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A discrete random variable is one that can only take on a finite number of different values.

A discrete random variable X with a set of potential values S and pmf p(x) has the following Expected Value (mean value):

\(\begin{aligned}{\rm{E(X) = }}{{\rm{\mu }}_{\rm{X}}}\\{\rm{ = }}\sum\limits_{{\rm{x}} \in {\rm{S}}} {{\rm{x \times p(x)}}} \end{aligned}\)

pmf with a random variable X

\({\rm{p(x;u) = }}{{\rm{e}}^{{\rm{ - \mu }}}}\frac{{{{\rm{\mu }}^{\rm{x}}}}}{{{\rm{x!}}}}\)

When for\({\rm{x = 0,1,}}.....\) has a Poisson Distribution with parameter \({\rm{\mu > 0}}\).

As a result, according to the definition, the following is correct:

\(\begin{aligned}E(X) &= \sum\limits_{{\rm{x = 0}}}^\infty {\rm{x}} {\rm{ \times }}{{\rm{e}}^{{\rm{ - \mu }}}}\frac{{{{\rm{\mu }}^{\rm{x}}}}}{{{\rm{x!}}}}\\&= {\rm{0 + }}\sum\limits_{{\rm{x = 1}}}^\infty {\rm{x}} {\rm{ \times }}{{\rm{e}}^{{\rm{ - \mu }}}}\frac{{{{\rm{\mu }}^{\rm{x}}}}}{{{\rm{x!}}}}\\&= \mu \sum\limits_{{\rm{x = 1}}}^\infty {\rm{x}} {\rm{ \times }}{{\rm{e}}^{{\rm{ - \mu }}}}\frac{{{{\rm{\mu }}^{{\rm{x - 1}}}}}}{{{\rm{x!}}}}\\&= {\rm{\mu }}\sum\limits_{{\rm{x = 1}}}^\infty {{{\rm{e}}^{{\rm{ - \mu }}}}} \frac{{{{\rm{\mu }}^{{\rm{x - 1}}}}}}{{{\rm{(x - 1)!}}}}\\\ &={\rm{\mu }}\sum\limits_{{\rm{y = 0}}}^\infty {{{\rm{e}}^{{\rm{ - \mu }}}}} \frac{{{{\rm{\mu }}^{\rm{y}}}}}{{{\rm{y!}}}}\\& = \mu {{\rm{e}}^{{\rm{ - \mu }}}}\sum\limits_{{\rm{y = 0}}}^\infty {\frac{{{{\rm{\mu }}^{\rm{y}}}}}{{{\rm{y!}}}}} \\ &= {\rm{\mu }}{{\rm{e}}^{{\rm{ - \mu }}}}{{\rm{e}}^{\rm{\mu }}}{\rm{n}}\\ &= {\rm{\mu }}\end{aligned}\)

(1) When \({\rm{a = 0}}\), the first term equals zero;

(2):We have a word for each term.

\(\frac{{\rm{x}}}{{\rm{x}}}{\rm{ = 1}}\)

\({\rm{(x - 1)!}}\)where there isn't any more;-

(3): make \({\rm{y = x - 1}}\). The total shifts to zero instead of one as a result of this.

(4): the total

\(\sum\limits_{{\rm{y = 0}}}^\infty {\frac{{{{\rm{\mu }}^{\rm{y}}}}}{{{\rm{y!}}}}} {\rm{ = }}{{\rm{e}}^{\rm{\mu }}}\)

Therefore, the value is: \({\rm{E(X) = \mu }}\).