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- A discrete variable ##K## is distributed according to the formula used for the normal distribution. How can the sum for all values of ##k## still be equal to one?

De normal distribution has the following form:

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

and it's integral is equal to one:

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

However, I now take with ##k## as an integer:

$$\displaystyle P \left( K=k \right) \, = \,1/2\,{\frac { \sqrt{2}}{\tau \, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( k-\nu \right) ^{2}}{{\tau }^{2}}}}}}$$

For example, for ##\nu = 0## and ##\tau:=25.0## I get:

$$\displaystyle \sum _{k=-\infty }^{\infty } 0.01595769121\,{{\rm e}^{- 0.0008000000000\,{k}^{2}}}= 1.0$$

and one would expect a value lower than 1. How is that possible?

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

and it's integral is equal to one:

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

However, I now take with ##k## as an integer:

$$\displaystyle P \left( K=k \right) \, = \,1/2\,{\frac { \sqrt{2}}{\tau \, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( k-\nu \right) ^{2}}{{\tau }^{2}}}}}}$$

For example, for ##\nu = 0## and ##\tau:=25.0## I get:

$$\displaystyle \sum _{k=-\infty }^{\infty } 0.01595769121\,{{\rm e}^{- 0.0008000000000\,{k}^{2}}}= 1.0$$

and one would expect a value lower than 1. How is that possible?

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