 '14 at 14:37

# Rate of votes compared to placement on page?

This is sort of a long-shot thinking-out-loud idea, but I feel that a "hotness" ranking should be determined by two things:

• The rate of votes that the answer is experiencing
• The answer's history of placement on the page

My logic is that the rate of votes is determined by two primary factors:

• The quality of that answer (which may be very roughly approximated by votes/viewer)
• The visibility of that answer (which determines the rate of new viewers)

Basically, we want to calculate votes/viewer = rate of votes / rate of new viewers.

I think the ranking system should try to determine which answers have the highest votes/viewer ratio. The ranking algorithm would look at the rate of votes and try to determine how much of that rate is due to answer quality and how much is due to visibility.

What we would need to do is take an empirical measurement of how an answer's placement on the page affects the rate of votes it is experiencing. I expect the visibility to decrease with lower placemement, with large decreases once the answers are on second/third pages.

Perhaps here is one way of calculating it. This is by no means the best way, just a possible realization of the main idea of this post. The idea of taking answer placement into account can be incorporated into several of the other suggestions.

A = answer, N = placement position number, votes(A) = all votes, score(A) = up - down

expectedrate(N) = average rate of votes experienced by all answers in Nth placement position
timespent(A,N) = the total time spent by answer A in placement position N
expectedvotes(A) = summation( timespent(A,N) * expectedrate(N) )
result(A) = score(A) * votes(A) / expectedvotes(A)

Basically, the answer's ranking is affected by how that answer's number of votes compares to the expected number of votes. The expected number of votes takes into account the how long the answer spent in each position and the rate of votes experienced by each position.

## Pros:

• High-quality answers with very low visibility will have that taken into account and rise to the top very quickly
• Highly-visible answers with the most upvotes will be penalized if the only reason for their upvotes is the visibility and not because of above average quality.

## Cons:

• Might not work when the pool of voters is very small and starts to diminish.
• The popularity of the question probably determines how influential answer placement is.
• Hard to calculate or measure.
• How do you work out the feedback loop of "placement on page" -> "new ranking" -> "placement on page" -> "new ranking" -> ...? :) – Martin Ender Aug 12 '14 at 14:49
• I tried to resolve that loop by taking into account the amount of time the answer spent at all answer positions in the past to determine the expected votes (I've changed it from current placement to placement history). – PhiNotPi Aug 12 '14 at 14:51
• I would use the views on the question while teh answer wa sin a certain position rather than the time spent there. – overactor Aug 12 '14 at 15:08

# Sort by votes minus amount of time since post

Here's a fancy userscript for your testing pleasures (doesn't work for multiple pages):

function rank(x) {
var score = parseInt($('.vote-count-post', x).text()); var milliseconds = new Date($('.post-signature:last-child .user-action-time span.relativetime', x).attr('title')) * 1,
hours = milliseconds / 1000 / 60 / 60;
return score - hours; // or some other factor
}

var ranked = $('.answer').sort(function(a, b) { return rank(a) < rank(b) ? 1 : -1; });$('#answers').empty().append(ranked);

This will rank a +20 answer posted 10 hours ago the same as a +15 answer posted 5 hours ago. I.e., it will subtract the amount of hours since post from the score. (Technically, hours since the epoch, but they work the same in practice.)

Of course, the amount that time has to do with the sorting could be changed, but 1 hour == 1 upvote seems like a good factor, from brief testing.

Advantages:

• Very simple.
• Can be tweaked easily.
• You see new answers when you come back in a day or two.
• Answers with a ridiculous amount of upvotes still stay at the top.
• Answers posted, say, a few months later will have a very large advantage.

Disadvantages:

• You see new answers when you come back in a day or two.
• Answers with a ridiculous amount of upvotes still stay at the top.
• Answers posted, say, a few months later will have a very large advantage.
• Not very fair for users who post on times like the weekend, whose posts sink down and don't have much of a chance at rising back up.
• Can still be affected by bias.
• You'd need an insane amount of upvotes to stay on top over an answer (even if downvoted) posted a month later. I don't think any linear relationship is going do here. But the userscript is really useful! :) – Martin Ender Aug 12 '14 at 15:41
• @MartinBüttner ... I probably should have thought of that. -_- On the other hand, people don't usually come back to challenges posted a month later, and maybe the sorting could be changed to pure votes after, say, no new answers in 3 days, or something along those lines. Also, isn't the "staying on top" thing exactly what you're trying to avoid? In fact, this could be a good thing, as it would mean that you can still get a few votes by answering a 3 month old question with a great answer. – Doorknob Aug 12 '14 at 15:45
• People coming back to old answers happens surprisingly often. I think I see it about once a week that some old challenge gets an answer and then another handful due to being bumped to the front page. What I'm saying is that a new answer on such an old question should go to the top, but should bubble down if it turns out to be worse than the previous top voted answer. With the linear relationship it's impossible to be voted below any top-voted answer after a month. – Martin Ender Aug 12 '14 at 15:50

highest = vote count for highest voted answer on the same question

hours = number of hours since answering

votes = number of votes on your answer

sorting: max(0, highest - votes - hours) + votes

By taking into account the highest voted answer, it ensures that all answers appear at the top, at least when first posted. "max" makes sure that the time since posting only benefits the post for a suitable period of time, during which users have the chance to vote on it. And in the end it still comes back to votes, but at least all answers would have had a chance to be upvoted.

note, when I say votes above, of course I mean upvotes - downvotes. Also, hours could be replaced with "number of hours since answering * n", to tweak it.

