Q.5.2 - A First Course in Probability

Question

Show that$E\left[Y\right]={\int }_0^{\infty }P\left\{Y>y\right\} dy-{\int }_0^{\infty }P\left\{Y<-y\right\} dy$.

Hint: Show that ${\int }_0^{\infty }P\left\{Y<-y\right\}dy=-{\int }_{-\infty }^{0}x{f}_y\left(x\right)dx$

${\int }_0^{\infty }P\left\{Y>y\right\}dy={\int }_0^{\infty }x{f}_y\left(x\right)dx$

Step-by-Step Solution

Answer

Therefore,

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

Hence Proved.

Step 1 Given Information.

$E\left[Y\right] ={\int }_0^{\infty }P\left\{Y>y\right\}dy-{\int }_{}^{}P\left\{Y<-y\right\}dy$

Step 2 Explanation.

$E\left[Y\right] ={\int }_{-\infty }^{\infty }x.f_y\left(x\right)dx={\int }_{-\infty }^{0}x.f_y\left(x\right)dx + {\int }_0^{\infty }x.f_y\left(x\right)dx$

$=I_1+I_2$

$I_1={\int }_{-\infty }^{0}x.f_y\left(x\right)dx=-{\int }_0^{\infty }x.f_y\left(x\right)dx$

$=-{\int }_0^{\infty }{\int }_{-x}^{0}dy.f_y\left(x\right)dx$

Step 3 Explanation.

$I_{1 , -\infty <-x<-y<0}$

$I_1 =-{\int }_0^{\infty }{\int }_{-\infty }^y{f}_y\left(x\right)dx dy$

$=-{\int }_0^{\infty }P(Y<-y)dy$

$I_{2 }={\int }_0^{\infty }x.f_y\left(x\right)dx$

$={\int }_0^{\infty }{\int }_0^{\infty }dy.f_y\left(x\right) dx , 0\le y\le \alpha \le \infty$

Step 4 Explanation.

$I_2={\int }_0^{\infty }{\int }_Y^{\infty }{f}_y\left(x\right) dx dy$

$={\int }_0^{\infty }P(Y>y)dy$

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