In every probability book I've looked at thus far, they talk about how the sum of $n$ independent and equally distributed random variables tends towards the normal distribution as $n$ tends towards infinity.

This is the Central Limit Theorem.

In the top answer to the following question...

Why do bell curves appear everywhere?

...the concept of convolutions of random variables is mentioned. Pretty much, the PMF of the sum of two independent random variables is the convolution of their individual PMFs.

I understand this, and its really nicely demonstrated in the answer to the following question:

density of sum of two uniform random variables $[0,1]$

Let's say we were to convolve the PMF of the sum of the two uniform random variables with another uniform random variable, and then with another, and then with another....

The Central Limit Theorem tells us we would arrive at the normal distribution.

Can someone please show this to me, and talk me through it?

I'd really like to understand why the convolution of the probability density functions of $n$ uniformly distributed independent random variables tends towards the normal as $n\rightarrow\infty$.


  • 1
    $\begingroup$ You are talking about rigor but are omitting crucial independence assumptions in many of your statements. The distribution of the sum of two random variables is the convolution of the distributions when they are independent, not in general. $\endgroup$ – Kavi Rama Murthy Nov 1 '19 at 0:11
  • $\begingroup$ Fixing it now @KaboMurphy thanks! $\endgroup$ – joshuaronis Nov 1 '19 at 0:15
  • $\begingroup$ The convolution of $n$ uniform $(0,1)$ distributions on $(0,1)$ does not tend to normal. You have to suitably normalize it to get normal distribution(as done in CLT). $\endgroup$ – Kavi Rama Murthy Nov 1 '19 at 5:38
  • $\begingroup$ @KaboMurphy could you pherhaps show me in an answer? Thanks! $\endgroup$ – joshuaronis Nov 1 '19 at 5:58

If $(X_i)$ is any i.i.d sequence with finite variance and $S_n=X_1+X_2+...+X_n$ then we can write the distribution of $S_n$ in terms of the distribution $\mu $ of $X_1$ s follows: define the convolution of two probability measures $\lambda $ and $\nu$ by $(\lambda * \nu) (E)=\int \lambda(E-x)d\nu(x)$. Now let $\mu^{0}=\delta_0$, the degenerate measure at $0$, $\mu^{1}=\mu$ and $\mu^{n+1}=\mu^{n}*\mu$ for $n \geq 1$. Then $\mu^{n}$ is the distribution of $S_n$. The distribution of $S_n-nm$ is $\delta_{-nm}*\mu^{n}$ (where $m$ is the mean of $X_1$). If $\sigma^{2}$ is the variance of $X_1$ then we can write the distribution of $\frac {S_n-nm} {\sqrt n \sigma}$ as the measure $\nu_n$ defined by $\nu_n (E)= (\delta_{-nm}*\mu^{n})(\sqrt n \sigma E)$. CLT says that this converges $N(0,1)$.

| cite | improve this answer | |

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.