- #1

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- 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

$$\displaystyle \sum _{k=-\infty }^{\infty }1/2\,{\frac { \left( k-\mu \right) ^{2} \sqrt{2} }{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( k-\mu \right) ^{2}}{{\sigma}^{2}}}}}}={\sigma}^{2}$$

$$\displaystyle P(K = k) = \frac{1}{2}~\frac{\sqrt{2}~e^{-\frac{1}{2}~\frac{\left(k -\mu \right)^{2}}{\sigma ^{2}}}}{\sigma ~\sqrt{\pi }}$$

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

$$\displaystyle \sum _{k=-\infty }^{\infty }1/2\,{\frac { \left( k-\mu \right) ^{2} \sqrt{2} }{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( k-\mu \right) ^{2}}{{\sigma}^{2}}}}}}={\sigma}^{2}$$