# Convergence problem

I practiced some problem (not homework) as below.

Suppose that $X_1,X_2,\cdots$ are independent and identically distributed real-valued random variables with $E|X_1|=\infty$. Show that $\sum_{n=1}^\infty P({|X_n| \geq a\cdot n})=\infty$ for each $a >0$.

I obtained that $\infty = E(|X_n|) = \int_{0}^\infty P(|X_n| >x)dx \leq \sum_{n=0}^\infty P(|X_n|>n)$, but I couldn't get the final step $\sum_{n=1}^\infty P({|X_n| \geq an})=\infty$ for each $a >0$ ( or can I use the statment to imply the result?) Please provide some hints, thanks!

• Consider $Y_n = X_n/a$. Sep 13 '12 at 10:49
• Looks like homework. Add the homework tag. Sep 13 '12 at 11:32
• $a$ or $a_n$ ?.. If this is $a$ there is nothing to do since $P(|X_n| \geq a)$ does not depend on $n$. Hence this is surely $(a_n)$ but what are the hypotheses about the sequence $(a_n)$ ? Sep 13 '12 at 12:15
• Hi, @StéphaneLaurent. Perhaps you misread the problem? :) Sep 13 '12 at 12:51
• @Stéphane: Yes, without LaTeX rendering, it would be difficult to parse! Your boss should really update your computer's browser capabilities so that you can waste time more efficiently. ;-) Sep 13 '12 at 14:44

Remember that $E[Y]=\int_0^\infty P\{Y\geq t\}\,dt$, for nonnegative $Y$.
We see that the $X_i$'s don't need to be independent. Identically distributed is enough.
P.S. Of course, $p_0=1$.