Suppose we have an unending source of random numbers from the interval [0,1] coming from the uniform distribution. Think of each number as the result of an experiment.


Fix a value $s$ (say, $s=100$).


We are going to create a random function $f\_s(t)$ for $t \in \mathbb{N}$ from this sequence of random numbers.


We do the following. We take the average of $t$ of the random numbers. This should be approximately $0.5$, since the numbers are uniformly distributed. Next, we do the same thing $s$ times, collecting the $s$ averages. We end up with $s$ numbers, all of which are fairly close to $0.5$.


Now, one would expect that the averages would all be closer to $0.5$ the higher the value of $t$. Thus we set $f\_s(t)$ to be the standard deviation of these $s$ values.


Next we let $s$ go to infinity, or at least, be reasonably large, to get the function $f(t)$. We can use the following MATLAB code to graph the function:


function w=randstd(trials,s)


for n=[1:trials]
 for r=[1:s]
 z(r,1)=mean(rand(n,1));
 end
 w(n,1)=std(z);
end
plot(w)


The result appears to be a fairly smooth curve, even as I try many times. My questions are:


1) What is the curve? Is it polynomial, logarithmic, or exponential?
2) Is there a mathematical explanation for this? Should we expect the curve to look like this for general random sets?
3) If not, is the curve I've graphed simply a result of the nature of whatever pseudorandom number generator MATLAB is using? If so, is the curve I've graphed at least the result of some mathematically quantifiable property of that pseudorandom number generator?


I understand there might not be a rigorous mathematical answer to this question, but I'm asking because there might be.



Standard Deviation of Averages of Random Numbers
Let $Y$ the be mean of $t$ samples of the distribution. $Y$ has mean $1/2$ and variance $1/(12 t)$. Then $f\_s(t)$ is the sample standard deviation of $s$ draws from $Y$. As the sample standard deviation is a consistent estimator, $f\_s(t)$ converges in probability to $\sqrt{1/(12 t)}$.


