Expected Value with certain probability in absolute value

Question

Let $X$ be a random variable and $M$ a constant such that $P[\lvert X\rvert\leq M]=1$. Show that $\lvert E(X) \rvert\leq E(\lvert X \rvert)\leq M$

For the first part, let $g(X)=\lvert X \rvert$ then $g$ is a convex function and then, by jensen inequality \begin{equation} g(E(X))\leq E(g(x)) \Rightarrow \lvert E(X) \rvert\leq E(\lvert X \rvert) \end{equation} How do I prove the second inequality? thanks a lot for your help

score 2 · Answer 1 · answered May 05 '18 at 20:28

2

For non-negative random variables, $X$,

$$ E[X] = \int_0^\infty P(X > x)dx $$

Since $|X| \leq M$ almost surely, $\{|X| > x\} \subseteq \{M > x\}$. Thus,

$$ E[|X|] = \int_0^\infty P(|X| > x)dx \leq \int_0^\infty P(M > x)dx = \int_0^\infty \chi_{\{M > x\}}dx = M$$

answered May 05 '18 at 20:28

I must assumed $X$ is non negative – Karen May 05 '18 at 20:31
No, because $|X|$ is. – May 05 '18 at 20:33
True, did not see that, but how do I prove your second line? i can't see it clearly – Karen May 05 '18 at 20:38
@Karen it is a classic trick, useful to know: $$\int_{0}^\infty x P(X \in dx) = \int_{0}^\infty \int_0^x dy P(X \in dx) = \int_0^\infty dy \int_{x}^\infty P(X \in dx) = \int_0^\infty P(X>y)dy$$ – Clement C. May 05 '18 at 22:27

score 0 · Accepted Answer · answered May 05 '18 at 22:07

You have, writing a tautology and using a crude bound, $$ |X| = |X|\mathbb{1}_{\{|X|\leq M\}} + |X|\mathbb{1}_{\{|X|> M\}} \leq M\mathbb{1}_{\{|X|\leq M\}} + |X|\mathbb{1}_{\{|X|> M\}} \tag{1} $$ so that, by linearity of expectation, $$ \mathbb{E}[|X|] \leq \mathbb{E}[M\mathbb{1}_{\{|X|\leq M\}}] + \mathbb{E}[|X|\mathbb{1}_{\{|X|> M\}}] = M\mathbb{E}[\mathbb{1}_{\{|X|\leq M\}}] + \mathbb{E}[|X|\mathbb{1}_{\{|X|> M\}}] \tag{2} $$ Finally, $$ \mathbb{E}[\mathbb{1}_{\{|X|\leq M\}}] = \mathbb{P}\{|X|\leq M\} = 1 $$ while $ \mathbb{E}[\lvert X\rvert\mathbb{1}_{\{|X|> M\}}] = 0$ since $\mathbb{P}\{|X|> M\}=0$.

Expected Value with certain probability in absolute value

2 Answers2