# Slightly weighted random sorting

This has in some form or another been suggested, but the weight was a little too inflexibly strong in all of them IMO. How about this algorithm (in Python pseudocode)?

# an arbitrary constant signifying the weight given to either the votes
# (greater than 1), the time (0 < n < 1), or neither (1).
ratio_weight = 1

# an arbitrary constant signifying the weight given to the statistics of
# each entry (greater than 1), the random element (0 < n < 1), or
# neither (1).
stats_weight = 1

def order_list(entry_list):
total_votes = reduce(entry_list, lambda x, y: x + y)

# this is the main weighting of the sorting. entry is a tuple in the
# form of (answer_entry, random_index_to_use).
def get_sorting_index(entry):
e = entry[0]
return entry[1] + stats_weight * (1 -
(ratio_weight * (e.votes/e.time)) / total_votes)

# get a list of random indices to work with
random_indices = list(range(len(entry_list)))
random.shuffle(random_indices)

# get a zipped list with the random indices
entry_list = zip(entry_list, random_indices)

# sort using the weighted, randomized entry indices
entry_list.sort(key=lambda entry: get_sorting_index(entry))
return [x[0] for x in entry_list] # strip random sorting index

Please comment if you think something doesn't look quite right (either bug or proposal).

# Merge Active + Votes + Random Rankings using a Condorcet method

I think each of these has its benefits, which can be merged.

Active sorting means that newer answers are more visible and given a chance to be viewed. This helps to counteract the bias towards older answers. It allows a sort of kick-start to new answers.

Votes sorting means that new users will see primarily see quality posts as opposed to downvoted posts. We don't want to place too much faith in here, but we can't simply ignore it.

Random sorting helps to scramble up the order. It is blind to quality, which can be both a pro and a con.

This algorithm would take the three rankings and use them as the votes in a Condercet election. There would be X copies of the active ranking, Y copies of the votes ranking, and Z copies of the random ranking (each copy is the same ordering). The X, Y, and Z are the weights. There are several Condercet-compliant methods to merge these, but something like Ranked Pairs could work decently.

Pros:

• The weightings are easy flexible, offering a continuum of options all the way to complete randomness.
• With the correct weights, the new posts with few votes are on the same footing as the old posts with many votes.
• This system favors the newer posts with many votes and is against the older posts with no votes.

Cons:

• Finding the correct weightings would be an empirical thing.
• The active ranking can be kinda chaotic sometimes when older posts are edited. It can give bad posts a second chance at greatness after being fixed, but could motivate people to create pointless edits. We could replace Active with Newest if this is a concern with little impact on the algorithm.


text:
# Conditional distribution of binomial random variables is hypergeometric

Let's say $X$ and $Y$ are binomial random variables with parameters $n$ and $p$ and $X+Y=m$. I want to show that the conditional distribution of $X$ if $X+Y=m$ is a hypergeometric distribution.

I'm thinking about putting these in terms of coin flips with the $m^\text{th}$ head occurring after $2n$ tosses. $X$ and $Y$ have $n$ tosses with probability of heads $p$. I'm not sure how to relate this to the hypergeometric distribution however.

## 1 Answer

We prove a somewhat more general result. Let $X$ and $Y$ be independent binomially distributed random variables with the same probability $p$ of "success" and number of trials equal to $a$ and $b$ respectively. (In your problem, we have $a=b=n$.)

Let us find the probability that $X=k$ given that $X+Y=m$. By the usual formula for the conditional probability, this is the probability that $X=k$ and $Y=m-k$ divided by the probability that $X+Y=m$.

The probability that $X=k$ and $Y=m-k$ is $\binom{a}{k}p^k(1-p)^{a-k}\binom{b}{m-k}p^{m-k} (1-p)^{b-(m-k)}$. This is $\binom{a}{k}\binom{b}{m-k}p^m(1-p)^{a+b-m}$.

The probability that $X+Y=m$ is $\binom{a+b}{m} p^m(1-p)^{a+b-m}$.

Divide. We get $$\frac{\binom{a}{k}\binom{b}{m-k}}{\binom{a+b}{m}}.\tag{1}$$

This is precisely the probability of getting $k$ red balls if we choose, without replacement, $m$ balls from $a+b$ balls $a$ of which are red and $b$ of which are blue. Formula (1) is a version of the familiar hypergeometric distribution function.


text:
# How to get the boolean value of an inequality involving an InterpolatingFunction?

Here's the code:

yan = FunctionInterpolation[x^2, {x, -1, 1}];
FullSimplify[yan[x] > -1, -1 < x < 1]

Needless to say, what I expect to see in the output is "True", but FullSimplify doesn't seem to work. What function should I turn to?

@J.M. @belisarius @acl

yan = FunctionInterpolation[x^2, {x, -1, 1}];
MinValue[{yan[x], -1 < x < 1},x]>-1

• Huh? "seems not to work" is correct? I just delete it because of my language sense…OK, let me add it back. Aug 6, 2012 at 5:55
• "doesn't seem to work" as you added is more standard, but in my opinion "seems not to work" is also acceptable and understandable. Aug 6, 2012 at 6:10
• In fact I've become confused after I searched the Internet, so I turned to the standard form to be on the safe side 囧. Aug 6, 2012 at 6:47

You can do this by explicitly constructing the InterpolatingPolynomial corresponding to yan, and then using FullSimplify:

yin = InterpolatingPolynomial[Transpose[Flatten /@ {yan[[3]],yan[[4]]}],x];
FullSimplify[yin > -1, -1 < x < 1]
(*True*)

Why does this work? Because yan actually has a list of points:

FullForm[yan]

so I can extract them with Transpose[Flatten /@ {yan[[3]],yan[[4]]}] and use them to construct a polynomial, which does the same thing as the interpolation function but which FullSimplify can now handle.

