To gain deeper insight to the Poisson and exponential random variables, I found that I could derive the random variables as follows:

I consider an experiment which consists of a continuum of trials on an interval $[0,t)$. The result of the experiment takes the form of an ordered $n$-tuple $\forall n \in \mathbb{N}$ containing distinct points on the interval. Every outcome is equally likely and I measure the size of the set containing tuples of $n$ different points by $I_n$ as:

$$ I_n = \int_0^{t} \int_0^{x_{n}} \int_0^{x_{n-1}} \cdots \int_0^{ x_2 } dx_1 dx_{2} dx_{3} \dots dx_{n-1} dx_{n} = \frac{ t^n } { n! }$$

It follows that on some interval, $[0,t)$, the probability that the experiment results in an $k$-tuple, $k \in \mathbb{N}$ is

$$P(X(t) = k) = \frac{I_k}{\sum_{n = 0}^{\infty} I_n} = \frac{e^{-t} t^k}{k!}$$

And for $k = 0$, we have $P(X(t) = 0) = e^{-t}$.


I was wondering if some similar intuition can applied to derive the Gaussian:

$$\frac{1}{\sqrt{2 \pi \sigma^2}} \exp \big(-\frac{(x-\mu)^2}{2\sigma^2} \big) \ \text{ or the standard normal, }\ \frac{e^{-x^2}}{\sqrt{\pi}}$$

I think that such an intuition might be obtained by gaining more insight into each term in the expansion of $\text{erf}(x)$ as is done for Poisson: $$ \begin{align} \text{erf}(x) &= \frac{1}{\sqrt \pi } \int_{-x}^{x} e^{-t^2} dt\\ &= \frac{2}{\sqrt \pi } \big( \sum_{n=0}^{\infty} \frac{ x^{2n+1} }{ n! (2n+1) } \big)^{-1} \end{align} $$

Any ideas aside from dismissal of the question are much appreciated!

  • $\begingroup$ I don't think there is much intuition beyond the intuition underlying the proof of the CLT. However, you can get a lot of nice visualization by explicitly computing the PDF of the sum of $n$ iid uniform $(-\sqrt{3},\sqrt{3})$ random variables (which is a piecewise polynomial that starts looking more and more like a standard Gaussian even for $n$ as small as $4$). $\endgroup$
    – Ian
    May 4, 2018 at 18:18
  • $\begingroup$ yes, central limit theorem is extremely interesting: en.wikipedia.org/wiki/Central_limit_theorem . Worth reading! Every distribution when iterated tends to the Gaussian in the end... $\endgroup$
    – anonymous
    Jun 18, 2018 at 4:04
  • $\begingroup$ You may find this video interesting. $\endgroup$
    – B. Mehta
    Aug 13, 2018 at 0:18

1 Answer 1


Take an experiment that has two outcomes, success (S) and failure (F), of probability $p$ and $q=1-p$ respectively. The probability of S after one trial is $p$. The probability of two S after two trials is $p^2$, for one S and one F is $2pq$ and for two F is $q^2$. In general, for $N$ trials, the probability of having k S is given by the Binomial distribution $P_S(k,N)= \frac{N!}{k!(N-k)!}p^k q^{N-k}$. These are just the coefficients in front of the terms in $(p+q)^N$ after multiplying them out

What happens if we take the limit of large $N$? The Binomial coefficients at large $N$ approximate a Gaussian, which can be seen visually by looking at a low row Pascal's Triangle. We can derive it from the above formula. To make this simple let's work with the symmetric case $p=q=1/2$, so that we have $$ P_S(k,N)= \frac{N!}{k!(N-k)!}\frac{1}{2^N} $$ An easy way to get the desired result is to exchange the index $k$ for one that starts from the center of the triangle where values are largest $ x\in (-N/2,N/2),\; k=x+N/2$. Then we swap this in and apply Stirling's approximation $n!=\sqrt{2\pi}n^ne^{-n}$:

$$ P_S(x,N)= \frac{N!}{(N/2+x)!(N/2-x)!}\frac{1}{2^N} \approx \frac{2}{\sqrt{2\pi N}}\frac{1}{(1-\frac{4 x^2}{N^2})^{(N+1)/2}}\left(\frac{1-\frac{2x}{N}}{1+\frac{2x}{N}}\right)^x $$

Finally exponentiation and taking the log we get $$ P_S(x,N) \approx \frac{2}{\sqrt{2\pi N }} e^{-\frac{1}{2}(N+1)\log(1-\frac{4 x^2}{N^2}) +x(\log(1-\frac{2x}{N})-\log(1+\frac{2x}{N})) } $$

Using the expansion $\log(1+x)=x-x^2/2+ \cdots$, keep everything to order $1/N$ $$ P_S(x,N) \approx \frac{2}{\sqrt{2\pi N}} e^{-\frac{2 x^2}{N} } $$

Now you can feel free to tack a on $dx$ and use a transformation of variables to scale $x \rightarrow \sqrt N x/2$ -- then $x$ is an "implicit" variable.

$$ P_S(x,N)dx \approx \frac{1}{\sqrt{2\pi}} e^{-\frac{ x^2}{2}} dx $$

For general $p,q$, the derivation is similar, see for example http://scipp.ucsc.edu/~haber/ph116C/NormalApprox.pdf

This isn't exactly an infinite expansion like in your example, but there's a similar vein of thought in the conclusion that $N$ choose $k$ limits to a Gaussian type shape for large $N$.

  • $\begingroup$ thanks for the answer. I'm looking for a derivation in which is enlightening to the terms of infinite series expansion of the exponential, or the cdf of the normal. $\endgroup$
    – jaslibra
    Aug 16, 2018 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.