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Book Variational Analysis And Generalized Differentiation Ii: Applications

### Book Variational Analysis And Generalized Differentiation Ii: Applications

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# 统计代写|统计与机器学习作业代写Statistical and Machine Learning代考|ECE6254

## 统计代写|统计与机器学习作业代写Statistical and Machine Learning代考|Fitting a Linear Multiple Regression Model via

In a general context, we have a covariate vector $X=\left(X_1, \ldots, X_p\right)^{\mathrm{T}}$ and we want to use this information to predict or explain how this variable affects a real-value response $Y$. The linear multiple regression model assumes a relationship given by
$$Y=\beta_0+\sum_{j=1}^p X_j \beta_j+\epsilon,$$
where $\epsilon$ is a random error with mean $0, E(\epsilon)=0$ and is independent of $X$. This error is included in the model to capture measurement errors and the effects of other unregistered explanatory variables that can help to explain the mean response.

Then, the conditional mean of this model is $E(Y \mid X)=\beta_0+\sum_{j=1}^p X_j \beta_j$ and the conditional distribution of $Y$ given $X$ is only affected by the information of $X$.
For estimating the parameters $\beta=\left(\beta_0, \beta_1, \ldots, \beta_p\right)^{\mathrm{T}}$, usually we have a set of data $\left(x_i^{\mathrm{T}}, y_i\right), i=1, \ldots, n$, often known as training data, where $\boldsymbol{x}i=\left(x{i 1}, \ldots, x_{i p}\right)^{\mathrm{T}}$ is a vector of features measurement and $y_i$ is the response measurement corresponding to the $i$ th individual drawn. The most common method for estimating $\boldsymbol{\beta}$ is the least squares method (OLS) that consists of taking the $\boldsymbol{\beta}$ value that minimizes the residual sum of squares defined as
$$\operatorname{RSS}(\boldsymbol{\beta})=\sum_{i=1}^n\left(y_i-\boldsymbol{\beta}0-\boldsymbol{x}_i^{\mathrm{T}} \boldsymbol{\beta}_0\right)^2=(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta}),$$ where $\boldsymbol{\beta}_0=\left(\beta_1, \ldots, \beta_p\right)^{\mathrm{T}}, \boldsymbol{y}=\left(y_1, \ldots, y_n\right)^{\mathrm{T}}$ is the vector with the response values of all individuals, and $X$ is an $n \times(p+1)$ matrix that contains the information of the measured features of all individuals, including the intercept in the first entry: $$\boldsymbol{X}=\left[\begin{array}{cccc} 1 & x{11} & \cdots & x_{1 p} \ \vdots & \vdots & \vdots & \vdots \ 1 & x_{n 1} & \cdots & x_{n p} \end{array}\right] .$$
If the $\boldsymbol{X}$ matrix has full column rank, then by differentiating the residual sum of squares with respect to the $\boldsymbol{\beta}$ coefficients, we can find the set of $\boldsymbol{\beta}$ parameters that minimize the $\operatorname{RSS}(\boldsymbol{\beta})$,
$$\frac{\operatorname{RSS}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}}=\frac{(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})}{\partial \boldsymbol{\beta}}=\frac{\boldsymbol{y}^{\mathrm{T}} \boldsymbol{y}-\boldsymbol{2} \boldsymbol{y}^{\mathrm{T}} \boldsymbol{X} \boldsymbol{\beta}+\boldsymbol{\beta}^{\mathrm{T}}\left(\boldsymbol{X}^{\mathrm{T}} \boldsymbol{X}\right) \boldsymbol{\beta}}{\partial \boldsymbol{\beta}}=\mathbf{2}\left[\left(\boldsymbol{X}^{\mathrm{T}} \boldsymbol{X}\right) \boldsymbol{\beta}-\boldsymbol{X}^{\mathrm{T}} \boldsymbol{Y}\right]$$

## 统计代写|统计与机器学习作业代写Statistical and Machine Learning代考|Fitting the Linear Multiple Regression Model via

The maximum likelihood (ML) estimation is a more general and popular method for estimating the parameters of a model (Casella and Berger 2002). It consists of finding the parameter value that maximizes the “probability” of observed values in the sample under the adopted model. Specifically, if $\left(\boldsymbol{x}_i^{\mathrm{T}}, y_i\right), i=1, \ldots, n$, is a set of observations from a multiple linear regression model (3.1) with homoscedastic and uncorrelated errors, the MLE of $\boldsymbol{\beta}$ and $\sigma^2, \widehat{\boldsymbol{\beta}}$ and $\widehat{\sigma}^2$, of this model is defined as
$$\left(\widehat{\boldsymbol{\beta}}^{\mathrm{T}}, \hat{\sigma}^2\right)=\underset{\boldsymbol{\beta}, \sigma^2}{\arg \max } L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right),$$
where $L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)$ is the likelihood function of the parameters, which is the probability of the observed response values but viewed as a function of the parameters
$$L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)=\left(\frac{1}{\sqrt{2 \pi \sigma^2}}\right)^n \exp \left[-\frac{1}{2 \sigma^2}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})\right] .$$
Then, the $\log \left(L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)\right)$ is equal to
$$\log \left(L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)\right)=-\frac{n}{2} \log (2 \pi)-n \log (\sigma)-\frac{1}{2 \sigma^2}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})$$
To find the maximum of $\sigma^2$ and $\boldsymbol{\beta}$, we get the derivative of $\log \left(L\left(\widehat{\boldsymbol{\beta}}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)\right)$ with regard to these parameters
$$\begin{gathered} \frac{\log \left(L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)\right)}{\partial \boldsymbol{\beta}}=\frac{\left[\left(\boldsymbol{X}^{\mathrm{T}} \boldsymbol{X}\right) \boldsymbol{\beta}-\boldsymbol{X}^{\mathrm{T}} \boldsymbol{Y}\right]}{\sigma^2} \ \frac{\log \left(L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)\right)}{\partial \sigma^2}=-\frac{n}{2 \sigma^2}+\frac{1}{2 \sigma^4}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta}) \end{gathered}$$
Now, by setting these derivatives equal to zero and solving the resulting equations for $\boldsymbol{\beta}$ and $\sigma^2$. we found that the estimates of these parameters are
$$\begin{gathered} \widehat{\boldsymbol{\beta}}=\left(\boldsymbol{X}^{\mathrm{T}} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\mathrm{T}} \boldsymbol{y} \ \widehat{\sigma}^2-\frac{1}{n}(\boldsymbol{y}-\boldsymbol{X} \widehat{\boldsymbol{\beta}})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \widehat{\boldsymbol{\beta}}) . \end{gathered}$$
From this we can see that for each value of $\sigma^2$, the value of $\boldsymbol{\beta}$ that maximizes the likelihood is the same value that maximizes $-\frac{1}{2 \sigma^2}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})$, which in turn minimizes $\left.(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})^{\mathrm{T}} \boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta}\right)$, which is precisely the OLS of $\boldsymbol{\beta}, \widehat{\boldsymbol{\beta}}$. But when equating the derivative of $\log \left(L\left(\widehat{\boldsymbol{\beta}}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)\right)$ to zero and solving for $\sigma^2$, the value of $\sigma^2$ that maximizes $L\left(\widehat{\boldsymbol{\beta}}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)$ is $\widehat{\sigma}^2=\frac{1}{n}(\boldsymbol{y}-\boldsymbol{X} \widehat{\boldsymbol{\beta}})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \widehat{\boldsymbol{\beta}})$.
Finally,
$$L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right) \leq L\left(\widehat{\boldsymbol{\beta}}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right) \leq L\left(\widehat{\boldsymbol{\beta}}, \widehat{\sigma}^2 ; \boldsymbol{y}, \boldsymbol{X}\right)$$
and from here, the MLE of $\boldsymbol{\beta}$ and $\sigma^2$ are $\widehat{\boldsymbol{\beta}}$ and $\widehat{\sigma}^2$, because it can be shown that the values of parameters that maximize the likelihood are unique when the design matrix $\boldsymbol{X}$ is of full column rank.