Maybe there's a better way to construct the InterpolatingPolynomial but this works.

The following is basically the same @acl did, but using the package InterpolatingFunctionAnatomy which (in principle) will behave better than peeking at the internal structures when Mma version changes.

Needs["DifferentialEquationsInterpolatingFunctionAnatomy"];
yan = FunctionInterpolation[x^2, {x, -1, 1}];

yin = InterpolatingPolynomial[Transpose[Flatten /@
{InterpolatingFunctionCoordinates@yan,
InterpolatingFunctionValuesOnGrid@yan}], x];
FullSimplify[yin > -1, -1 < x < 1]
(*
True
*)

• very nice. how come I have less votes than you though, despite being first and explaining details?! Not fair! :)
– acl
Aug 5, 2012 at 19:38
• @acl That was because I forgot to upvote your answer. Easy to correct! :D Aug 5, 2012 at 19:40
• really, I was joking!
– acl
Aug 5, 2012 at 19:48
• In fact, you don't need to load the package if you remember the actual syntax being used internally; in this case, it's yan["Coordinates"] and yan["ValuesOnGrid"] that can be used directly. Still, this works only because the original function was well approximated by a polynomial. In general, a polynomial interpolant can be more oscillatory than the piecewise polynomial interpolant used by InterpolatingFunction[]; be careful! Aug 5, 2012 at 23:48
• @J.M. yes, it was intended just for this case Aug 5, 2012 at 23:52

Today I ocasionally recall this question I asked 10 years ago, and notice nowadays ResourceFunction["InterpolatingFunctionToPiecewise"] or InterpolationToPiecewise in this post seems to be the best choice to resolve the problem:

yan = FunctionInterpolation[x^2, {x, -1, 1}];
yanAnalytic = ResourceFunction["InterpolatingFunctionToPiecewise"][yan, x]
FullSimplify[yanAnalytic > -1, -1 < x < 1]
(* True *)

This method perfectly handles the troublesome case shown by J.M., too:

func = FunctionInterpolation[1/(1 + x^2), {x, -2, 2}];
funcAnalytic = ResourceFunction["InterpolatingFunctionToPiecewise"][func, x];
Simplify[funcAnalytic < 3/2, -2 < x < 2]
(* True *)

BTW, I don't evaluate my MinValue method in the question as the best, because it's essentially a numeric method. MinValue has secretly called NMinimize internally!:

Trace[MinValue[{yan[x], -1 < x < 1}, x], __NMinimize, TraceInternal -> True]


text:
Definitions

# Folded normal distribution

The folded normal distribution is a probability distribution related to the normal distribution. Given a normally distributed random variable X with mean μ and variance σ2, the random variable Y = |X| has a folded normal distribution. Such a case may be encountered if only the magnitude of some variable is recorded, but not its sign. The distribution is called Folded because probability mass to the left of the x = 0 is "folded" over by taking the absolute value.

The cumulative distribution function (CDF) is given by

$F_Y\left(y; mu, sigma\right) = int_0^y frac\left\{1\right\}\left\{sigmasqrt\left\{2pi\right\}\right\} , exp left\left(-frac\left\{\left(-x-mu\right)^2\right\}\left\{2sigma^2\right\} right\right), dx$
+ int_0^{y} frac{1}{sigmasqrt{2pi}} , exp left(-frac{(x-mu)^2}{2sigma^2} right), dx.

Using the change-of-variables z = (x − μ)/σ, the CDF can be written as

$F_Y\left(y; mu, sigma\right) = int_\left\{-mu/sigma\right\}^\left\{\left(y-mu\right)/sigma\right\} frac\left\{1\right\}\left\{sqrt\left\{2pi\right\}\right\} , exp left\left(-frac\left\{1\right\}\left\{2\right\}left\left(z + frac\left\{2mu\right\}\left\{sigma\right\}right\right)^2right\right) dz$
+ int_{-mu/sigma}^{(y-mu)/sigma} frac{1}{sqrt{2pi}} , exp left(-frac{z^2}{2} right) dz.

The expectation is then given by

$E\left(y\right) = sigma sqrt\left\{2/pi\right\} exp\left(-mu^2/2sigma^2\right) + muleft\left[1-2Phi\left(-mu/sigma\right)right\right],$

where Φ(•) denotes the cumulative distribution function of a standard normal distribution.

The variance is given by

$operatorname\left\{Var\right\}\left(y\right) = mu^2 + sigma^2 - left\left\{ sigma sqrt\left\{2/pi\right\} exp\left(-mu^2/2sigma^2\right) + muleft\left[1-2Phi\left(-mu/sigma\right)right\right] right\right\}^2.$

Both the mean, μ, and the variance, σ2, of X can be seen to location and scale parameters of the new distribution.

