Discrete type normal distribution

In summary: Because the result of a Fourier transform with a real argument will always be complex, and you can't generally even compare complex numbers unless they are equal.I just wanted to point out that the fact that the probabilities sum to 1 is a consequence of the fact that the Fourier transform of a gaussian is another gaussian, plus parseval's theorem.This probably works for the other parts too, adding polynomial multipliers just yields polynomial multipliers on the other end, e.g.https://www.wolframalpha.com/input?i=fourier+transform+x^2e^(-x^2)So dear Office_Schredder,So what could be your final conclusion?In summary, the given formula describes a
  • #1
169
13
TL;DR Summary
How to prove that a certain discrete type normal distribution has as expectation ##\mu## and variance ##\sigma^2##.
The following is given:
$$\displaystyle P(K = k) = \frac{1}{2}~\frac{\sqrt{2}~e^{-\frac{1}{2}~\frac{\left(k -\mu \right)^{2}}{\sigma ^{2}}}}{\sigma ~\sqrt{\pi }}$$
How can you prove that the following equalities are correct?
$$\displaystyle \sum _{k=-\infty }^{\infty }1/2\,{\frac { \sqrt{2}}{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( k-\mu \right) ^{2}}{{\sigma}^{2}}}}}}=1,$$
$$\displaystyle \sum _{k=-\infty }^{\infty }1/2\,{\frac {k \sqrt{2}}{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( k-\mu \right) ^{2}}{{\sigma}^{2}}}}}}=\mu,$$
and
$$\displaystyle \sum _{k=-\infty }^{\infty }1/2\,{\frac { \left( k-\mu \right) ^{2} \sqrt{2} }{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( k-\mu \right) ^{2}}{{\sigma}^{2}}}}}}={\sigma}^{2}$$
 
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  • #2
Are you assuming that the formula you are given is a probability distribution with that mean and expectation, or are you trying to prove that it works?
 
  • #3
Office_Shredder said:
Are you assuming that the formula you are given is a probability distribution with that mean and expectation, or are you trying to prove that it works?
I assume that the formula I have given describes a discrete probability distribution with expectation ##\mu## and standard deviation ##\sigma## and my question is whether that assumption is correct.

Reference: https://www.physicsforums.com/threads/discrete-type-normal-distribution.1046309/
 
  • #5
Office_Shredder said:
I don't think this actually works?
For example when ##\mu=0## and ##\sigma=1##, unless I typo'd the probabilities only add up to about 0.57

https://www.wolframalpha.com/input?i=sum_{k=-infty}^{infty}+sqrt(2)/(2pi)+e^(-(k^2)/2)
Sorry, but with ##\mu = 0## and ##\sigma = 1## I obtain with Maple:
$$\displaystyle \sum _{x=-1000}^{1000}1/2\,{\frac { \sqrt{2}{{\rm e}^{-1/2\,{x}^{2}}}}{ \sqrt{\pi }}}=1,$$
$$\displaystyle \sum _{x=-1000}^{1000}1/2\,{\frac {x \sqrt{2}{{\rm e}^{-1/2\,{x}^{2}}}}{ \sqrt{\pi }}}=0$$
and
$$\displaystyle \sum _{x=-1000}^{1000}1/2\,{\frac {{x}^{2} \sqrt{2}{{\rm e}^{-1/2\,{x}^{2}}}}{ \sqrt{\pi }}}=0.99999$$
 
  • #6
Sorry, I missed a square root around the pi in my attempt.

Fascinating. I'll think about it.Edit: the probabilities sum to 1 I think is a consequence of the fact that the Fourier transform of a gaussian is another gaussian, plus parseval's theorem.

This probably works for the other parts too, adding polynomial multipliers just yields polynomial multipliers on the other end, e.g.

https://www.wolframalpha.com/input?i=fourier+transform+x^2e^(-x^2)
 
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  • #7
Office_Shredder said:
Sorry, I missed a square root around the pi in my attempt.

Fascinating. I'll think about it.Edit: the probabilities sum to 1 I think is a consequence of the fact that the Fourier transform of a gaussian is another gaussian, plus parseval's theorem.

This probably works for the other parts too, adding polynomial multipliers just yields polynomial multipliers on the other end, e.g.

https://www.wolframalpha.com/input?i=fourier+transform+x^2e^(-x^2)
So dear Office_Schredder,

So what could be your final conclusion?
 
