- #1

Ad VanderVen

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- TL;DR Summary
- How can the result of an integral of a normal distribution be the same as the result of a sum?

I have the following function for the normal distribution:

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

How can the following integrals be equal to their sums?

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

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

and

$$\displaystyle \int_{-\infty }^{\infty }\!1/2\,{\frac { \left( x-\mu \right) ^{2} \sqrt{2}

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

}{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}}$$

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

How can the following integrals be equal to their sums?

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

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

and

$$\displaystyle \int_{-\infty }^{\infty }\!1/2\,{\frac { \left( x-\mu \right) ^{2} \sqrt{2}

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

}{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}}$$

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