## References

• Leone FC, Nottingham RB, Nelson LS (1961). "The Folded Normal Distribution". Technometrics 3 (4): 543–550.
• Johnson NL (1962). "The folded normal distribution: accuracy of the estimation by maximum likelihood". Technometrics 4 (2): 249–256.
• Nelson LS (1980). "The Folded Normal Distribution". J Qual Technol 12 (4): 236–238.
• Elandt RC (1961). "The folded normal distribution: two methods of estimating parameters from moments". Technometrics 3 (4): 551–562.
• Lin PC (2005). "Application of the generalized folded-normal distribution to the process capability measures". Int J Adv Manuf Technol 26 825–830.

Search another word or see folded overon Dictionary | Thesaurus |Spanish


text:
# Finding all $d$-dimensional indecomposable representations

Given a connected quiver algebra $$A$$ over a finite field $$K$$.

Question : Is there an effective/quick method to obtain all $$d$$-dimensional indecomposable representations for a fixed $$d$$ with a computer algebra system such as the GAP package QPA?

Now this is implemented in QPA version 1. There is an error above, one needs to consider both ExtOverAlgebra(S[i],iso[j]) and ExtOverAlgebra(iso[j],S[i]) in order to get all modules of the given length. It looks like this in QPA:

gap> A := KroneckerAlgebra(GF(3),2);;
gap> AllModulesOfLengthAtMost( A, 4 );
[ <[ 0, 0 ]>, <[ 1, 0 ]>, <[ 0, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]>,
<[ 1, 1 ]>, <[ 1, 1 ]>, <[ 2, 1 ]>, <[ 2, 1 ]>, <[ 2, 1 ]>, <[ 2, 1 ]>,
<[ 2, 1 ]>, <[ 2, 1 ]>, <[ 1, 2 ]>, <[ 1, 2 ]>, <[ 1, 2 ]>, <[ 1, 2 ]>,
<[ 1, 2 ]>, <[ 1, 2 ]>, <[ 3, 1 ]>, <[ 3, 1 ]>, <[ 3, 1 ]>, <[ 3, 1 ]>,
<[ 3, 1 ]>, <[ 2, 2 ]>, <[ 3, 1 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>,
<[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>,
<[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>,
<[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>,
<[ 2, 2 ]>, <[ 2, 2 ]>, <[ 1, 3 ]>, <[ 1, 3 ]>, <[ 1, 3 ]>, <[ 1, 3 ]>,
<[ 1, 3 ]>, <[ 1, 3 ]> ]
gap> AllIndecModulesOfLengthAtMost( A, 4 );
[ <[ 1, 0 ]>, <[ 0, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]>,
<[ 2, 1 ]>, <[ 1, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]>,
<[ 2, 2 ]>, <[ 2, 2 ]>, <[ 2, 2 ]> ]

The first one finds all non-isomorphic modules of length at most 4, and the second one finds all non-isomorphic indecomposable modules of length at most 4.

Best regards, the QPA-team.

I don't know if it would be quick, but you can inductively construct all modules of dimension $$d$$ using QPA. First you can construct all modules of dimension $$2$$, then all modules of dimension $$3$$, and so on. The command ExtOverAlgebra(S[i],S[j]) would compute a basis for all the extensions between the different simple modules S[i] and S[j]. The basis vectors are maps from the first syzygy of S[i] to S[j]. Taking all linear combinations of these and then taking the pushout along the inclusion of the first syzygy of S[i] to the projective cover of S[i], would produce all modules of dimension $$2$$. It would possibly be a lot of duplicates on this list of modules, but those can be first reduced by applying the command Unique(list) and then doing actually isomorphism test with the command IsomorphicModules.

For the Kronecker algebra over a field of three elements it would look like something like this:

gap> A := KroneckerAlgebra(GF(3),2);;
gap> S := SimpleModules(A);;
gap> len2 := List( [1..2], i -> List( [1..2], j -> ExtOverAlgebra(S[i],S[j])));;
gap> len2[1][1];len2[2][1];len2[2][2];
[ [  ], [  ], [  ] ]
[ [  ], [  ], [  ] ]
[ [  ], [  ], [  ] ]
gap> len2 := len2[1][2];
[ <<[ 0, 2 ]> ---> <[ 1, 2 ]>>, [ <<[ 0, 2 ]> ---> <[ 0, 1 ]>>, <<[ 0, 2 ]> ---> <[ 0, 1 ]>> ], function( map ) ... end ]
gap> test := [];
[  ]
gap> for a in Elements(GF(3)) do for b in Elements(GF(3)) do Add(test, a*len2[2][1] + b*len2[2][2]); od; od;
gap> temp := List( test, t -> Range(PushOut(t,len2[1])[2]));
[ <[ 1, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]> ]
gap> temp := Unique(temp);
[ <[ 1, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]>, <[ 1, 1 ]> ]
gap> iso := [];;
gap> for i in [1..5] do for j in [i+1..5] do Add(iso, [i,j,IsomorphicModules( temp[i], temp[j])]); od;od;
gap> iso;
[ [ 1, 2, false ], [ 1, 3, false ], [ 1, 4, false ], [ 1, 5, false ], [ 2, 3, false ], [ 2, 4, false ], [ 2, 5, false ], [ 3, 4, false ], [ 3, 5, false ],
[ 4, 5, false ] ]

The variable iso now contain all the five non-isomorphic modules of dimension $$2$$. Repeating the above procedure for all elements on the list iso and each simple, that is, do ExtOverAlgebra(S[i], iso[j]) for all $$i$$ and $$j$$, constructing all extensions: Then you would have a list of all modules of length $$3$$, and so on.


