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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!

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    $\begingroup$ Consider $Y_n = X_n/a$. $\endgroup$ – Douglas Zare Sep 13 '12 at 10:49
  • $\begingroup$ Looks like homework. Add the homework tag. $\endgroup$ – Michael R. Chernick Sep 13 '12 at 11:32
  • $\begingroup$ $a$ 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 $(a_n)$ ? $\endgroup$ – Stéphane Laurent Sep 13 '12 at 12:15
  • $\begingroup$ Hi, @StéphaneLaurent. Perhaps you misread the problem? :) $\endgroup$ – cardinal Sep 13 '12 at 12:51
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    $\begingroup$ @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. ;-) $\endgroup$ – cardinal Sep 13 '12 at 14:44
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Remember that $E[Y]=\int_0^\infty P\{Y\geq t\}\,dt$, for nonnegative $Y$.

enter image description here

We see that the $X_i$'s don't need to be independent. Identically distributed is enough.

P.S. Of course, $p_0=1$.

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    $\begingroup$ (+1) For most readers, the very first line under your graph should be more than enough. I was just about to mention the oddity of the iid assumption in the question statement; I'm glad to see you made an edit to point this out! :-) $\endgroup$ – cardinal Sep 13 '12 at 16:46
  • $\begingroup$ Sharp as usual, Mr. cardinal. $\endgroup$ – Zen Sep 13 '12 at 16:49

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