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Suppose I sample $n$ (some number) numbers from a Normal distribution with zero mean and standard deviation $\sigma$. That is, from $\mathcal{N}(0,\sigma)$. We will call this list of numbers $L$. We will take $L(i)$ to mean the $i$th element of $L$.

I now want to create a function using these numbers. It will be a very simple function. It will be piecewise-constant, with each constant segment having the same length and the constant value corresponding to the values in $L$. A picture of this function for $n=30$ is given underneath,

enter image description here

If I now sampled this function finely (say 10 samples per constant segment), and I made a histogram of the sampled values, I believe that the histogram will reproduce a $\mathcal{N}(0,\sigma)$ distribution.

Now, we consider another function, where the elements of $L$ are instead the vertices of a piecewise-linear function. For $n = 30$, this function looks like enter image description here

That is, there is a linear interpolation between each of the vertices, which are the elements of $L$. If now now sampled this function finely, and I make a histogram of the sampled values, would the histogram also be like $\mathcal{N}(0,\sigma)$? Would it be for large $n$?

K L
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  • Of course it will not be .. you have a discrete distribution now. – Ahmad Bazzi Sep 07 '18 at 18:50
  • Sure, the histograms will of course be discrete. However, once normalized accordingly, the question I wanted to ask is if those discrete histograms would have the same behavior (look like) as $\mathcal{N}(0,\sigma)$. – K L Sep 07 '18 at 18:56
  • The distribution will be a function of sampling interval size (heights of the samples in fig. 1) and number of discrete samples. – Ahmad Bazzi Sep 07 '18 at 19:01
  • As long as you sample finely enough, would it be a function of sampling interval size? For instance, in the limit of infinitely fine sampling, the creation of the histogram would simply be to take either the first or second figure, and put it on its side, and let gravity drop those values down. Asked in this way, the question is, would these gravity-created histograms look like $\mathcal{N}(0,\sigma)$? – K L Sep 07 '18 at 19:06
  • In the limit, you'd have a normal of course. – Ahmad Bazzi Sep 07 '18 at 19:07

1 Answers1

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Let's say we have $$L \sim \mathcal{N}(0,\sigma^2) $$And you have another random discrete variable $S$, which consists of discretization levels. One scheme is as such (PS: I'm just trying to mimic your figure because you did not provide any extra information on the sampling scheme) \begin{align} f(L = l) &= 0 \qquad \text{ if } \vert l \vert < \frac{W}{2} \\ f(L = l) &= A \qquad \text{ if } (1-\frac{1}{2})W< l < (1+\frac{1}{2})W \\ f(L = l) &= -A \qquad \text{ if } - (1+\frac{1}{2})W< l < - (1-\frac{1}{2})W \\ f(L = l) &= 2A \qquad \text{ if } (2-\frac{1}{2})W< l < (2+\frac{1}{2})W \\ f(L = l) &= -2A \qquad \text{ if } - (2+\frac{1}{2})W< l < - (2-\frac{1}{2})W \end{align} So in general the sampling scheme is \begin{equation} f(L = l) = kA, \qquad \text{ if } (k-\frac{1}{2})W< l < (k+\frac{1}{2})W \end{equation} and \begin{equation} f(L = l) = -kA, \qquad \text{ if } -(k+\frac{1}{2})W< l < -(k-\frac{1}{2})W \end{equation} Then your distribution is just \begin{equation} Pr(S = kA) = Pr((k-\frac{1}{2})W< l < (k+\frac{1}{2})W ) \end{equation} which by the definition of the Normal distribution is \begin{equation} Pr(S = kA) = \int\limits_{(k-\frac{1}{2})W}^{(k+\frac{1}{2})W } \Phi_{\sigma}(x) \ dx \qquad k = 0 \ldots (K-1) \end{equation} Similary which by the definition of the Normal distribution is \begin{equation} Pr(S = -kA) = -\int\limits_{-(k-\frac{1}{2})W}^{-(k+\frac{1}{2})W } \Phi_{\sigma}(x) \ dx \qquad k = -(K-1) \ldots -1 \end{equation} where $\Phi_{\sigma}(x)$ is the PDF of a zero mean normal distribution with standard deviation $\sigma$ and $KA$ is the largest sampling level.

NOTE: There is one exception on this sampling distribution, which is the largest sampling level: $$f(L=l) = KA \qquad \text{ if } (K-\frac{1}{2})W< l$$then \begin{equation} Pr(S = KA) = \int\limits_{(K-\frac{1}{2})W}^{+\infty} \Phi_{\sigma}(x) \ dx \end{equation} and \begin{equation} Pr(S = -KA) = \int\limits_{-\infty}^{-(K+\frac{1}{2})W} \Phi_{\sigma}(x) \ dx \end{equation}

Ahmad Bazzi
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