text:
## A new look at condition A.(English)Zbl 1246.53078

An isoparametric hypersurface in $$S^n$$ can be obtained as the intersection of $$S^n$$ with the level set of a homogeneous polynomial on $$\mathbb {R}^{n+1}$$, of degree $$g=1$$, $$2$$, $$3$$, $$4$$ or $$6$$. These isoparametric hypersurfaces have $$g$$ principal curvatures, where only $$2$$ different multiplicities $$m_i$$, $$i=1,2$$, of the principal curvatures occur. For the construction of inhomogeneous isoparametric hypersurfaces with $$4$$ different principal curvatures, H. Ozeki and M. Takeuchi [Tohoku Math. J., II. Ser. 27, 515–559 (1975; Zbl 0359.53011)] imposed conditions (“A” and “B”) on the focal submanifolds of the isoparametric hypersurface, the first of which appeared to be crucial in subsequent work. In particular, J. Dorfmeister and E. Neher [Osaka J. Math. 20, 145–175 (1983; Zbl 0543.53046)] showed that the first of these conditions suffices to characterize a particular class of isoparametric hypersurfaces [D. Ferus, H. Karcher and H.-F. Münzner, Math. Z. 177, 479–502 (1981; Zbl 0443.53037)]. The author provides a simpler and more geometric proof of this result, aiming to shed light on the geometric meaning of the condition for the construction of classes of examples, paying particular attention to the exceptional multiplicity pairs $$(3,4)$$ and $$(7,8)$$. In particular, the author analyses a construction of isoparametric hypersurfaces using octonians in order to provide a better understanding of the relation between the algebraic and geometric structures involved.

### MSC:

 53C42 Differential geometry of immersions (minimal, prescribed curvature, tight, etc.) 53C40 Global submanifolds

### Citations:

Zbl 0359.53011; Zbl 0543.53046; Zbl 0443.53037
Full Text:

### References:

 [1] E. Cartan: Sur des familles remarquables d’hypersurfaces isoparamétriques dans les espaces sphériques , Math. Z. 45 (1939), 335-367. · Zbl 0021.15603 [2] T.E. Cecil, Q.-S. Chi and G.R. Jensen: Isoparametric hypersurfaces with four principal curvatures , Ann. of Math. (2) 166 (2007), 1-76. · Zbl 1143.53058 [3] J. Dorfmeister and E. Neher: An algebraic approach to isoparametric hypersurfaces in spheres , I, Tôhoku Math. J. (2) 35 (1983), 187-224. · Zbl 0507.53038 [4] J. Dorfmeister and E. Neher: An algebraic approach to isoparametric hypersurfaces in spheres , II, Tôhoku Math. J. (2) 35 (1983), 225-247. · Zbl 0507.53039 [5] J. Dorfmeister and E. Neher: Isoparametric triple systems of algebra type , Osaka J. Math. 20 (1983), 145-175. · Zbl 0543.53046 [6] J. Dorfmeister and E. Neher: Isoparametric triple systems of FKM-type , II, Manuscripta Math. 43 (1983), 13-44. · Zbl 0531.53049 [7] D. Ferus, H. Karcher and H.F. Münzner: Cliffordalgebren und neue isoparametrische Hyperflächen , Math. Z. 177 (1981), 479-502. · Zbl 0443.53037 [8] D. Husemoller: Fibre Bundles, third edition, Graduate Texts in Mathematics 20 , Springer, New York, 1994. · Zbl 0202.22903 [9] K. McCrimmon: Quadratic forms permitting triple composition , Trans. Amer. Math. Soc. 275 (1983), 107-130. · Zbl 0505.17004 [10] R. Miyaoka: The Dorfmeister-Neher theorem on isoparametric hypersurfaces , Osaka J. Math. 46 (2009), 695-715. · Zbl 1185.53059 [11] R. Miyaoka: Isoparametric hypersurfaces with $$(g,m)=(6,2)$$ , · Zbl 1263.53049 [12] H.F. Münzner: Isoparametrische Hyperflächen in Sphären , Math. Ann. 251 (1980), 57-71. · Zbl 0417.53030 [13] H.F. Münzner: Isoparametrische Hyperflächen in Sphären , II, Math. Ann. 256 (1981), 215-232. · Zbl 0438.53050 [14] H. Ozeki and M. Takeuchi: On some types of isoparametric hypersurfaces in spheres , I, Tôhoku Math. J. (2) 27 (1975), 515-559. · Zbl 0359.53011 [15] \begingroup H. Ozeki and M. Takeuchi: On some types of isoparametric hypersurfaces in spheres , II, Tôhoku Math. J. (2) 28 (1976), 7-55. \endgroup · Zbl 0359.53012 [16] S. Stolz: Multiplicities of Dupin hypersurfaces , Invent. Math. 138 (1999), 253-279. · Zbl 0944.53035
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# Finding/Updating Parameters

PX4 behaviour can be configured/tuned using parameters (e.g. Multicopter PID gains, calibration information, etc.).

The QGroundControl Parameters screen allows you to find and modify any of the parameters associated with the vehicle. The screen is accessed by clicking the top menu Gear icon and then Parameters in the sidebar.

Most of the more commonly used parameters are more conveniently set using the dedicated setup screens described in the Basic Configuration section. The Parameters screen is needed when modifying less commonly modified parameters - for example while tuning a new vehicle.