  • #8
Ad VanderVen said:
So dear Office_Schredder,

So what could be your final conclusion?
Have you learned about Fourier series?
 
  • #9
Ad VanderVen said:
Summary: How to prove that a certain discrete type normal distribution has as expectation ##\mu## and variance ##\sigma^2##.

The following is given:
$$\displaystyle P(K = k) = \frac{1}{2}~\frac{\sqrt{2}~e^{-\frac{1}{2}~\frac{\left(k -\mu \right)^{2}}{\sigma ^{2}}}}{\sigma ~\sqrt{\pi }}$$
How can you prove that the following equalities are correct?
$$\displaystyle \sum _{k=-\infty }^{\infty }1/2\,{\frac { \sqrt{2}}{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( k-\mu \right) ^{2}}{{\sigma}^{2}}}}}}=1,$$
$$\displaystyle \sum _{k=-\infty }^{\infty }1/2\,{\frac {k \sqrt{2}}{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( k-\mu \right) ^{2}}{{\sigma}^{2}}}}}}=\mu,$$
and
$$\displaystyle \sum _{k=-\infty }^{\infty }1/2\,{\frac { \left( k-\mu \right) ^{2} \sqrt{2} }{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( k-\mu \right) ^{2}}{{\sigma}^{2}}}}}}={\sigma}^{2}$$

Have I not already addressed these questions here:

https://www.physicsforums.com/threa...iscrete-random-variable.1013035/#post-6657061

using Poisson's summation formula? They are not exactly correct, but they are correct to a very good approximation.

EDIT: When I said "They are not exactly correct, but they are correct to a very good approximation." I was referring to your formulas, not my own. Sorry for the confusion.
 
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  • #10
Office_Shredder said:
Have you learned about Fourier series?
I heard about Fourier series. However, I know nothing about it.
 
  • #12
Ad VanderVen said:
I am not interested in appoximations. I simply want direct prove.

When I said "They are not exactly correct, but they are correct to a very good approximation." I was referring to your formula, not my own. Sorry for the confusion.

The expressions that I arrived at using Poisson's summation formula were exact! Using these expressions, we considered some examples and showed, in those cases, that the formula you think are equalities are not actually equalities! They were only approximately equal!

There are going to be a lot more cases where they are not actually equalities.
 
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  • #13
The following expressions (derived using Poisson's summation formula) are exact:

$$
\sum_{k=-\infty}^\infty 1/2 \dfrac{\sqrt{2}}{ \sigma\sqrt{\pi}}e^{-(k - \mu)^2/2 \sigma^2} = 1 + 2 \sum_{m=1}^\infty e^{ - m^2 \pi^2 2 \sigma^2} \cos 2 \pi m \mu .
$$

and

\begin{align*}
\sum_{k=-\infty}^\infty 1/2 \dfrac{k \sqrt{2}}{\sigma\sqrt{\pi}} e^{- (k - \mu)^2 / 2 \sigma^2} &= \mu +
2 \mu \sum_{m=1}^\infty e^{ - m^2 \pi^2 2 \sigma^2} \cos 2 \pi m \mu
\nonumber \\
& - 4 \sigma^2 \pi \sum_{m = 1}^\infty m e^{- m^2 \pi^2 2 \sigma^2} \sin 2 \pi m \mu .
\end{align*}

and

\begin{align*}
\sum_{k=-\infty}^\infty 1/2 \dfrac{(k - \mu)^2 \sqrt{2}}{\sigma \sqrt{\pi}} e^{- (k - \mu)^2 / 2 \sigma^2} &= \sigma^2 - \; 2 \sigma^2 \sum_{m = 1}^\infty (2 \pi m^2 - 1) e^{- m^2 \pi^2 2 \sigma^2} \cos 2 \pi m \mu
\end{align*}

Your formula are only correct if the sums on the right hand side happen to add up to zero. But that will only happen for some values of ##\mu## and ##\sigma## (sorry, I'm too preoccupied at the moment to look into that issue in any detail. Maybe others can help you do that.)

To see the proof of the above expressions see posts #15 and #16 of the other thread:

https://www.physicsforums.com/threa...iscrete-random-variable.1013035/#post-6657061
 
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