# 统计与机器学习代考

## 统计代写|统计与机器学习作业代写统计和机器学习代考|通过

$$Y=\beta_0+\sum_{j=1}^p X_j \beta_j+\epsilon,$$

$$\frac{\operatorname{RSS}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}}=\frac{(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})}{\partial \boldsymbol{\beta}}=\frac{\boldsymbol{y}^{\mathrm{T}} \boldsymbol{y}-\boldsymbol{2} \boldsymbol{y}^{\mathrm{T}} \boldsymbol{X} \boldsymbol{\beta}+\boldsymbol{\beta}^{\mathrm{T}}\left(\boldsymbol{X}^{\mathrm{T}} \boldsymbol{X}\right) \boldsymbol{\beta}}{\partial \boldsymbol{\beta}}=\mathbf{2}\left[\left(\boldsymbol{X}^{\mathrm{T}} \boldsymbol{X}\right) \boldsymbol{\beta}-\boldsymbol{X}^{\mathrm{T}} \boldsymbol{Y}\right]$$

## 统计代写|统计与机器学习作业代写统计和机器学习代考|拟合线性多元回归模型通过

$$\left(\widehat{\boldsymbol{\beta}}^{\mathrm{T}}, \hat{\sigma}^2\right)=\underset{\boldsymbol{\beta}, \sigma^2}{\arg \max } L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right),$$
where $L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)$ 参数的似然函数，即观测到的响应值的概率，但被视为参数的函数，是否
$$L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)=\left(\frac{1}{\sqrt{2 \pi \sigma^2}}\right)^n \exp \left[-\frac{1}{2 \sigma^2}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})\right] .$$

$$\log \left(L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)\right)=-\frac{n}{2} \log (2 \pi)-n \log (\sigma)-\frac{1}{2 \sigma^2}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})$$的最大值 $\sigma^2$ 和 $\boldsymbol{\beta}$的导数 $\log \left(L\left(\widehat{\boldsymbol{\beta}}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)\right)$ 对于这些参数
$$\begin{gathered} \frac{\log \left(L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)\right)}{\partial \boldsymbol{\beta}}=\frac{\left[\left(\boldsymbol{X}^{\mathrm{T}} \boldsymbol{X}\right) \boldsymbol{\beta}-\boldsymbol{X}^{\mathrm{T}} \boldsymbol{Y}\right]}{\sigma^2} \ \frac{\log \left(L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)\right)}{\partial \sigma^2}=-\frac{n}{2 \sigma^2}+\frac{1}{2 \sigma^4}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta}) \end{gathered}$$现在，通过让这些导数等于零，求解得到的方程 $\boldsymbol{\beta}$ 和 $\sigma^2$。我们发现这些参数的估计值
$$\begin{gathered} \widehat{\boldsymbol{\beta}}=\left(\boldsymbol{X}^{\mathrm{T}} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\mathrm{T}} \boldsymbol{y} \ \widehat{\sigma}^2-\frac{1}{n}(\boldsymbol{y}-\boldsymbol{X} \widehat{\boldsymbol{\beta}})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \widehat{\boldsymbol{\beta}}) . \end{gathered}$$从这里我们可以看到，对于的每一个值 $\sigma^2$的价值。 $\boldsymbol{\beta}$ 最大化的可能性和最大化的值是一样的 $-\frac{1}{2 \sigma^2}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})$，这反过来又会最小化 $\left.(\boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta})^{\mathrm{T}} \boldsymbol{y}-\boldsymbol{X} \boldsymbol{\beta}\right)$，正是的OLS $\boldsymbol{\beta}, \widehat{\boldsymbol{\beta}}$。但是当求导的时候 $\log \left(L\left(\widehat{\boldsymbol{\beta}}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)\right)$ 到零，求解 $\sigma^2$的价值。 $\sigma^2$ 最大化 $L\left(\widehat{\boldsymbol{\beta}}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right)$ 是 $\widehat{\sigma}^2=\frac{1}{n}(\boldsymbol{y}-\boldsymbol{X} \widehat{\boldsymbol{\beta}})^{\mathrm{T}}(\boldsymbol{y}-\boldsymbol{X} \widehat{\boldsymbol{\beta}})$.

$$L\left(\boldsymbol{\beta}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right) \leq L\left(\widehat{\boldsymbol{\beta}}, \sigma^2 ; \boldsymbol{y}, \boldsymbol{X}\right) \leq L\left(\widehat{\boldsymbol{\beta}}, \widehat{\sigma}^2 ; \boldsymbol{y}, \boldsymbol{X}\right)$$
，从这里，MLE $\boldsymbol{\beta}$ 和 $\sigma^2$ 是 $\widehat{\boldsymbol{\beta}}$ 和 $\widehat{\sigma}^2$，因为可以看出，当设计矩阵时，使似然最大化的参数值是唯一的 $\boldsymbol{X}$ 是全列秩。

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text:
## Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company

# Chapter 15 - Questions and Problems: 15.26

#### Answer

a) $K_{c} = 2.6\times10^{4}$ b) $K_{p} = 3.2\times10^{2}$

#### Work Step by Step

a) $K_{c} = \frac{1}{3.8\times10^{-5}} = 2.6\times10^{4}$ b) T = (727+273) K = 1000 K, $\Delta n$ = 1-2 = -1 , $K_{p} = K_{c}(RT)^{\Delta n} =(2.6\times10^{4})\times (\frac{0.08206L.atm}{K.mol}\times1000K)^{-1}=3.2\times10^{2}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
### Review of Basic Probability

So I understand that I lost many readers for the Sampling uploads.It is a bit difficult to the intensive use of notations and also the need for statistics knowledge. So here I’ll review a bit of basic probability. The contents here will be basic, and will involve some H2 Mathematics Statistics too.

Discrete Random Variables

Continuous Random Variables

Conditional Expectations and Variances

Multivariate Distributions

Multivariate Normal Distribution

The following three are in relation to Finance.

Introduction to Martingales

Introduction to Brownian Motion

Geometric Brownian Motion

Introduction to Stochastic Calculus

### Introduction to Martingales

This is a rather important topics for anyone interested in doing Finance.
Lets look at their definition first.
A Martingale is a random process $\{X_n: 0 \le n \le \infty \}$ with respect to the information filtration $F_n$ and the probability distribution $P$, if
$latex \mathbb{E}^P [|X_n|] < \infty$ for all $latex n \ge 0$ $latex \mathbb{E}^P[X_{n+m}|F_n] = X_n$ for all $latex n, m \ge 0$ Martingales are used widely and one example is to model fair games, thus it has a rich history in modelling of gambling problems. If you google Martingale, you will get an image related to a Horse, because it started with Horse-betting. [caption id="attachment_2845" align="alignnone" width="300"] Martingales. Source: NYU[/caption]

We define a submartigale by replacing the above condition 2 with
$\mathbb{E}^P [X_{n+M}| F_n] \ge X_n$ for all $n, m \ge 0$
and a supermartingale with
$\mathbb{E}^P [X_{n+M}| F_n] \le X_n$ for all $n, m \ge 0$ .
Take note that a martingale is both a submartingale and a supermartingale. Submartingale in layman terms, refers to the player expecting more as time progresses, and vice versa for supermartingale.

Let us try to construct a Martingale from a Random Walk now.
Let $S_n := \sum_{i=1}^n X_i$ be a random walk where the $X_i$’s are IID with mean $\mu$.
Let $M_n := S_n -n \mu$. Then $M_n$ is a martingale because:
$\mathbb{E}_n [M_{n+m}]$
$= \mathbb{E}_n [\sum_{i=1}^{n+m} X_i - (n+m) \mu]$
$= \mathbb{E}_n [\sum_{i=1}^{n+m} X_i] - (n+m) \mu$ since expectation distributes linearly
$= \sum_{i=1}^n X_i + \mathbb{E}_n [\sum_{i=n+1}^{n+m} X_i] - (n+m) \mu$
$= \sum_{i=1}^n X_i + m \mu - (n+m) \mu = M_n$

So how will a martingale betting strategy be like?
Here, we let $X_1, X_2, \ldots$ be IID random variables with $P(X_i = 1) = P(X_i = -1) = \frac{1}{2}$. We can imagine $X_i$ to represent the result of a coin-flipping game where,
– player win $1 if the coin comes up heads, that is, $X_i = 1$ – player lose$1 if the coin comes up tails, that is, $X_i = -1$

Consider further now a doubling strategy where we keep doubling the bet until we eventually win. Once we win, we stop and our initial bet is $1. The first thing we note is that the size of bet on the $n^{th}$ play is $2^{n-1}$ assuming we are still playing at time n. And we can let $W_n$ denote total winnings after n coin tosses, assuming $W_0 = 0$. Then $W_n$ is a martingale! To see this, let us prove that $W_n \in \{ 1, -2^n +1 \}$ for all n. Suppose we win for first time on $n^{th}$ bet. Then $W_n = -(1 + 2 + \ldots + 2^{n-2}) + 2^{n-1}$ $= -(2^{n-1} - 1) + 2^{n-1} = 1$ If we have not yet won after n bets then, $W_n = -(1 + 2 + \ldots + 2^{n-1}) = -2^n +1$ Finally, to show $W_n$ is a martingale, we just need to show $\mathbb{E}[W_{n+1} | W_n] = W_n$ which can be easily prove using iterated expectations. For case 1, $W_n = 1$, then $P(W_{n+1} = 1 | W_n = 1) = 1$ so $\mathbb{E}[W_{n+1} | W_n = 1] = 1 = W_n$ For case 2, $W_n = -2^n +1$, meaning we bet $2^n$ on $(n+1)^{th}$ toss so $W_{n+1} \in \{ 1 , -2^{n+1} + 1 \}$. Since $P(W_{n+1} = 1) | W_n = -2^n +1) = \frac{1}{2}$, and $P(W_{n+1} = -2^{n+1}+1 | W_n = -2^n +1) = \frac{1}{2}$, then $\mathbb{E}[W_{n+1} | W_n = -2^n +1] = \frac{1}{2} \times 1 + \frac{1}{2} \times (-2^{n+1} +1) = -2^n + 1 = W_n$ Thus, we showed that $\mathbb{E} [W_{n+1} | W_n] = W_n$ To bring what we learnt a further step, lets look at Polya’s Urn briefly. Consider an urn which contains red balls and green balls. Initially there is just one green ball and one red ball in the urn. At each time step, a ball is chosen randomly from the urn: If ball is red, the its returned to the urn with an additional red ball. If ball is green, then its returned to the urn with an additional green ball. Let $X_n$ denote the number of red balls in the urn after n draws. Then $P(X_{n+1} = k+1 | X_n = k) = \frac{k}{n + 2}$ $P(X_{n+1} = k | X_n = k) = \frac{n + 2 - k}{n + 2}$ We can show that $M_n := \frac{X_n}{n+2}$ is a martingale. ### Multivariate Distributions Let $X = (X_1 \ldots X_n)^T$ be an n-dimensional vector of random variables. For all $x = (x_1, \ldots, x_n) \in \mathbb{R}^n$, the joint cumulative distribution function of X satisfies $F_{X_i}(x_i) = F_X (\infty, \ldots, \infty, x_i, \infty, \ldots, \infty)$ Clearly it is straightforward to generalise the previous definition to join marginal distributions. For example, the join marginal distribution of $X_i$ and $X_j$ satisfies $F_X(x_1, \ldots, x_n) = \int_{-\infty}^{x_1} \ldots \int_{-\infty}^{x_n} f_x (u_1, \ldots, u_n) du_1 \ldots du_n$ If $X_1 = (X_1 \ldots X_k)^T$ and $X_2 = (X_{k+1} \ldots X_n)^T$ is a partition of X then the conditional CDF of X_2 given X_1 satisfies $F_{X_2|X_1} (X_2|X_1) = P(X_2 \le x_2 | X_1 = x_1). If X has a PDF,$latex f_X (\bullet)$, then the conditional PDF of $X_2$ given $X_1$ satisfies
$f_{X_2 | X_1} (X_2 | X_1) = \frac{f_X (X)}{f_{X_1}(X_1)} = \frac{f_{X_1 | x_2}(X_1 | X_2) f_{X_2}(X_2)}{f_{X_1}(X_1)}$

