【MT&PT】Ex,Dx prove

I Possion,Gaussion,Exponential distribution

II Ex,Ex^2,Dx prove

$-Possion\ distribution:$

$X~\pi(\lambda)$

$E(x)=\sum_{k=0}^{n}k\frac{\lambda^{k}e^{-\lambda}}{k!}$

$=\lambda e^{-\lambda}\sum^{\infty}_{(k-1)!}=\lambda e^{-\lambda}e^{\lambda}$

$=\lambda$

$E(X^{2})=E[X(X-1)+X]=E[X(x-1)]+E(X)$

$=\sum_{k=0}^{\infty}k(k-1)\frac{\lambda^{k}e^{-\lambda}}{k!}+\lambda$

$=\lambda^{2}e^{-\lambda}\sum_{k=2}^{\infty}\frac{e^{k-2}}{(k-2)!}+\lambda$

$=\lambda^{2}+\lambda$

$D(X)=\lambda^{2}$

$-Normality\ distribution:$

$-Exponential\ distribution:$

$E(X)=\int_{-\infty}^{\infty}x\frac{1}{\theta}e^{-\frac{x}{\theta}}dx$

$=-xe^{-\frac{x}{\theta}}|^{\infty}_{0}+\int_{0}^{\infty}e^{-\frac{x}{\theta}}dx$

$=\theta$

$E(X^{2})=\int_{-\infty}^{\infty}x^{2}\frac{1}{\theta}e^{\frac{x}{\theta}}dx$

$=2\theta^{2}$

$D(X)=\theta^{2}$