Let $\mu$ and $\sigma^2$ denote the mean and variance of the random variable x. determine $e[(x-\mu)/\sigma]$ and $e{[((x-\mu)/\sigma)^2]}$

Respuesta :

[tex]\mathbb E\left(\dfrac{X-\mu}{\sigma}\right)=\dfrac1\sigma\mathbb E(X)-\dfrac\mu\sigma=\dfrac{\mu-\mu}\sigma=0[/tex]

[tex]\mathbb E\bigg(\left(\dfrac{X-\mu}\sigma\right)^2\bigg)=\dfrac1{\sigma^2}\mathbb E\left(X^2-2\mu X+\mu^2\right)=\dfrac{\mathbb E(X^2)-2\mu\mathbb E(X)+\mu^2}{\sigma^2}[/tex]
[tex]=\dfrac{\mathbb E(X^2)-2\mu^2+\mu^2}{\sigma^2}=\dfrac{\mathbb E(X^2)-\mu^2}{\sigma^2}=\dfrac{\mathbb E(X^2)-\mathbb E(X)^2}{\sigma^2}[/tex]
[tex]=\dfrac{\mathbb V(X)}{\sigma^2}=\dfrac{\sigma^2}{\sigma^2}=1[/tex]