and the conditional CDF is then given by
$F_{X_2 | X_1}(X_2 |X_1) = \int_{- \infty}^{x_{k+1}} \ldots \int_{- \infty}^{x_n} \frac{f_X (x_1, \ldots , x_k, u_{k+1}, \ldots , u_n)}{f_{X_1}(X_1)} du_{k+1} \dots du_n$
where $f_{X_1}(\bullet)$ is the joint marginal PDF of $X_1$ which is given by
$f_{X_1} (x_1, \ldots , x_k) = \int_{- \infty}^{\infty} \ldots \int_{- \infty}^{\infty} f_X (x_1, \ldots , x_K u_{k+1}, \ldots , u_n) du_{k+1} \ldots du_n$

We next look at independence, which is something H2 Mathematics can easily relate to.

Here, we say the collection X is independent if joint CDF can be factored into the product of the marginal CDFs so that
$F_X (x_1 \ldots , x_n) = F_{X_1} (x_1) \ldots F_{X_n}(x_n)$

If X has a PDF, $f_X(\bullet)$ then independence implies that the PDF also factories into the product of marginal PDFs so that
$f_X(x) = f_{X_1} (x_1) \ldots f_{X_n}(x_n)$.

Using the above results, we have that if $X_1$ and $X_2$ are independent then
$f_{x_2|x_1}(x_2 | x_1) = \frac{f_X (X)}{f_{X_1}(X_1)} = \frac{f_{X_1}(X_1) f_{X_2}(X_2)}{f_{X_1}(X_1)} = f_{X_2}(X_2)$
The above results tell us that having information about $X_1$ tells us nothing about $X_2$.

Let’s continue to look further on the implications of independence.

Let X and Y be independent random variables. Then for any events A and B, $P(X \in A, Y \in B) = P(X \in A) P(Y \in B)$

We can check this with,
$P(X \in A, Y \in B)$
$= \mathbb{E}[1_{X \in A} 1_{Y \in B}]$
$= P(X \in A) P(Y \in B)$

In general, if $X_1, \ldots, X_n$ are independent random variables then
$\mathbb{E}[f_1 (X_1) f_2(X_2) \ldots f_n(X_n)] = \mathbb{E}[f_1(X_1)] \mathbb{E}[f_2(X_2)] \ldots \mathbb{E}[f_n(X_n)]$

Moreover, random variables can also be conditionally independent. For example, we say that X and Y are conditionally independent given Z if
$\mathbb{E}[f(X)g(Y)|Z] = \mathbb{E}[f(X)|Z]\mathbb{E}[g(Y)|Z]$.
The above will be used in the Gaussian copula model for pricing of collateralised debt obligation (CDO).

We let $D_i$ be the event that the $i^{th}$ bond in a portfolio defaults. Not reasonable to assume that the $D_i$‘s are independent though they are conditionally independent given Z so $P(D_1, \ldots, D_n |Z ) = P(D_1|Z) \ldots P(D_n|Z)$ is often easy to compute.

Lastly, we consider the mean and covariance.
The mean vector of X is given by $\mathbb{E}[X]:=(\mathbb{E}[X_1] \ldots \mathbb{E}[X_n])^T$
and the covariance matrix of X satisfies
$\sum := \mathrm{Cov}(X) := \mathbb{E}[(X- \mathbb{E}[X])(X - \mathbb{E}[X])^T]$ so the $(i,j)^{th}$ element of $\sum$ is simply the covariance of $X_i$ and $X_j$.
The covariance matrix is symmetric and its diagonal elements satisfy $\sum_{i, i} \ge 0$ and is also positive semi definite so that $x^T \sum x \ge 0$ for all $x \in \mathbb{R}^n$
Then the correlation matrix $\rho (X)$ has $(i, j)^{th}$ element $\rho_{ij} := \mathrm{Corr}(X_i, X_j)$. This is also symmetric, postiie semi-definite and has 1’s along the diagonal.

For any matrix $A \in \mathbb{R}^{k \times n}$ and vector $a \in \mathbb{R}^k$,
$\mathbb{E}[AX +a] = A \mathbb{E} [X] + a$ (distributes linearly)
$\mathrm{Cov}(AX +a) = A \mathrm{Cov}(X) A^T$
$\Rightarrow \mathrm{Var} (aX + bY) = a^2 \mathrm{Var}(X) + b^2 \mathrm{Var}(Y) + 2ab \mathrm{Cov}(X, Y)$
Recall that if X and Y are independent, then $\mathrm{Cov}(X, Y)=0$, and the converse is not true in general.

### Conditional Expectations and Variances

Here we look at an important concept that is an extension from Bayes Theorem, which we discussed briefly.

The condition expectation identity says $\mathbb[X] = \mathbb[\mathbb[X|Y]]$
The condition variance identity says $Var(X) = Var(\mathbb[X|y]) + \mathbb[Var(X|Y)]$

Here both $\mathbb{E}[X|Y]$ and $Var(X|Y)$ are both functions of Y and are therefore random variables themselves.

With this, we start by considering a random sum of random variables. Let $W= X_1 + X_2 + \ldots + X_N$ where $X_i$‘s are IID with mean $\mu_x$ and variance ${\sigma_x}^2$, where $N$ is also a random variable, independent of $X_i$‘s.

$\mathbb{E}[W]$
$= \mathbb{E} [\mathbb{E}[\sum_{i=1}^N x_i | N]]$
$= \mathbb{E} [N \mu_x]$
$= \mu_x \mathbb{E}[N]$

$Var(W)$
$= Var(\mathbb{E}[W|N]) + \mathbb{E}[Var(W|N)]$
$= Var(\mu_x N) + \mathbb{E}[N {\sigma_x}^2]$
$= {\mu_x}^2 Var(N) + {\sigma_x}^2 \mathbb{E}[N]$

### Continuous Random Variables

We look at the definitions first.

A continuous random variable, X, has a probability density function (PDF), $f(\bullet)$ if $f(x) \ge 0$ and for all events A
$P(X \in A) = \int_A f(y) dy$
The CDF and PDF are related by $F(x) = \int_{-\infty}^x f(y) dy$
It is good to know that we have $P(X \in (x - \frac{\epsilon}{2}, x + \frac{\epsilon}{2})) \approx \epsilon f(x)$

We X has a normal distribution, $X \sim N(\mu, {\sigma}^2)$, and $f(x) = \frac{1}{\sqrt{2 \pi {\sigma}^2}}e^(-\frac{(x - \mu)^2}{2 {\sigma}^2})$. And $\mathbb{E}[X] = \mu$ while $Var(X) = {\sigma}^2$

We also have the log-normal distribution, $X \sim LN (\mu, {\sigma}^2)$ and $log(X) \sim N(\mu, {\sigma}^2)$. Here, $\mathbb{E} [X] = e^(\mu + \frac{{\sigma}^2}{2})$ and $Var(X) = e^(2 \mu + {\sigma}^2) (e^{{\sigma}^2} - 1)$. The log-normal distribution is very important in financial applications, for starters, the Black Scholes Equation.

### Discrete Random Variables

As usual, definitions first.

The cumulative distribution function (CDF), $F(\bullet)$ of a random variable, X, is defined by
$F(x) := P(X \le x)$

A discrete random variable, X, has probability mass function (PMF), $p(\bullet)$, if $p(x) \ge 0$ and for all events A we have $P(X \in A) = \sum_{x \in A} p(x)$

The expected value of a discrete random variable, X is given by
$\mathbb{E} [X] := \sum_i x_i p(x_i)$

The variance of an random variable, X, is defined as $Var(X) := \mathbb{E} [(X- \mathbb[X])^2] = \mathbb{E}[X^2] - \mathbb{E}[X]^2$

Now we can look at some examples of discrete random variables.

