Question

Show that if $X$has density function$f$, then$E\left[g\right(X\left)\right]={\int }_{-\infty }^{\infty }g\left(x\right)f\left(x\right)dx$

Hint: Using Theoretical Exercise$5.2$, start with$E\left[g\right(X\left)\right]={\int }_0^{\infty }P\{g\left(X\right)>y\}dy-{\int }_0^{\infty }P\{g\left(X\right)<-y\}dy$and then proceed as in the proof given in the text when$g\left(X\right)\ge 0.$

Short answer

Therefore,

$E\left[g\right(X\left)\right]={\int }_{x:g\left(x\right)>0}g\left(x\right)f\left(x\right)dx$

Step-by-step solution

Given Information.

$E\left[g\right(X\left)\right]={\int }_{-\infty }^{\infty }g\left(x\right)f\left(x\right)dx$

Explanation.

$E\left[g\right(X\left)\right]={\int }_0^{\infty }P\{g\left(X\right)>y\}dy-{\int }_0^{\infty }P\{g\left(X\right)<-y\}dy$

$g\left(x\right)>0$

$E\left[g\right(X\left)\right]=P\left(g\right(x)>y)dy$

$={\int }_0^{\infty }{\int }_{x:g\left(x\right)>0}^{\infty }f\left(x\right)dx$

Explanation.

$={\int }_{x:g\left(x\right)>0}{\int }_0^{g\left(x\right)}dyf\left(x\right)dx$

$={\int }_{x:g\left(x\right)>0}{{\left(y\right)}^{g\left(x\right)}}_0.f\left(x\right)dx$

$={\int }_{x:g\left(x\right)>0}g\left(x\right)f\left(x\right)dx$