# Math Help - Cumulative distrubution function question

1. ## Cumulative distrubution function question

Can anyone provide any insight in how I could show the following (attached below)

2. $0\le nP(X>n) =n\int_n^{\infty}f(x)dx =\int_n^{\infty}nf(x)dx\le \int_n^{\infty}xf(x)dx\to 0$

as $n\to \infty$ since the first moment is finite.

3. Hello,

$1-F(x)=P(X>n)=P(X\cdot \bold{1}_{X>n}>n)=P(Y_n>n)\leq \frac{E(Y_n)}{n}$ by Markov's inequality.
where $Y_n=X\cdot \bold{1}_{X>n}$
As n goes to infinity, $Y_n$ obviously goes to 0 (X is almost surely finite since it's integrable)
and $|Y_n|$ is bounded by |X|, which is integrable.
So we can apply the dominated convergence theorem, and we have $n(1-F(x))\leq E(Y_n) \to 0 \quad \quad \square$