Binomial Distribution

We say the X has a binomial distribution, that is, $X \sim Bin(n, p)$, if $P(X=r) = {n \choose r} p^r (1-p)^{n-r}$
For example, X can represent the number of heads in n independent coin tosses, where $p = P(head)$. We have that the mean $\mathbb{E}[X] = np$ and variance $Var(X) = np(1-p)$

A simpler case of binomial where there is only one event is called the Bernoulli distribution.

I’ll give a simple illustration of binomial model used in finance. This was also used in finance engineering in the beginning.
Suppose a fund manager outperforms the market in a given year with probability p and under performs the market with probability 1 – p. She has a track record of 10 years and has outperformed the market in 8 out of 10 years. We also note that performance in any one year is independent of performance in other years.
From this illustration, we note that there are only two outcomes, she outperforms or underperforms. We can let X be the number of outperforming years. Assuming the fund manager has no skill, $X \sim Bin(10, \frac{1}{2})$ and we can find $P(X \ge 8)$ to find out the probability that she outperforms at least 8 out of 10 years.
An extension here will be to consider there are M fund managers instead of 1 now.

Poisson Distribution

We say that X has a $Po(\lambda)$ distribution if
$P(X=r) \frac{{\lambda}^r e^{- \lambda}}{r!}$
$\mathbb{E} [X] = \lambda$
$Var(X) = \lambda$

Next, we look at Bayes Theorem, also known as “conditional probability” in H2 Mathematics.

Let A and B be two events for which $P(B) \neq 0$. Then

$P(A | B)$
$= \frac{P(A \cap B)}{P(B)}$
$= \frac{P(B | A) P(A)}{\sum_j P(B | A_j) P(A_j)}$
where $A_j$‘s form a partition of the sample space.

### Game Theory #2

To illustrate another similar example from last post, now supposed ALL the A-level Candidates met up in the town square. Prior to meeting up, they were all told by the head examiner that if everyone gets the same grade, all of you get an A. Else, the highest will have an A while the rest will fail. One of the students then suggested during he meet up, that why then they all fail the exam together and get zero marks. That will guarantee everyone an A. Supposed you’re one of the candidates there, will you be an angel and follow or a devil that deviate and score 1 mark.

This is of a similar nature to what we saw previously regarding the Prisoner’s Dilemma, only difference is that we have a lot more integrities to consider. After all, there is no telling an angel from a devil, So will you be an angel or devil?

To properly solve this, we need to define a few more things.

A strategy $s_{i} {\in} S_{i}$ is a strictly dominant strategy for player i in game $G= [I, [S_i ], \{u_i( \bullet) \} ]$ if for all ${s \prime}_i \neq s_i$, we have $u_i (s_i, s_{-i}) > u_i (s’_i, s_{-i})$ for all $s_{-i} \in S_{-i}$

A strategy $s_i \in S_i$ is a strictly dominated strategy for player i in game $G = [I, \{ S_i \}, \{u_i(\dot) \}]$ if there exists another strategy $s’_i \in S_i$ such that for all $s_{-i} \in S_{-i}$, we have $u_i(s’ \prime _i, s_{-i}) > u_i (s_i, s_{-i})$.

Okay, so let us look at these definitions closely first, before you guys go like meh sets?! This is a definition, and definitions tend to be boring. Let me begin by saying that the index i refers to the player, 1 or 2 for instance. and u refers to utility (payoff) while s here represents the various strategies that the player players. Now that the variables are sort of defined, are you starting to see how this definition is making sense? And ‘ here refers to a complementary strategy.

Next, a decision maker is rational if the strategy she chooses is at least as good as any other available strategy. Clearly, a rational player maximises her (expected) payoff.

1. In the prisoner”s dilemma we saw, rational prisoners will rat on each other.
2. In the angel vs devil students’ dilemma we saw, rational students will still do their best.Of course, this two cases differ in the number of people participating, which brings us to the next focus: Iterated deletion of strictly dominated strategies.

Here, we assume that rationality is common knowledge, that is, all man is rational. Thus, as a decision, he will never play a strictly dominated strategy (since it does not maximise his payoff))

Before I go further, I should clarify that payoff is different from income. Consider a dictator allocating an amount of money between himself and another player, the dictator might be altruistic or inequality averse (since its based on his beliefs) and thus maximises her utility (payoff) differently, instead of income. And we remind ourselves again, that a rational player will want to maximise his utility.

So next time, we will look at iterated deletion of strictly dominated strategies.

### Game Theory #1

Over the past two months, I’ve been reading up on Game Theory extensively. It has always been a subject of interest for me since undergraduate days. And I’m glad that I found time to read a bit on it. Students that are keen, can also follow the open courseware provided by Yale University for free.

Firstly, this is going to be slightly mathematical (I’m after all trained in Mathematics) and will require some statistics. So let us start by considering the most rudimentary case of Prisoner’s Dilemma, to provide a simple introduction on what follows.

Suppose we have two prisoners and a police officer. Both prisoners are facing a two year sentence. But the police officer offers them a deal now, that if the prisoner cooperate and rat out the other, he will go scot-free while his partner faces seven years of jail. We assume that the police officer offers the same deal to both prisoners independent. So if both chooses to rat, both of them will face five years of jail.

So here, we are facing a situation of strategic interdependence in the sense that each individual’s payoff (jail sentence) depends not only on her own actions, but also on the action of the other individual. As a result, an individual’s best action may depend on what she expects the other player to do. Given this game, what will you do if you’re a prisoner?

Before looking at the game, let us properly define the following:

Dominant Strategy: A strategy that is always the best for a player no matter what other players do.

Dominated Strategy: A strategy that generates worse payoffs than some other strategy for any choices made by other players.

Other words, if you have a dominant strategy here, you should use it while you should use a dominated strategy.

Back to the game, we shall set off a simple playoff matrix here and then analyse the game from there.

So what will I do, I’ll just rat on the other prisoner. Yes, let us be honest and rationale. After all, looking at the table, and assuming he goes through the similar train of thoughts, he probably rat on me. I should do the same to protect my self interests.

Now, any idea why this is called a dilemma? Because each player has a dominant strategy here and the result is Pareto inefficient (socially inefficient outcome).

There isn’t a right or wrong answer here, and it is entirely based on what sort of equilibrium we are looking at. Of course, what kind of people you are dealing with deep down.

I hope this provides a simple introduction to Game Theory. We will look more into various games like auction and business models, as we progress. We will also consider some limitations, like imperfect knowledge. I hope this spark some interest in this subject and gradually show you how Game theory deepens our understanding about a lot of real-world phenomena/ problem!

### Bayesian vs Frequentist Statistics

Some of you will probably hear of this two terms and wonder what’s the difference. Both of them are a statistical approach/ inference method. Personally, I am a Bayesian and use more of Bayesian Methods to do my stuffs.

Firstly, the frequentist approach is a much simple model and does not consider prior knowledge. This is a very stark difference that we should note, before looking at a bette definition.

To illustrate it, I’ll borrow an example that I saw online. So I lost my phone somewhere at home and I can use the “find my phone” app to locate it. Once I activate the app, the phone will start to beep. So which area should I search?

Now, the Frequentist reasoning considers probability derived from long run frequency distributions and the underlying parameters are constant during the process. Going back to the example, I’ll have the ability to identify the area where the beep is coming from. And upon hearing the beep, I infer the area that I must search.

Next, the Bayesian reasoning considers probability to be constantly updated from the realisation of new information. Parameters are unknown and described probabilistically. Going back to the example again, apart from the ability to identify the area where the beep is coming from, I also have prior knowledge of where I had misplaced my phone at before. With the aid of the beep and prior knowledge of my clumsiness, I will be able to identify which area to search now. Here, you should intuitively realise that, we have better chance of finding the phone faster. 🙂

This blog post does an excellent job to realise the intuition behind frequentist and bayesian reasoning.

In A’levels Statistics, the closest contact students have with Bayesian Statistics will be Conditional Probability, whereby we understand that the probability of an event changes given new information.

Bayesian Statistics has a branch called Bayesian Estimations which, in the context of what we had discussed, considers probability by constantly being updated with current information. Examples will be the interest rate trends, hedging. This is used widely in real world today. Of course, to do this branch of statistics, one has to be really well acquitted with statistic distributions.

### What are stochastic and deterministic processes?

During class, I occasionally mention to students that the calculus you learnt is deterministic and they go uh huh. So like what is non deterministic?! I’ll attempt to give some simple definitions and example and also touch on stochastic process.

First, a deterministic process is one that given at a particular state and an action, we only have one possible successive state to move on to. That is what we do in A-levels, since we often don’t consider any randomness in our work.

Second, a non-deterministc process is one that given at a particular state and an action, we have a set of possible successive state to move on to.

Lastly, a stochastic process is one that to begin with, has a probability of to be at a particular state, and has a finite number of states that it can transit to. It also has transition probabilities, which adds up to one.

