Evaluating the following expression Why is the value of the following expression equal to 0? I have a feeling that I need to apply L'Hopital's rule, but I do not know where.
$[-x(1-F_X(x)]\Big|_0^{\infty}$, where $F_X(x)$ is the cumulative distribution function of the random variable $X$. 
Thanks!
 A: Assuming your notation means
$$\left[-x(1-F_X(x))\right]_{x=0}^{x=\infty},$$
this can be written in terms of an arbitrary parameter, say $b$, as
$$\lim\limits_{b\rightarrow\infty}\left[-x(1-F_X(x))\right]_{x=0}^{x=B}.$$
The argument of the limit simply evaluates to
$$-b(1-F_X(b))-0=-b(1-F_X(b).$$
To evaluate the original expression, we now need to calculate
$$\lim\limits_{b\rightarrow\infty}-b(1-F_X(b)),$$
recalling that $\lim\limits_{x\rightarrow\infty}F_X(x)=1$ since $F_X$ is a cumulative density function. This is done using L'Hopital's Rule.
A: The limit evaluates to $0$ for $x=0$. We need to use L'Hopital's Rule to evaluate the limit at $\infty$. 
$lim_{x→\infty}[-x(1-F_X(x)]$ = $lim_{x→\infty}[-x]/lim_{x→\infty}\frac{1}{1-F_X(x)}$ = $lim_{x→\infty}[-1]/lim_{x→\infty}\frac{f_X(x)}{[1-F_X(x)]^2}$ = $lim_{x→\infty}\frac{[1-F_X(x)]^2}{f_X(x)}$ = $\frac{0}{\infty}$ = $0$.
A: To give a probabilist answer, let us assume that there is a positive $t$, such that $E(e^{t X})<\infty$. The limit that you want to calculate is: 
$$
\lim_{y\to\infty}y\Pr(X>y)
$$
Now using Markov inequality, you get:
 $$
\Pr(X>y)\leq e^{-ty}E(e^{tX})\implies y\Pr(X>y)\leq ye^{-ty}E(e^{tX})
$$
Now take the limit $y\to\infty$ and the RHS goes to zero and so does the LHS. 
You can also have the weaker assumptions of finite mean and variance and use the Chebyshev inequality to get the same result.