While some parameters can be changed in flight, this is not recommended (except where explicitly stated in the guide).

## Finding a Parameter

You can search for a parameter by entering a term in the Search field. This will show you a list of all parameter names and descriptions that contain the entered substring (press Clear to reset the search).

You can also browse the parameters by group by clicking on the buttons to the left (in the image below the Battery Calibration group is selected).

If you can't find an expected parameter, see the next section.

## Missing Parameters

Parameters are usually not visible because either they are conditional on other parameters, or they are not present in the firmware (see below).

### Conditional Parameters

A parameter may not be displayed if it is conditional on another parameter that is not enabled.

You can usually find out what parameters are conditional by searching the full parameter reference and other documentation. In particular serial port configuration parameters depend on what service is assigned to a serial port.

### Parameter Not In Firmware

A parameter may not be present in the firmware because you're using a different version of PX4 or because you're using a build in which the associated module is not included.

New parameters are added in each PX4 version, and existing parameters are sometimes removed or renamed. You can check whether a parameter should be present by reviewing the full parameter reference for the version you're targeting. You can also search for the parameter in the source tree and in the release notes.

The other reason that a parameter might not be in firmware is if its associated module has not been included. This is a problem (in particular) for FMUv2 firmware, which omits many modules so that PX4 can fit into the 1MB of available flash. There are two options to solve this problem:

• Check if you can update your board to run FMUv3 firmware, which includes all modules: Firmware > FMUv2 Bootloader Update
• If your board can only run FMUv2 firmware you will need to rebuild PX4 with the missing modules enabled. You can see these commented out in boards/px4/fmu-v2/default.cmake:
    DRIVERS
#barometer # all available barometer drivers
barometer/ms5611
#batt_smbus
#camera_capture

You may also need to disable other modules in order to fit the rebuilt firmware into 1MB flash. Finding modules to remove requires some trial/error and depends on what use cases you need the vehicle to meet.

## Changing a Parameter

To change the value of a parameter click on the parameter row in a group or search list. This will open a side dialog in which you can update the value (this dialog also provides additional detailed information about the parameter - including whether a reboot is required for the change to take effect).

When you click Save the parameter is automatically and silently uploaded to the connected vehicle. Depending on the parameter, you may then need to reboot the flight controller for the change to take effect.

## Tools

You can select additional options from the Tools menu on the top right hand side of the screen.

Refresh
Refresh the parameter values by re-requesting all of them from the vehicle.

Reset all to defaults
Reset all parameters to their original default values.

Load from file / Save to file
Load parameters from an existing file or save your current parameter settings to a file.

Clear RC to Param
This clears all associations between RC transmitter controls and parameters. For more information see: Radio Setup > Param Tuning Channels.

Reboot Vehicle
Reboot the vehicle (required after changing some parameters).


text:
# How to draw a horizontal tree with branches on the right and left?

I need to draw trees but they should extend (from the root) on both sides. TikZ seems a good solution but do not how to manage it.

-

The key is the grow option (See Section 18.5.2 Default Growth Function of the manual). A little example:

\documentclass{article}
\usepackage{tikz}

\begin{document}

\begin{tikzpicture}[level distance=10mm,sibling distance=10mm,every node/.style={fill=blue!30,circle,inner sep=5pt}
]
\node {0}
child[grow=left] {
child {node{10}} child {node{20}} child {node{30}}
}
child[grow=right] {
child {node{40}} child {node{50}} child {node{60}}
};
\end{tikzpicture}

\end{document}

Arrows can be added to some edge(s) using the edge from parent. If arrows need to be added to many edges, the best thing to do is to define a style (as Alan Munn suggested in a comment):

\documentclass{article}
\usepackage{tikz}

\begin{document}

\begin{tikzpicture}[level distance=10mm,
sibling distance=10mm,
every node/.style={fill=blue!30,circle,inner sep=5pt},
arrow/.style={edge from parent/.style={draw,-latex}}
]

\node {0}
child[grow=left] {
child {node{10}} child {node{20}} child[arrow] {node{30}}
}
child[grow=right] {
child {node{40}} child {node{50}} child[arrow] {node{60}}
};
\end{tikzpicture}

\end{document}

-
Sorry one more thing, how can I edit the edges and apply a style for the arrows? –  Andry Apr 2 '12 at 1:28
@Andry: TikZ offers so many possibilities for what you want that you will have to narrow a bit the scope of your question. How do you want the edges to look like and what kind of style do you want for the arrows? –  Gonzalo Medina Apr 2 '12 at 1:30
Well I just want edges to have a arrow, in your perfect example from 0 to 60 I want an arrow in 60's direction... However thankyou for your kindness. –  Andry Apr 2 '12 at 1:42
@Andry: you can use edge from parent to change the attributes for some of the edge(s): try this: \begin{tikzpicture}[level distance=10mm,sibling distance=10mm,every node/.style={fill=blue!30,circle,inner sep=5pt} ] \node {0} child[grow=left] { child {node{10}} child {node{20}} child {node{30}} } child[grow=right] { child {node{40}} child {node{50}} child {node{60} edge from parent[-latex]} }; \end{tikzpicture} –  Gonzalo Medina Apr 2 '12 at 1:48
@GonzaloMedina This would be better made into its own style attached to the main tikzpicture: arrow/.style={edge from parent/.style={draw,-latex}} and then for each node that requires an arrow, just use child[arrow]. –  Alan Munn Apr 2 '12 at 1:52

Depending on the complexity of your tree it might be simpler to put two trees together. Here's an example using tikz-qtree to draw the trees.