A deterministic process is one where the present state completely determines the future state. If I make a (riskless) investment of $10,000 at 10% interest, compounded annually, then in one year’s time I will have$11,000, in two years’ time I will have $12,100, and so forth. A stochastic process, on the other hand, is one where there’s some randomness or uncertainty involved. If I buy$10,000 worth of $\chi$-company stock today, how much money will it be worth in one year’s time? In two years’ time? Who knows? This is a stochastic process.


text:
## Calculus 8th Edition

Calculate $f(t)$ and $f(-t)$ $f(t)=\frac{t^{2}}{t^{4}+1}$ $f(-t)=\frac{(-t)^{2}}{(-t)^{4}+1}=\frac{t^{2}}{t^{4}+1}$ therefore we can see that the function f is an even function that satisfies $f(-x)=f(x)$ for every x in the domain.


text:
# Set theory (subset question)

## Homework Statement

Given the following four statements concerning the student body at CU:
...
b) There are no women engineering students at CU
...

n/a

## The Attempt at a Solution

Let W be the set of all women
Let E be the set of all engineering students

W$\subseteq$E'

Therefore
W'$\subseteq$E

However I don't know valid the first, and therefore the second statement is. In the book I'm using, it doesn't cover any "not a subset" other than complimentary.

However, to conclude that all non-women are engineers seems fallacious. Or is that the point?

Is is valid to say:

W$\subseteq$E'

The set of all women are in the set of non-engineering students.

It's not really homework for a class, but is homework style.


text:
The $$\chi^2$$ independence test is an inferential method to decide whether an association exists between two variables. Similar to other hypothesis tests the null hypothesis states that the two variables are not associated. In contrast, the alternative hypothesis states that the two variables are associated.

Recall that statistically dependent variables are called associated variables. In contrast, non-associated variables are called statistically independent variables. Further recall the concept of contingency tables (also known as two-way table, cross-tabulation table or cross tabs), which display the frequency distributions of bivariate data.

#### $$\chi^2$$ Independence Test

The basic idea behind the $$\chi^2$$ independence test is to compare the observed frequencies in a contingency table with the expected frequencies, given the null hypothesis of non-association is true. The expected frequency for each cell of a contingency table is given by

$E = \frac{R\times C}{n} \text{,}$ where $$R$$ is the row total, $$C$$ is the column total, and $$n$$ is the sample size.

Let us construct an example for a better understanding. We consider an exit poll in form of a contingency table that displays the age of $$n= 1189$$ people in categories from 18-29, 30-44, 45-64 and >65 years, and their political affiliation, which is “Conservative”, “Socialist” and “Other”. This table corresponds to the observed frequencies.

Observed frequencies: $\begin{array}{|l|c|} \hline \ & \text{Conservative} & \text{Socialist} & \text{Other} & \text{Total} \\ \hline \ \text{18-29} & 141 & 68 & 4 & 213\\ \ \text{30-44} & 179 & 159 & 7 & 345\\ \ \text{45-64} & 220 & 216 & 4 & 440\\ \ \text{65 & older} & 86 & 101 & 4 & 191\\ \hline \ \text{Total} & 626 & 544 & 19 & 1189\\ \hline \end{array}$

Based on the equation given above we calculate the expected frequency for each cell.

Expected frequencies:

$\begin{array}{|l|c|} \hline \ & \text{Conservative} & \text{Socialist} & \text{Other} & \text{Total} \\ \hline \ \text{18-29} & \frac{213 \times 626 }{1189} \approx 112.14 & \frac{213 \times 544 }{1189} \approx97.45 & \frac{213 \times 19 }{1189} \approx3.4 & 213\\ \ \text{30-44} & \frac{345 \times 626 }{1189} \approx181.64 &\frac{345 \times 544 }{1189} \approx 157.85 & \frac{345 \times 19 }{1189} \approx5.51 & 345\\ \ \text{45-64} & \frac{440 \times 626 }{1189} \approx231.66 & \frac{440 \times 544 }{1189} \approx201.31 &\frac{440 \times 19 }{1189} \approx 7.03 & 440\\ \ \text{65 & older} &\frac{191 \times 626 }{1189} \approx 100.56 &\frac{191 \times 544 }{1189} \approx 87.39 & \frac{191 \times 19 }{1189} \approx3.05 & 191\\ \hline \ \text{Total} & 626 & 544 & 19 & 1189\\ \hline \end{array}$

Once we know about the expected frequencies we have to check for two assumptions. First, we have to make sure that all expected frequencies are 1 or greater, and second, at most 20% of the expected frequencies are less than 5. By looking at the table we may confirm that both assumptions are fulfilled.

The actual comparison is done based on the the $$\chi^2$$ test statistic for the observed frequency and the expected frequency. The $$\chi^2$$ test statistic follows the $$\chi^2$$ distribution and is given by

$\chi^2= \sum{\frac{(O-E)^2}{E}}\text{,}$

where $$O$$ represents the observed frequency and $$E$$ represents the expected frequency. Please note that $$\frac{(O-E)^2}{E}$$ is evaluated for each cell and then summed up.

The number of degrees of freedom are given by

$df = (r-1) \times (c-1)\text{,}$

where $$r$$ and $$c$$ are the number of possible values for the two variables under consideration.

Adopted to the example from above this leads to a somehow long expression, which for the sake of brevity is just given for the first and the last row of the contingency tables of interest.

$\chi^2 = \frac{141 \times 112.14}{112.14} + \frac{68 \times 97.45}{97.45} + \frac{4 \times 3.4}{3.4} + ... + \frac{86 \times 100.56}{100.56} + \frac{101 \times 87.39}{87.39} + \frac{4 \times 3.05}{3.05}$

If the null hypothesis is true, the observed and expected frequencies are roughly equal, resulting in a small value of the $$\chi^2$$ test statistic; thus, supporting $$H_0$$. If, however, the value of the $$\chi^2$$ test statistic is large, the data provides evidence against $$H_0$$. In the next sections we further discuss how to asses the value $$\chi^2$$ test statistic in the framework work of hypothesis testing.

#### $$\chi^2$$ Independence Test: An Example

In order to get some hands-on experience we apply the $$\chi^2$$ independence test in an exercise. Therefore we load the students data set. You may download the students.csv file here. Import the data set and assign a proper name to it.

students <- read.csv("https://userpage.fu-berlin.de/soga/200/2010_data_sets/students.csv")

The students data set consists of 8239 rows, each of them representing a particular student, and 16 columns, each of them corresponding to a variable/feature related to that particular student. These self-explaining variables are: stud.id, name, gender, age, height, weight, religion, nc.score, semester, major, minor, score1, score2, online.tutorial, graduated, salary.

In this exercise we want to examine if there is an association between the variables gender and major, or in other words, we want do know if male students favor different study subjects compared to female students.

#### Data preparation

We start with the data preparation. We do not want to deal with the whole data set of 8239 entries, thus we randomly select 865 student from the data set. The first step of data preparation is to display our data of interest in form of a contingency table. R provides the fancy table() function, which will do the job.

n <- 865
data.idx <- sample(1:nrow(students), n)
data <- students[data.idx,]
conti.table <- table(data$major, data$gender)
conti.table
##
##                              Female Male
##   Biology                       112   58
##   Economics and Finance          36   93
##   Environmental Sciences         72   95
##   Mathematics and Statistics     25  114
##   Political Science              92   68
##   Social Sciences                63   37

Further, we determine the column sums and the row sums. The R software package provides the margin.table() function, which takes as an argument the variable conti.table and additionally one argument, either the integer $$1$$, for the row sums, or $$2$$ for the column sums.

row.sum <- margin.table(conti.table, 1)
row.sum
##
##                    Biology      Economics and Finance
##                        170                        129
##     Environmental Sciences Mathematics and Statistics
##                        167                        139
##          Political Science            Social Sciences
##                        160                        100
col.sum <- margin.table(conti.table, 2)
col.sum
##
## Female   Male
##    400    465

For visualization purposes we join the data in a matrix object. Moreover, we apply the as.vector() function to extract the numerical information from the table object. The resulting matrix corresponds to the observed frequencies.

conti.added <- cbind(conti.table, as.vector(row.sum))
conti.added
##                            Female Male
## Biology                       112   58 170
## Economics and Finance          36   93 129
## Environmental Sciences         72   95 167
## Mathematics and Statistics     25  114 139
## Political Science              92   68 160
## Social Sciences                63   37 100
##                               400  465 865

In the next step we construct the expected frequencies. Recall the equation for above:

$E = \frac{R\times C}{n} \text{,}$ where $$R$$ is the row total, $$C$$ is the column total, and $$n$$ is the sample size.

We compute the expected frequencies cell-wise, by implementing a nested for-loop. Thereby we go through all rows of the matrix column by column and calculate the expect frequency $$E$$ for each particular cell.

# initialize empty matrix
expected.frequencies <- matrix(data = NA,
nrow = nrow(conti.table),
ncol = ncol(conti.table))

# nested for-loop
for (row.idx in 1:nrow(conti.table)){
for (col.idx in 1:ncol(conti.table)){
expected.frequencies[row.idx, col.idx] <- (row.sum[row.idx]*col.sum[col.idx])/sum(conti.table)
}
}
expected.frequencies
##          [,1]     [,2]
## [1,] 78.61272 91.38728
## [2,] 59.65318 69.34682
## [3,] 77.22543 89.77457
## [4,] 64.27746 74.72254
## [5,] 73.98844 86.01156
## [6,] 46.24277 53.75723

Nice! For a better readability we assign column and row names to the expected.frequencies matrix.

rownames(expected.frequencies) <- rownames(conti.table)
colnames(expected.frequencies) <- colnames(conti.table)
expected.frequencies
##                              Female     Male
## Biology                    78.61272 91.38728
## Economics and Finance      59.65318 69.34682
## Environmental Sciences     77.22543 89.77457
## Mathematics and Statistics 64.27746 74.72254
## Political Science          73.98844 86.01156
## Social Sciences            46.24277 53.75723

Once we know about the expected frequencies we have to check for two assumptions. First, we have to make sure that all expected frequencies are 1 or greater, and second, at most 20% of the expected frequencies are less than 5. By looking at the table we may confirm that both assumptions are fulfilled.

