5
$\begingroup$

How do I prove that $\int_0^\infty Pr(Y\geq y) dy = E[Y]$ if $Y$ is a non-negative random variable?

$\endgroup$
12
$\begingroup$

Assuming we have a continuous random variable with an existant probability density function $f_Y$.

$\begin{align} \int_0^\infty \Pr(Y \geqslant y) \operatorname d y & = \int_0^\infty \int_y^\infty f_Y(z)\operatorname d z\operatorname d y \\[1ex] & = \int_0^\infty \int_0^z f_Y(z)\operatorname d y\operatorname d z \\[1ex] & = \int_0^\infty f_Y(z)\int_0^z 1\operatorname d y\;\operatorname d z \\[1ex] & = \int_0^\infty z f_Y(z)\operatorname d z \\[1ex] & = \mathsf E[Y] \end{align}$

$\endgroup$
  • $\begingroup$ Could you please elaborate on how you changed the order of integration to arrive at the limits of 0 and z in the inner integral? $\endgroup$ – user2510050 Oct 5 '14 at 6:18
  • 2
    $\begingroup$ $\begin{align} & \text{The double integral is performed over the interval:} \\ & \{(y,z):z\in[0,\infty)\cap y\in[z,\infty)\} \\ \equiv & \{(y,z):0\leqslant z\leqslant y<\infty\} \\ \equiv & \{(y,z):y\in [0,\infty)\cap z\in[0,y]\} \end{align}$ @user2510050 $\endgroup$ – Graham Kemp Oct 5 '14 at 6:48
  • $\begingroup$ Got it, thanks! $\endgroup$ – user2510050 Oct 5 '14 at 17:26
6
$\begingroup$

This proof assumes a background in measure theory.

Let $1_{Y\ge y}$ the indicator function for the set $\{Y\ge y\}$. Then $$ \int_0^\infty P(Y\ge y)\,dy = \int_0^\infty \int 1_{Y\ge y}\,dP\,dy=\int\int_0^\infty1_{Y\ge y}\,dy\,dP=\int Y\,dP=E[Y] $$ The middle equality follows from the Fubini-Tonelli theorem. The third equality follows since $Y\ge0$, so the function $1_{Y\ge y}$ is $1$ on the interval $[0,Y]$, and $0$ elsewhere, so its integral over the real line is $Y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.