Solutions - 5130-02

PMF of a Poisson distribution:

$$P_X(k)=e^{-\alpha }\frac{{\alpha }^k}{k!}$$

Expected value:

$$\begin{array}{c}E[X]=\sum _{k=0}^{\infty }k{P}_X(k)\\ \\ \gg \gg \sum _{k=0}^{\infty }{e}^{-\alpha }\frac{{\alpha }^k}{(k-1)!}\\ \\ \gg \gg \alpha e^{-\alpha }\sum _{k=0}^{\infty }\frac{{\alpha }^{k-1}}{(k-1)!}\\ \\ \gg \gg \alpha e^{-\alpha }{e}^{\alpha }\gg \gg \alpha \end{array}$$