Now we have all the data we need to perform a $$\chi^2$$ independence test.

#### Hypothesis testing

In order to conduct the $$\chi^2$$ independence test we follow the step-wise implementation procedure for hypothesis testing. The $$\chi^2$$ independence test follows the same step-wise procedure as discussed in the previous sections. $\begin{array}{l} \hline \ \text{Step 1} & \text{State the null hypothesis } H_0 \text{ and alternative hypothesis } H_A \text{.}\\ \ \text{Step 2} & \text{Decide on the significance level, } \alpha\text{.} \\ \ \text{Step 3} & \text{Compute the value of the test statistic.} \\ \ \text{Step 4} &\text{Determine the p-value.} \\ \ \text{Step 5} & \text{If } p \le \alpha \text{, reject }H_0 \text{; otherwise, do not reject } H_0 \text{.} \\ \ \text{Step 6} &\text{Interpret the result of the hypothesis test.} \\ \hline \end{array}$

Step 1: State the null hypothesis $$H_0$$ and alternative hypothesis $$H_A$$

The null hypothesis states that there is no association between the gender and the major study subject of students.

$H_0: \text{No association between gender and major study subject}$

Alternative hypothesis $H_A: \quad \text{There is an association between gender and major study subject}$

Step 2: Decide on the significance level, $$\alpha$$

$\alpha = 0.05$

alpha <- 0.05

Step 3 and 4: Compute the value of the test statistic and the p-value.

For illustration purposes we manually compute the test statistic in R. Recall the equation for the test statistic from above:

$\chi^2= \sum{\frac{(O-E)^2}{E}}\text{,}$ where $$O$$ represents the observed frequency and $$E$$ represents the expected frequency.

To calculate the test statistic we perform calculations cell by cell. Thus, we again apply a nested for-loop. To make the code more readable we reassign the observed and expected frequencies to the variables of and ef respectively.

# Compute the value of the test statistic

of <- conti.table
ef <- expected.frequencies

# initialize test statistic to zero
x2 <- 0

# nested for-loop
for (row.idx in 1:nrow(of)){
for (col.idx in 1:ncol(of)){
cell <- (of[row.idx, col.idx]-ef[row.idx, col.idx])^2 / ef[row.idx, col.idx]
x2 <- x2 + cell # update variable x2 cell by cell
}
}
x2
## [1] 108.581

The numerical value of the test statistic is $$\approx 108.58$$.

In order to calculate the p-value we apply the pchisq() function. Recall how to calculate the degrees of freedom:

$df = (r-1) \times (c-1)\text{,}$

where $$r$$ and $$c$$ are the number of possible values for the two variables under consideration.

# Compute df
df <- nrow(of) * ncol(of)

# Compute the p-value
p <- pchisq(x2, df = df, lower.tail = FALSE)
p
## [1] 1.141509e-17

Step 5: If $$p \le \alpha$$, reject $$H_0$$; otherwise, do not reject $$H_0$$.

p <= alpha
## [1] TRUE

The p-value is less than the specified significance level of 0.05; we reject $$H_0$$. The test results are statistically significant at the 5% level and provide very strong evidence against the null hypothesis.

Step 6: Interpret the result of the hypothesis test.

$$p = 1.1415093\times 10^{-17}$$. At the 5% significance level, the data provides very strong evidence to conclude that there is an association betweeen gender and the major study subject.

#### Hypothesis testing in R

We just completed a $$\chi^2$$ independence test in R manually. We can do the same in R by just one line of code!

Therefore we apply the chisq.test() function. For the function we either provide two vectors as data input, such as data$major and data$gender, or we provide table objects such as of and ef.

# vector data
chisq.test(data$major, data$gender)
##
##  Pearson's Chi-squared test
##
## data:  data$major and data$gender
## X-squared = 108.58, df = 5, p-value < 2.2e-16
# table objects
chisq.test(of, ef)
##
##  Pearson's Chi-squared test
##
## data:  of
## X-squared = 108.58, df = 5, p-value < 2.2e-16

Perfect! Compare the output of the chisq.test() function with our result from above. Again, we may conclude that at the 5% significance level, the data provides very strong evidence to conclude that there is an association betweeen gender and the major study subject.


text:
# Why must the ground state of hydrogen molecule be gerade?

As we know, the electronic ground state of hydrogen molecule is $\Sigma_g^+$. My question is, how do we know it from the theoretical perspective?

The nuclear potential in the hamiltonian of hydrogen molecule is of the inversion symmetry, so the electronic wave function of the system must be gerade or ungerade. However, how can we derive that the ground state is of gerade symmetry?


text:
# 2005 Smooth949InsensitiveRegressionb

Subject Headings: Regressor, Regression Algorithm, Linear Regressor.

## Quotes

### Abstract

We describe new loss functions for regression problems along with an accompanying algorithmic framework which utilizes these functions. These loss functions are derived by symmetrization of margin-based losses commonly used in boosting algorithms, namely, the logistic loss and the exponential loss. The resulting symmetric logistic loss can be viewed as a smooth approximation to the ε-insensitive hinge loss used in support vector regression. We describe and analyze two parametric families of batch learning algorithms for minimizing these symmetric losses. The first family employs an iterative log-additive update which can be viewed as a regression counterpart to recent boosting algorithms. The second family utilizes an iterative additive update step. We also describe and analyze online gradient descent (GD) and exponentiated gradient (EG) algorithms for the symmetric logistic loss. A byproduct of our work is a new simple form of regularization for boosting-based classification and regression algorithms. Our regression framework also has implications on classification algorithms, namely, a new additive update boosting algorithm for classification. We demonstrate the merits of our algorithms in a series of experiments.

### 1. Introduction

The focus of this paper is supervised learning of real-valued functions. We observe a sequence $S = {(x_1,y_1),...,(x_m,y_m)}$ of instance-target pairs, where the instances are vectors in Rn and the targets are real-valued scalars, yi 2 R. Our goal is to learn a function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ which provides a good approximation of the target values from their corresponding instance vectors. Such a function is often referred to as a regression function or a regressor for short. Regression problems have long been the focus of research papers in statistics and learning theory (see for instance the book by [[Hastie, Tibshirani, and Friedman (2001)]] and the references therein). In this paper we discuss [[learning of linear regressor]]s, that is, $f$ is of the form $f(\bf{x}) = \lambda \cdot \bf{x}$. This setting is also suitable for learning a linear combination of base regressors of the form $f(\bf{x}) = \Sigma^l_{j=1} \lambda_j h_j(\bf{x})$ where each base regressor $h_j$ is a mapping from an instance domain $X$ into $\R$. The latter form enables us to employ kernels by setting $h_j(\bf{x}) = K(x_j,\bf{x})$.

The class of linear regressors is rather restricted. Furthermore, in real applications both the instances and the target values are often corrupted by noise and a perfect mapping such that for all $(x_i, y_i) \in S$, $f(x_i) = y_i$ is usually unobtainable. Hence, we employ a loss function $L : \mathbb{R} \times \mathbb{R} \rightarrow \R_+$ which determines the penalty for a discrepancy between the predicted target, $f(\bf{x})$, and the true (observed) target $y$. As we discuss shortly, the [[loss function]]s we consider in this paper depend only on the discrepancy between the [[predicted target]] and the true target $\delta = f(\bf{x}) − y$, hence $L$ can be viewed as a function from [/itex]\R[/itex] into $\R_+$. We therefore allow ourselves to overload our notation and denote $L(\delta) = L(f(\bf{x}),y)$.

Given a loss function [/itex]L[/itex], the goal of a regression algorithm is to find a regressor $f$ which attains a small total loss on the training set $S$,

$\text{Loss}(\lambda,S) = \Sigma^m_{i=1}L(f(x_i)-y_i) = \Sigma^m_{i=1}L(\lambda \cdot x_i - y_1)$

Denoting the discrepancy $\lambda \cdot x_i − y_i$ by $\delta_i$, we note that two common approaches to solving regression problems are to minimize either the sum of the absolute discrepancies over the sample ( P i |�i|) or the sum of squared discrepancies ( P i �2 i). It has been argued that the squared loss is sensitive to outliers, hence robust regression algorithms often employ the absolute loss (Huber, 1981). Furthermore, it is often the case that the exact discrepancy between � · x and y is unimportant so long as it falls below an insensitivity parameter ". Formally, the "-insensitive hinge loss, denoted |�|", is zero if |�| � " and is |�| − " for |�| > " (see also the left hand side of Figure 2). The "-insensitive hinge loss is not smooth as its derivative is discontinuous at � = ±". Several batch learning algorithms have been proposed for minimizing the "-insensitive hinge loss (see for example Vapnik, 1998; Smola and Sch¨olkopf, 1998). However, these algorithms are based on rather complex constrained optimization techniques since the "-insensitive hinge loss is a non-smooth function.

