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The following image is from robert collins' ppt from http://www.cse.psu.edu/~rcollins/CSE486/

How to calculate sigma ?

It shows [1 4 6 4 1] is corresponding to sigma=1.

I learned that usually 3*sigma is best to represent. So kernel size is 7*7 is suited. But its size is 5*5.

I don't know why [1 4 6 4 1] is matched to sigma=1.

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The well-known twentieth-century American columnist Dorothy Parker once reviewed a book saying

"This is not a book to be put down lightly; it should be thrown with great force."

Be that as it may, Professor Collins is suggesting that a bar plot of the binomial coefficients on the $n$-th row of Pacal's triangle, after suitable normalization, will look from far away like a Gaussian distribution with mean $n/2$ and standard deviation $\sqrt{n}/2$. He then further obfuscates this by using the fact that for even values of $n$, we can write the standard deviation as $\sqrt{k/2}$ where $k = n/2$ while for odd $n$, it is simply $\sqrt{n}/2$, inviting the reader to arrive at the mistaken conclusion that there is a fundamental difference between the odd and even rows.

To get a better understanding of where the $n/2$ and $\sqrt{n}/2$ comes from, look for reading material on the DeMoivre-LaPlace approximation which is a special case of the Central Limit Theorem.

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  • $\begingroup$ Thank you very much. I'm a novice and your advice is impressing for me. I am somewhat confused but I'll study! $\endgroup$ – jakeoung Jul 21 '13 at 15:15
  • $\begingroup$ $\sigma^2$ =(approximate) npq, where p=1/2 and q=1/2. So $\sigma = \sqrt{npq} = \sqrt n / 2$ Am I right? $\endgroup$ – jakeoung Jul 21 '13 at 23:45
  • $\begingroup$ $\sigma^2$ equals $npq$, not $\sigma^2 \approx npq$, but yes, that is why we get $\sqrt{n}/2$. $\endgroup$ – Dilip Sarwate Jul 22 '13 at 1:56
  • $\begingroup$ Thank you. There are still many things I don't understand. It is hard for me. $\endgroup$ – jakeoung Jul 22 '13 at 15:51

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