I'm looking at the exercise: Show that, for a non-negative random variable $X$ with cdf $F(x)$, we have $E[X] = \int_0^\infty 1 - F(x) dx$. The hint suggests using integration by parts.
After some searching online, I've found a few derivations of this fact, using various techniques, but I don't see the method that the author of my textbook seems to have had in mind. I think I've got a good solution using the hint, but there's a step where I'm shaky on the justification. I wonder if anyone can tell me whether I've got it right:
$\begin{align}
E[X] &= \int_0^\infty x f(x) dx \\
&=\lim_{b\to\infty}\int_0^bxf(x)dx \\
&=\lim_{b\to\infty}\left(\left[xF(x)\right]_0^b - \int_0^bF(x)dx\right) \text{(used integration by parts at this step.)}\\
&=\lim_{b\to\infty}\left(bF(b) - \int_0^bF(x)dx\right) \\
&=\lim_{b\to\infty}\left(\int_0^bF(b)dx - \int_0^bF(x)dx\right) \\
&=\lim_{b\to\infty}\left(\int_0^bF(b) - F(x)dx\right) \\
&=\int_0^\infty 1 - F(x)dx
\end{align}$
It's that last step that worries me. Why is it justified to take the limit inside the integral, while at the same time rewriting the limit of the definite integral as an improper integral? Am I even doing that right?
Thanks in advance.
Answer
The equality
$$\lim_{b \to \infty} \int_0^b F(b)-F(x) dx = \int_0^\infty 1-F(x) dx$$
follows from the monotone convergence theorem, since $g(b,x)=(F(b)-F(x))1_{[0,b]}(x)$ is an increasing function of $b$ for each fixed $x$. This is probably a bit difficult to handle if you don't have this machinery (or similar machinery like the dominated convergence theorem).
As for doing integration by parts from the start (without already knowing the answer), it's not a big deal: you take $dv=dx,u=1-F(x),du=-f(x)dx$, then just perform the integration by parts on $[0,b]$ and send $b \to \infty$ in the end.
No comments:
Post a Comment