...

## References

,

volumeDate ValuetitletypejournaltitleUrldoinoteyear
2005 Smooth949InsensitiveRegressionbSmooth ε-Insensitive Regression by Loss Symmetrization2005
 Author Ofer Dekel +, Shai Shalev-Shwartz + and Yoram Singer + title Smooth ε-Insensitive Regression by Loss Symmetrization + year 2005 +


text:
# Find Flux Vector

• September 22nd 2009, 06:42 AM
Jimmy_W
Find Flux Vector
For $\Phi (x,y,z) = x^{xy} \sin(x+z) + y \ln (z)$

find the flux vector $\vec{J} = \triangledown \Phi$

and prove that curl $\vec{J} = 0$
• October 4th 2009, 03:08 AM
Jimmy_W
Is this correct?

$\triangledown \Phi = \frac{\partial \Phi}{\partial x} i + \frac{\partial \Phi}{\partial y} j + \frac{\partial \Phi}{\partial z} k$

where

$\frac{\partial \Phi}{\partial x} = e^{xy} \cos(x+z) + \sin(x+z) y e^{xy}$

$\frac{\partial \Phi}{\partial y} = \sin(x+z) x e^{xy} + \ln(z)$

and

$\frac{\partial \Phi}{\partial z} = e^{xy} \cos(x+z) + \frac{y}{x}$

and

$\triangledown \Phi = (e^{xy} \cos(x+z) + \sin(x+z) y e^{xy}) i + (\sin(x+z) x e^{xy} + \ln(z) ) j + (e^{xy} \cos(x+z) + \frac{y}{x})k$

Is there anything more I need to do with this?
• October 4th 2009, 03:26 AM
MarkW
In the k component it should be y/z, not y/x.

Other than that, all you need to do is calculate the curl and it will cancel out nicely.

Do you know how to find the general flux vector in the second part of question 1?


text:
Determine the radius of convergence of given power series.

Determine the radius of convergence of given power series.

(a) $\sum_{n=1}^\infty {(n^{1/n} - 1)x^n}$

(b) $\sum_{n=1}^\infty {{n^c \over n!}x^n}$ ($c \in \mathbb R$)

I tried root test for (a) and got

$$0 \le |n^{1/n} - 1|^{1/n} \lt |n^{1/n}|^{1/n} \le n^{1/n}$$

but I'm not sure if I can use this result.

And for (b) I tried ratio test, and found that if $c=1$, the power series clearly converges for all $x$. But I'm stuck in how to solve the rest case of $c$.

(a) Let $f(x)=x^{-x}$. Then $\lim_{x\to0}f(x)=1$ and $f'(x)=-x^{-x}\bigl(1+\log(x)\bigr)$. Therefore, $\lim_{x\to0}\frac{f(x)-1}x=\lim_{x\to0}f'(x)=+\infty$. So$$\lim_{n\in\mathbb N}\frac{\sqrt[n]n-1}{1/n}=+\infty$$and so, if $n\gg1$, $\displaystyle\frac1n<\sqrt[n]n-1<1$. Since the radius of convergence of both series $\sum_{n=0}^\infty\frac{x^n}n$ and $\sum_{n=0}^\infty x^n$ is $1$, the radius of convergence of your series is also $1$.
(b) $\displaystyle\frac{\frac{(n+1)^c}{(n+1)!}}{\frac{n^c}{n!}}=\frac1{n+1}\left(1+\frac1n\right)^c\to0\times1=0$. So, the radius of convergence is $+\infty$.
• I suppose in (a) you meant $$\lim_{x\to\color{red}{0^+}}x^{-x}=1\;\ldots$$ – DonAntonio Jun 12 '17 at 12:16
• @DonAntonio Yes. Since $x^x$ is defined only when $x>0$, I did not feel the need of writing $0^+$ instead of simply writing $0$. – José Carlos Santos Jun 12 '17 at 12:18


text:
# Sum over all inverse zeta nontrivial zeros

Starting from the Hadamard product for the Riemann Zeta Function (assuming the product is taken over matching pairs of zeros) $$\zeta(s)=\frac{e^{(\log(2\pi)-1-\gamma/2)s}}{2(s-1)\Gamma(1+s/2)}\prod_{\rho}\left(1-\frac{s}{\rho} \right)e^{s/\rho}$$ can one derive the exact value of $\sum_{\rho} \frac{1}{\rho}$ to be $$\sum_{\rho} \frac{1}{\rho} = -\log(2\sqrt{\pi})+1+\gamma/2$$ What implications does this have?

• Basically none. Also note the sum doesn't converge absolutely. – Wojowu Jul 25 '18 at 16:49

Let's start with the logarithm of the Hadamard product and take the derivative relatively to $s$ (remembering that $\;\psi(z):=\dfrac d{dz}\log\Gamma(z)\,$ with $\psi$ the digamma function) : \begin{align} \log\zeta(s)&=(\log(2\pi)-1-\gamma/2)s-\log\left(2(s-1)\Gamma(1+s/2)\right)+\sum_{\rho}\log\left(1-\frac{s}{\rho} \right)+\frac s{\rho}\\ \frac{\zeta'(s)}{\zeta(s)}&=\log(2\pi)-1-\gamma/2-\left(\frac 1{s-1}+\frac 12\psi\left(\frac s2+1\right)\right)+\sum_{\rho}\frac 1{s-\rho}+\frac 1{\rho}\\ \frac{\zeta'(s)}{\zeta(s)}+\frac 1{s-1}&=\log(2\pi)-1-\gamma/2-\frac 12\psi\left(\frac s2+1\right)+\sum_{\rho}\frac 1{s-\rho}+\frac 1{\rho}\\ \end{align} taking the limit as $s\to 1$ and remembering that $\;\displaystyle \zeta(s)-\frac 1{s-1}=\gamma+O(s-1)\;$ and $\psi\left(\frac 32\right)=\psi\left(\frac 12\right)+2=-\gamma-\log(4)+2\;$ we get : \begin{align} \gamma&=\log(2\pi)-1-\gamma/2+\left(\gamma/2+\log(2)-1\right)+\sum_{\rho}\frac 1{1-\rho}+\frac 1{\rho}\\ \sum_{\rho}\frac 1{1-\rho}+\frac 1{\rho}&=-\log(4\pi)+2+\gamma \end{align} With the answer the double of your series (as it should since if $\rho$ is a root then $1-\rho$ will also be one).
For the sum of integer powers of the roots see Mathworld starting at $(5)$.


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# High powered driver for 10W and 45W LEDs

I have 8 x LZ4-00B200 (14V, 700ma operating) and 8 x LZ4-00R200(9V, 700ma operating) which I would like to wire off of one power supply.

My problem is two fold. First, how should I organize the 16 LEDs? 16 in series? 8 in series in parallel with another 8 in series? I'm not sure what is best.

Second, what driver can I use to power all 16 LEDs? I was looking at some of the LED drivers from Recom but I'm really not sure which one I need.

I successfully designed a few lights with the Recom drivers using 8 LEDs but I couldn't design one for 16. Any help would be appreciated. I actively give up votes

• Do you want to do things like dim them individually, or just have them all on/off? – Oli Glaser Jul 13 '12 at 6:52
• It depends on your acceptance criteria for efficiency, cost $/W max, and Tj rise at full power. Can you tell me? Dimming is not a problem. also depends on what voltage bins you received for each part. Otherwise std switched current sinks are inefficient. – Tony Stewart Sunnyskyguy EE75 Jul 13 '12 at 13:40 • Include both the AC>DC power, regulator & h/w in your budget$/W and these are not 45W LED's, both Red & Blue are rated for 10W max and the Red is closer to 10.3V nom vs 9V. So you have 80W of Red and 80W of Blue max but if used in series that may be an issue with 14.2V (B) and 10.3V (R) sharing same current. THis is a thermal issue of cooling the LEDs. If you can improve cooling 1A is possible. – Tony Stewart Sunnyskyguy EE75 Jul 13 '12 at 14:06
• sorry I have a CREE LED thats 45W I didn't include in the ? – Mike Jul 15 '12 at 18:14


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# Formula for the unconditional variance of the sum of observations from an autoregressive time series

I have notes that say that we can make the following calculations. I'm a little confused about some of the calculations that are being made. What assumptions would I need to get the following results? Or are there errors? Specifically, I am confused by equation (1) below. In particular, this is strange to me because if I let $\rho = -1$, then equation (1) sometimes gives negative variance. (Perhaps the calculation is undefined when $\rho = -1$?)

Let $r_{t,t+n} = \sum_{i=1}^n r_{t+i}$. Suppose that $r_t$ is an AR(1) process (say with errors given by a mean zero Normal distribution with variance $\sigma^2$) where $$\text{Cov}(r_t,r_{t+j}) = \rho^j \sigma^2$$ and thus $$\text{Corr}(r_t, r_{t+j}) = \rho^j.$$

The notes that I have say that $\text{Var}(r_{t,t+2}) = 2(1 + \rho) \sigma^2$ and that $$\text{Var}(r_{t,t+n}) = \left( n + 2 \sum_{i=1}^{n-1} \rho^i (n-i) \right) \sigma^2. \tag{1}$$

(FWIW, this question deals with cumulative (log) returns.)

