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?

  • $\begingroup$ 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. $\endgroup$ – robert bristow-johnson Mar 6 '18 at 8:56
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    $\begingroup$ I think this is the reverse of the De Moivre-Laplace Theorem . $\endgroup$ – Shukant Pal Feb 8 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.

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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.

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