\documentclass{article}

\usepackage{tikz}
\usepackage{tikz-qtree}
\usetikzlibrary{trees} % this is to allow the fork right path

\begin{document}

\begin{tikzpicture}[level distance=1.25in,sibling distance=.25in,scale=.75]
\tikzset{edge from parent/.style=
{thick, draw,
edge from parent fork right},every tree node/.style={draw,minimum width=1in,text width=1in, align=center},grow'=right}
\Tree
[. parent
[.{nice child0}
[.{grandchild0-0 } ]
[.{grandchild0-1 } ]
[.{grandchild0-2 } ]
[.{grandchild0-3 with a really long name } ]
]
[.child1
[.{grandchild1-0 } ]
[.{grandchild1-1 } ]
[.{grandchild1-2 } ]
]
[.child2 ]
[.child3 ]
]
\begin{scope}
\tikzset{edge from parent/.style=
{thick, draw,
edge from parent fork left},every tree node/.style={draw,minimum width=1in,text width=1in, align=center},grow'=left}

\Tree
[.\node[draw=none]{};
[.{nice child0}
[.{grandchild0-0 } ]
[.{grandchild0-1 } ]
[.{grandchild0-2 } ]
[.{grandchild0-3 with a really long name } ]
]
[.child1
[.{grandchild1-0 } ]
[.{grandchild1-1 } ]
[.{grandchild1-2 } ]
]
[.child2 ]
[.child3 ]
]

\end{scope}

\end{tikzpicture}
\end{document}

-
Interesting solution however... :) –  Andry Apr 2 '12 at 1:52
As I said, when your trees get complicated, the tikz-qtree method of input becomes helpful. It wasn't clear from your question exactly what kind of tree you wanted. –  Alan Munn Apr 2 '12 at 1:54


text:
# $g^{-1}$ in the function of a random variable

Function of a random variable

$\Omega\overset{X}{\rightarrow}\mathbb{R}\overset{g}{\rightarrow}\mathbb{R}$

In the next example of a function of a random variable, $P_Y(y)= P_X(g^{-1}(y))$.

Example: A system has three output states $\Omega_X = \left \{ -1,0,1 \right \}$ with probabilities $P_X(-1)=\frac{1}{3}$, $P_X(0)=\frac{1}{3}$ and $P_X(1)=\frac{1}{3}$.

What's the probability function of $Y=X^2$?

$\Omega_Y = \left \{ 0,1 \right \}$.

$P_Y(0)= P_X(g^{-1}(0))=P_X(0) =\frac{1}{3}$ and $P_Y(1)= P_X(g^{-1}(1))=P_X(\left \{ -1,1 \right \}) =\frac{2}{3}$ (End of the example)

If $P_Y(1)= P_X(g^{-1}(1))$ then the inverse function $g^{-1}(y)$, has for an element of its domain, two elements of its codomain $P_X(\left \{ -1,1 \right \})$. By the definition of a function

"In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output",

then $g^{-1}(y)$ is not a function.

One can say that the image of $P_X(g^{-1}(1))$ is the addition ($P_X(-1) + P_X(1)$), but the addition ($P_X(-1) + P_X(1)$) is not the image of $P_X(g^{-1}(1))$. The images of $P_X(g^{-1}(1))$ are two: $P_X(-1)$ and $P_X(1)$, the addition would be a completely different element.

Is it or isn't $g^{-1}(y)$ a function? And if it is not, why do they use $g^{-1}(y)$as a notation of this relation?

$g^{-1}(A)$ doesn't mean the inverse function in this setting but the preimage of $A$ under $g$ so $g^{-1}(A) = \{ x \in \Bbb R | g(x) \in A \}$.

If $g$ is a measurable function then $g^{-1}(A)$ is measurable so $P_X(A)$ is well defined.

And because you are in a discrete setting ofc it holds that $g^{-1}(A)$ can be written as a disjoint union of elements of $g^{-1}(A)$ so $$g^{-1}(A) = \bigcup_{x \in g^{-1}(A)}\{x\}$$ and because it's countable and disjoint it also holds $$P_X\left(g^{-1}(A)\right) = \sum_{x \in g^{-1}(A)} P_X(\{x\})$$

Setting $A = \{1\}$ you get your result.

Indeed it's a bit "handwaving" to write $P_Y(1)$ instead of $P_Y(\{1\})$ but it's clear in this setting what is meant…

