$X$ and $Y$ are independent r.v. such that $E|X|^p < \infty$ and $E|Y|^p < \infty$ for $p\ge 1$. Assume $EY=0$. Prove $E|X|^p \le E|X+Y|^p$.
I have no idea how to proceed.
How to prove $E|X|^p \le E|X+Y|^p$ given certain conditions?
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probability-theory
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0Start with expectations conditional on $X$. – 2017-01-12
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0Can you provide more details? – 2017-01-12
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1Another hint: use Jensen's inequality – 2017-01-12
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0It doesn't seem to lead to the conclusion. $E|X+Y|^p \ge |E(X+Y)|^p=|EX|^P$ – 2017-01-12
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0It's because you haven't considered conditional on $X$ expectations first. – 2017-01-12
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0Thanks. Got it. $E|X+Y|^p=E[E(|X+Y|^p|X)] \ge E[|E(X+Y|X)|^P]=E|X|^p$ – 2017-01-12
1 Answers
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With the hints of Sergei Golovan:
Consider conditional expectation $ E(|X+Y|^p|\ X)$ and apply Jensen's inequality to it: $$E(|X+Y|^p|\ X )\geq |E(X+Y|\ X)|^p. $$ Moreover, since $X$ and $Y$ are independent and $EY=0$, we have $E(X+Y|\ X)=X+EY=X$. This gives $$E(|X+Y|^p|\ X )\geq |X|^p, a.s. $$ Now we take expectations on both sides in the last inequality and obtain: $$E|X+Y|^p=E(E(|X+Y|^p|\ X ))\geq E|X|^p, $$ as desired.