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Simple question: Why does $E(|X|) < \infty$ imply $E(|X|I_{|X|>a} )$ tends to $0$ as $a$ tends to infinity?

I've seen it in a few proofs and I can't see why this is the case, I've tried a proof using Markov's inequality but am unconvinced by my reasoning.

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This follows from the dominated convergence theorem and the fact that the measure of the set where $|X|=\infty$ is zero. –  Chris Janjigian Nov 18 '12 at 16:58
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3 Answers 3

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No need to apply a monotone convergence theorem. First, fix $\delta>0$, and let $Y\leqslant X$ a simple function such that $E[|X|-Y]<\delta$ (such a function exists by definition of Lebesgue integral). We write $Y:=\sum_{j=1}^Nc_j\chi_{A_j}$, where $A_j$ are measurable and pairwise disjoin. This gives $$E[|X|\chi_{\{|X|> a}]<\delta+\sum_{j=1}^Nc_j\Bbb P(A_j\cap\{|X|>a\}).$$ As $a\Bbb P(A_j\cap\{|X|>a\})\leqslant \int_{A_j}|X|d\Bbb P$, we get $$E[|X|\chi_{\{|X|> a}]<\delta+\frac 1a\sum_{j=1}^Nc_j\int_{\Omega}\chi_{A_j}|X|d\Bbb P.$$ Taking $\limsup_{a\to +\infty}$ and using the fact that $\delta$ is arbitrary, we get the result.

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$\left| X \right| = \left| X \right|{I_{\left| X \right| > a}} + \left| X \right|{I_{\left| X \right| \leqslant a}}$

so by Lebesgue monotone convergence theorem

$E\left[ X \right] = \mathop {\lim }\limits_{a \to + \infty } E\left[ {\left| X \right|{I_{\left| X \right| > a}}} \right] + \mathop {\lim }\limits_{a \to + \infty } E\left[ {\left| X \right|{I_{\left| X \right| \leqslant a}}} \right] = $

$ = \mathop {\lim }\limits_{a \to + \infty } E\left[ {\left| X \right|{I_{\left| X \right| > a}}} \right] + E\left[ {\mathop {\lim }\limits_{a \to + \infty } \left| X \right|{I_{\left| X \right| \leqslant a}}} \right] = $

$ = \mathop {\lim }\limits_{a \to + \infty } E\left[ {\left| X \right|{I_{\left| X \right| > a}}} \right] + E\left[ {\left| X \right|} \right] $ $\Rightarrow \mathop {\lim }\limits_{a \to + \infty } E\left[ {\left| X \right|{I_{\left| X \right| > a}}} \right] = 0$

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Write $$E|X| = E[|X|1_{|X| > a}] + E[|X|1_{|X| \leq a}] $$ Take limits on both sides. By Monotone convergence theorem $E[|X|1_{|X| \leq a}] \rightarrow E|X|$ You can cancel E|X| on both sides as it is $< \infty$. Thus you arrive at your result.

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I've also used $1_{|X| \leq a}\rightarrow 1$. Thats because $|X|< \infty $a.s. Follows from $E|X| < \infty$ a.s. –  Gautam Shenoy Nov 18 '12 at 17:02
    
I used the continuous version of MCT. –  Gautam Shenoy Nov 18 '12 at 17:03
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