• Thanks a lot! Can you please expand on "it's a bit "handwaving" to write $P_Y(1)$ instead of $P_Y(\{1\})$, but it's clear in this setting what is meant..." $P_Y(\{1\})$ I think it means the element $1$ from the set $A$. What does $P_Y(1)$ mean? I can't see how this change in the notation makes any difference in the interpretation of $g^{-1}$ .Can you please explain? And finally how do you know that "$g^{-1}(A)$ doesn't mean the inverse function but the preimage of "A" under "g"? – match6 Nov 29 '16 at 20:08
• In your first posting you wrote "$P_Y(1)= P_X(g^{-1}(1))$" but $P_Y$ is a probability measure so as an argument only measurable sets are possible. Although 1 is a real number and not a set (and so $P_Y(1)$ is strictly speaking not well defined it's well known that for a single elemented set the set braces can be omnited. So $P_Y(x)$ means $P_Y(\{x\})$ for a real number x. – Gono Nov 30 '16 at 8:21
• To your $g^{-1}$ question: You claim, that $g^{-1}$ means the inverse function, so I could also ask you how do you know that? Just as well $g^{-1}$ could mean the reciprocal of g. But it doesn't. I just told you that your claim it's the inverse function is wrong. And here it's the same with the notation stuff above: For real numbers x and the preimage by $g^{-1}(x)$ always $g^{-1}(\{x\})$ is meant if you are technically correct. For a inverse function writing $g^{-1}(x)$ is totally ok because you have there one unique solution so you can write $y = g^{-1}(x)$. But not for measurable functions… – Gono Nov 30 '16 at 8:26
• … in general because the preimage is not always a unique element and if you consider the definition of a measurable function you will see, that it's defined by preimages not by the inverse function. – Gono Nov 30 '16 at 8:27
• Thanks. It helped me a lot. – match6 Nov 30 '16 at 13:22


text:
# [OS X TeX] What's in a page number?

Will Robertson will at guerilla.net.au
Mon Nov 15 21:48:50 EST 2004

On 16 Nov 2004, at 12:40 PM, Alain Schremmer wrote:

> I am not sure I understand the implication: Are you saying that the
> hyperref package would do it?

I was a little wrong, but in essence yes, to an extent. I got the
hyperref option wrong, though (should be [pdfpagelabels] instead of
[plainpages=false]).

If you run the appended minimal document through pdflatex and open the
output in Adobe Reader, the page numbering is shown as it appears in
TeX (i,ii,iii,iv,1,2,3).

Oh, but I get error messages when I try to view the arabic numbered

Anyway, the principle works, so maybe this is something that will also
be able to be used in TeXShop when Mac OS X 10.4 is released. (TeXShop
leverages Mac OS X technologies to display the PDF, so if Mac OS X
doesn't support something, then TeXShop won't either.)

Will

\documentclass{memoir}
\usepackage[pdfpagelabels]{hyperref}
\begin{document}
\frontmatter
page i
\newpage page ii
\newpage page iii
\mainmatter
\newpage page 1
\newpage page 2
\newpage page 3
\end{document}

--------------------- Info ---------------------
Mac-TeX Website: http://www.esm.psu.edu/mac-tex/
& FAQ: http://latex.yauh.de/faq/
TeX FAQ: http://www.tex.ac.uk/faq
List Post: <mailto:MacOSX-TeX at email.esm.psu.edu>


text:
# Why are Viking LR results not sufficient evidence of life on Mars?

Gil Levin, one of the scientists involved with the 1976 Viking mission to Mars, has claimed and continues to claim that there is life on Mars, based off the positive results of the Labeled Release (LR) experiment.

This was the only one of the landers' four life-detecting instruments that returned positive results, but prior to the mission it was expected that because of the instruments' radically different approaches, actual Martian organisms might still only return positive results in one of them. Furthermore, there is no evidence (to my knowledge) that the LR instrument was faulty, nor that LR's strong positive results could be explained by any non-biological process. (Levin details his fairly comprehensive explanation and argument for life on Mars in this paper).

Despite these findings, scientific consensus still posits that no life on Mars has ever been detected, by LR or any other experiment. My question is simply: Why?

Can anyone point me to a credible source, statement, or publication that addresses LR's results and explains why they don't constitute sufficient evidence for life on Mars?

• Extraordinary claims require extraordinary evidence. Viking did not provide that. You should ask the other question: why, with such poor and ambiguous evidence, would anyone believe there is life on Mars? Oct 15 '19 at 11:22
• This kind of attitude is exactly what I'm talking about. Where is the rationale? Yes, there's the extraordinary claims/evidence argument, but in this case, the extraordinary evidence is that an instrument specifically designed to test for life in Mars performed multiple successful tests without error and produced positive results every time, with no plausible alternative explanations. At what point do we say, "We set out to test for this thing, and our test says we found it"?
– ETL
Oct 16 '19 at 2:22
• What attitude? You mean scientific rigor? The experiment was NOT designed to find life, none so far is. It was designed to find released $^{14}CO_2$ gas and it did, albeit in a confusing and incoherent manner (your statement "produced positive results every time" is wrong, why would you lie on the internet), that is inconsistent with what we'd expect from life. The other experiments on the Vikings gave negative results. I don't see any strong evidence for life here. Also only parroting one guy that has a history of overselling his results is not a very credible approach to science. Oct 16 '19 at 9:50
• You're right, I should've said "produced positive results every time, except in control scenarios where negative results were expected and further supported the hypothesis." Beyond that, your logic is circular and your evidence nonexistent. Welcome to StackExchange, please provide sources for the "confusing and incoherent" results you claim were produced. I started this question by stating that I couldn't find sources that reliably refuted "the one guy." If you have one, please share and post an answer. Otherwise, you're just parroting what YOU have been told. Where's the scientific rigor?
– ETL
Oct 16 '19 at 17:51
• Short follow-up: The Yen+2000 paper science.sciencemag.org/content/289/5486/1909.full.pdf+html seems to explain the experiment+control well. It has been cited by a paper with Levin on it only once: koreascience.or.kr/article/JAKO201213549741763.page which I haven't read yet. Is that the explanation which you've referred to as 'disproven' in your comment to @MSalters answer? Oct 16 '19 at 20:40