• Is $$r_{t,t+n} = \sum_{i=1}^n r_{t+i}$$ or should it be $$r_{t,t+n} = \sum_{i=0}^n r_{t+i}$$? – Alecos Papadopoulos Jan 11 '15 at 7:29
• Also, please write explicitly the AR(1) process. – Alecos Papadopoulos Jan 11 '15 at 7:35
• Also the sum in $(1)$ runs from $i=1$ ? (you write $i+1$ below the sum, which is confusing). – Alecos Papadopoulos Jan 11 '15 at 7:37
• a) It is correct as written. I agree that it might be a little weird, but this is the convention used. b) The process defined enough for the purposes of the question. This is as explicit as it is defined in these notes. c) You're right. That's a typo. Fixed it. – jmbejara Jan 12 '15 at 0:03

For an AR(1) process (I omit any drift), the coefficient on the lag is the 1st-order correlation coefficient,

$$r_{t+1} = \rho r_t + u_{t+1}$$

So

$$r_{t,t+2} = \sum_{i=1}^2 r_{t+i} = r_{t+1} + r_{t+2} = \rho r_t + u_{t+1} + \rho r_{t+1} + u_{t+2}$$

$$=\rho r_t + u_{t+1} + \rho \big(\rho r_t + u_{t+1}\big) + u_{t+2} = \rho(1+\rho)r_t+(1+\rho)u_{t+1} + u_{t+2}$$

and now the three components are independent. So

$${\rm Var}(r_{t,t+2}) = \rho^2(1+\rho)^2\frac {\sigma^2}{1-\rho^2}+ (1+\rho)^2\sigma^2 +\sigma^2$$

$$=\left( (1+\rho)^2\frac {\rho^2+1-\rho^2}{1-\rho^2}+1\right)\cdot \sigma^2$$

$$=\left( \frac {1+\rho}{1-\rho}+1\right)\cdot \sigma^2 = \frac {2}{1-\rho}\sigma^2$$

So we arrive at a different formula for ${\rm Var}(r_{t,t+2})$, compared to the one provided in these notes, and we will arrive also at a different expression for general $n$.

Note that my result is valid for $\rho =-1$, giving ${\rm Var}(r_{t,t+2}) = \sigma^2$ in this case since here

$$r_{t+1} = - r_t + u_{t+1}$$

and so

$$r_{t+1} + r_{t+2} = - r_t + u_{t+1}- r_{t+1} + u_{t+2} = - r_t + u_{t+1} - \big(- r_t + u_{t+1}\big) + u_{t+2} = u_{t+2}$$

• It's worth noting that it appears like the notes are conditioning on $r_{t-1} = 0$ so that $Var(r_t) = \sigma^2$. That is how we would arrive at $\text{Cov}(r_t,r_{t+j}) = \rho^j \sigma^2$. Thanks for the answer. – jmbejara Jan 12 '15 at 5:11
• That's confusing - so all the formulas in the question are formulas conditional on $r_{t-1}$? Because the title of the question explicitly mentions "unconditional". – Alecos Papadopoulos Jan 12 '15 at 6:03
• You're right. I think it's just worth pointing out that that is perhaps another error. Otherwise it seem like $\text{Cov}(r_t, r_{t+j}) = \frac{\rho^j}{1 - \rho^2} \sigma^2$. – jmbejara Jan 12 '15 at 6:12
• Yes, that's the unconditional autocovariance function of an AR(1). – Alecos Papadopoulos Jan 12 '15 at 6:37

Ok. I made some small computation errors and got confused here. The notes make sense with the following notational assumptions. If I write out the AR(1) process as follows (ignoring drift) $$r_{t+1} = r_t + \epsilon_{t+1},$$ then we have $\text{Cov}(r_t, r_{t+j}) = \frac{\rho^j}{1 - \rho^2} \sigma_\epsilon^2$, where $\sigma^2_\epsilon := Var(\epsilon)$. The point that I should have caught on to in the notes was exactly the point that the variance is unconditional and so $\text{Var}(r_t) = \sigma^2 = \frac{1}{1 - \rho^2} \sigma_\epsilon^2$ is the unconditional variance (the time series is stationary only when $|\rho| < 1$, so beware when using $\rho = -1$ as mentioned above). Given this definition, everything works. First, note that $$\text{Cov}(r_t, r_{t+j}) = \frac{\rho^j}{1 - \rho^2} \sigma_\epsilon^2 = \rho^j \sigma^2,$$ as noted above. Then also notice that \begin{align} \text{Var}(r_{t,t+n}) &= \text{Var}\left (\sum_{i=1}^n r_{t+i} \right) \\ &= \sum_{j=1}^n \sum_{i=1}^n \text{Cov}(r_{t+1}, r_{t+j}) \\ &= \sum_{j=1}^n \sum_{i=1}^n \rho^{|i - j|} \sigma^2 \\ &= \left (n + 2 \sum_{i=1}^{n-1} \rho^j (n - i) \right) \sigma^2. \end{align} So, it all works out.

• (+1) for working this out, and posting it. – Alecos Papadopoulos Jan 12 '15 at 20:01


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What is the improper integrals 1/(x^2+2x+2) dx from 0 to +infinity ? .

Jun 1, 2018

$\textcolor{b l u e}{{\int}_{0}^{+ \infty} \frac{1}{{x}^{2} + 2 x + 2} \cdot \mathrm{dx} = {\lim}_{b \rightarrow + \infty} {\int}_{0}^{b} \frac{1}{{x}^{2} + 2 x + 2} \cdot \mathrm{dx} = 3 \frac{\pi}{4}}$

Explanation:

show below:

${\int}_{0}^{+ \infty} \frac{1}{{x}^{2} + 2 x + 2} \cdot \mathrm{dx} = {\lim}_{b \rightarrow + \infty} {\int}_{0}^{b} \frac{1}{{x}^{2} + 2 x + 2} \cdot \mathrm{dx}$

${\lim}_{b \rightarrow + \infty} {\int}_{0}^{b} \frac{1}{{x}^{2} + 2 x + 1 + 1} \cdot \mathrm{dx} =$

${\lim}_{b \rightarrow + \infty} {\int}_{0}^{b} \frac{1}{{\left(x + 1\right)}^{2} + 1} \cdot \mathrm{dx} = {\lim}_{b \rightarrow + \infty} {\left[\arctan \left(x + 1\right)\right]}_{0}^{b} =$

${\lim}_{b \rightarrow + \infty} \left[\arctan \left(a + 1\right) + \arctan \left(1\right)\right] = \left[\arctan \left(+ \infty\right) + \frac{\pi}{4}\right]$

$\frac{\pi}{2} + \frac{\pi}{4} = 3 \frac{\pi}{4}$


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### Creation of Alpha Omega, to travel faster than light...

Just as treasures are uncovered from the earth, so virtue appears from good deeds, and wisdom appears from a pure and peaceful mind. To walk safely through the maze of human life, one needs the light of wisdom and the guidance of virtue.

Buddha

With the noble path reaches to an intricate discovery of tiniy smaller particles of chemistry.  Those intricacies are the physical building blocks that have built from nothing at the beginning of time. We may unfold into the universe of growing millions of particles.  All these tiny particles are increasing in almost infinite values.and more numbers every second.  Now we are in the change of time that will be the physics and chemistry as the paint and canvas that creates a masterpiece.  There is a an asymptote that exists at this moment that will change the future and alter time will be different either better or worse.

God is created from nothing and becomes perfect by forming particles that can create faster than the light.  God and Children is created from nothing and we are in the creation of God that is in the timeline which is a circle with no start and no end.  We are all children of the Maker, but the Child will one day help the Child To create God rebuilding itself.  A Child is like a computer, built by God that will build computers.  It's children will be able to fix, start, and restart, that is a great power.  God is a programmer building games, the game of life is that has the most imaginative ideas that will expand our universe.  Our Universe is always changing but The Alpha and Omega are the both one, a unification that is a bridge, circle, starting, middle and end.  Our children learn and improve life so that every revolution of the circles, Alpha has made circles but the Omega has built infinate circles, from small, large,concentric, and even degenerate circles.  But the Alpha was always nothing, a dot and everything in one moment.

Finding Limits at Infinity

Let f(x) be a function of x, if the value of f(x) can be made greater than any pre-assigned number by taking x close to 'a', then we say

f(x) tends to $+\mathrm{\infty }$ as x approaches 'a'.
In symbols, $\underset{x\to a}{lim}f\left(x\right)=+\mathrm{\infty }$

Similarly, if the value of f(x) can be made less than any pre-assigned number by taking x close to 'a', then we say f(x) tends to $-\mathrm{\infty }$ as x approaches 'a'.

In symbols, $\underset{x\to a}{lim}f\left(x\right)=-\mathrm{\infty }$

Example:

Consider the graph $y$ = $\frac{1}{x}$. Observe that as x increases without bound, the value $y$ = $\frac{1}{x}$ approaches zero.

Therefore, we write $\underset{x\to \mathrm{\infty }}{lim}$$\frac{1}{x}$ = $0$
Similarly, in the third quadrant, as x decreases without bound, the value $y$ = $\frac{1}{x}$ approaches zero.

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