If X and Y are random variables on the standard probability space with X integrable and $Y|\le M$ almost surely, show that $X\cdot Y$ is integrable and that $|E[X\cdot Y]|\le ME[|X|]$.
Product of an integrable random variable and a bounded random variable
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probability-theory
random-variables
expectation
1 Answers
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Integrals are monotone and linear, so if $f\ge g$ then $\displaystyle\int f\ge \int g$ and $\displaystyle\int cf=c\int f$. So
$$\int_{\Omega} |XY|\,dP=\int_{\Omega}|X||Y|\,dP \le M\int_{\Omega}|X|\,dP=ME[|X|].$$
In particular $XY$ is integrable by the dominated convergence theorem.
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0Dominated convergence theorem.... why do you need this? – 2017-02-23
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0@saz to establish integrability. You can also use eg. Hölder's inequality. – 2017-02-23
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0... but your estimate $$\int_{\Omega} |XY| \, d\mathbb{P} \leq M \mathbb{E}(|X|)<\infty$$ already shows that $X \cdot Y$ is integrable. – 2017-02-23
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0@saz Yes, I know that these are functionally the same, but students sometimes have issues with knowing that estimates like that show integrability simultaneously with such bounds. I agree it is not necessary for the trained professional: it was a pedagogical choice. – 2017-02-23