# Binomial Approximation of Gaussian Distribution

It is said that we may use the binomial coefficients ( a layer from Pascal's triangle) to approximate the 1-D Gaussian kernel with certain $\sigma$, where $\frac{n}{4} = \sigma$ and $n$ is the index of the layer.

This works really nice when we want to generate a gaussian quickly. But is there a way to prove this? Or just coincidence?

And how do we decide the variance of certain layer of Pascal's triangle? Say the third layer is [1 3 3 1] (consider [1 1] as the first layer), and it is used to approximate gaussian kernel of $\sigma = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$. But how do we prove this?

• maybe should ask the math SE this question. we be just dumb electrical engineers. we don't know which end of the integral symbol is up. – robert bristow-johnson Mar 6 '18 at 8:56
• I think this is the reverse of the De Moivre-Laplace Theorem . – Shukant Pal Feb 8 '20 at 15:10

Let $$X \sim B(n, p)$$. At the third layer $$\begin{bmatrix}1 & 3 & 3 & 1\end{bmatrix}$$ of the Pascal's triangle, we have $$X \sim B(3, \frac{1}{2})$$. Which means $$P(X = k) = {3 \choose k} (\frac{1}{2})^k (\frac{1}{2})^{3-k} = \frac{{3 \choose k}}{2^3}$$. The variance is given by $$\sigma^2 = n p q = 3 \cdot \frac{1}{2} \frac{1}{2} = \frac{3}{4}$$ and the std by $$\sigma = \sqrt{\frac{3}{4}}$$. You can derive the variance by looking at $$n$$ independent Bernoulli trials.

In general, a 1d binomial filter/kernel of size $$n$$ with $$p = \frac{1}{2}$$ is given by $$\begin{bmatrix}P(X = 0) & P(X = 1) & \cdots & P(X = n-1)\end{bmatrix}$$. As $$n \to \infty$$, this kernel will approximate the normal distribution with mean $$\frac{n}{2}$$ and variance $$\sigma^2 = \frac{n}{4}$$. If you want a different variance/std, you should change $$p \neq 0.5$$.

As somebody already wrote in the comments, you should look at the De Moivre-Laplace theorem to prove this. However, this is only a special case of the more general central limit theorem. By showing that the binomial distribution is asymptotically equivalent to the normal distribution as $$n\to\infty$$, I would argue that we have shown already both directions.

Hi: The CLT for the sum says that

$$\sum_{i-1}^{n} X_{i} \rightarrow Norm(n \mu, n \sigma^2)$$ where $$\mu$$ is the mean of $$X_{i}$$ and $$\sigma^2$$ is the variance of $$X_{i}$$.

And, when $$X_{i}$$ is binomial, it can be shown that, $$\mu = p$$ and $$\sigma^2 = p(1-p)$$.

So, you put those into the CLT for the sum and that's why.