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Let $X$ and $Y$ be non negative i.i.d random variables such that $E[X]<\infty$. Without assuming that $E[X^2]<\infty$, show that $E[\min(X,Y)^2]<\infty$.

We defined the expectation as: Let $X:\Omega\rightarrow S$ be a random element of $(S,\mathcal{S})$ with distribution $\mu$ and let a measurable function $h:S\rightarrow \mathbb{R}$ then $$E[h(x)]=\int_{S}h(x)\mu(dx)$$ whenever LHS or RHS are well defined.

First of all I am actually clueless on how to proceed to solve the question. But this might be due to the fact that I'm not so comfortable with this definition. I don't really understand what $\mu$ is and what does $\mu(dx)$ represent.

I also feel like we should suppose that $E[Y]<\infty$ as well.

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1 Answer 1

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$E[Y] < \infty$ is implied because $X$ and $Y$ are i.i.d $\mu(dx)$ represents the measure of the infinitesimal segment $dx$ according to the probability distribution of $p$. Probability distributions cannot always be represented with a function, but if they can you might see this written as $p(x) dx$.

Finally, $\min(x,y)^2 \leq x y$, thus $E[\min(x,y)^2] \leq E[x y] = E[x]E[y] < \infty$.

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