Given a random variable $X$, if $E(X^2)< \infty$, i.e., the second moment exists and is finite, does this mean $\log(E(X^2))< \infty$ as well? My intuition is that $E(X^2)< \infty$ implies there exists a $K \in \mathbb{R}$ such that: $$ E(X^2) < K $$ Applying logs to both sides seems to make sense. However, there is the case that $E(X^2) = 0$. In this case, $\log(E(X^2)) = -\infty$. Would this be a problem? Thanks.
Given a random variable $X$, if $E(X^2)< \infty$, does this mean $log(E(X^2))< \infty$ as well?
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probability
probability-theory
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0As you see, if $X=0$ a.s., then $log(E(X^2))=-\infty$. If you agree that $-\infty<\infty$, then $log(E(X^2))<\infty$ is true. – 2017-02-09
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4This has absolutely nothing to do with probability or probability theory. You might have well asked, if $x<\infty$ then $\log x < \infty$? – 2017-02-09
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1What do you introduce K for? $E(X^2)$ is a constant itself which is $\in \mathbb R$. – 2017-02-09
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note that $E(X^2)\geq 0$. If $E(X^2)$ is finite then $\log(E(X^2)) < \infty$. Just note that if $X=0$ a.s. then $E(X^2)=0$ and $\log(E(X^2))=-\infty < \infty$.
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0When $E(X^2)$ is finite, does that mean there exists some constant $K$ such that $E(X^2)
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0When $E(X^2)$ is finite then there exists a constant $K$ such that $E(X^2)=K$. So the answer to you question is yes :) – 2017-02-09