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$X$ is real random variable such that $\mathbb{P}(X > 0) = 1$, $\mathbb{E}X^2 < \infty$, $\mathbb{E}X=1$. Let $Z$ be real random variable such that $\mathbb{P}(Z \in A)=\mathbb{E}X\mathbb{1}_{\ln X \in A}$ for all borel sets $A \subset \mathbb{R}$. Show that $\mathbb{E}|X\ln X| < \infty$ and $\mathbb{E}Z = \mathbb{E} X \ln X$.

Here's my unsuccessful try for the second part: Firstly let' note that: $$\mathbb{E}X \ln X = \int_{\mathbb{R}}x\ln x\mu_X(dx)$$ so let's try to transform $\mathbb{E} Z$ to this form. $$\mathbb{P}(X>0) = 1 \Rightarrow \forall_{A \in \mathcal{B}(\mathbb{R})}\mathbb{P}(X\mathbb{1}_{\ln X \in A} > 0) = 1 \Rightarrow \mathbb{E}X\mathbb{1}_{\ln X \in A} = \int_0^{\infty}\mathbb{P}(X\mathbb{1}_{\ln X \in A} \ge t)dt$$ $$\mathbb{E} Z = \int_{-\infty}^\infty\mathbb{P}(Z > t)dt = \int_{-\infty}^\infty \mathbb{E}X\mathbb{1}_{\ln X \in [t, \infty)}dt = \int_{-\infty}^\infty\mathbb{E}X\mathbb{1}_{X \in [e^t,\infty)}dt=\int_{-\infty}^\infty\int_0^\infty\mathbb{P}(X\mathbb{1}_{ X \in [e^t,\infty)} \ge s)dsdt=\int_{-\infty}^\infty\int_0^\infty\mathbb{P}(X \ge s, X \ge e^t)dsdt=\int_{-\infty}^\infty\int_{e^t}^\infty\mathbb{P}(X \ge s)dsdt + \int_{-\infty}^\infty\int_0^{e^t}\mathbb{P}(X \ge e^t)dsdt$$ I tried to transform last equation but without any effect. Is this even a good way to solve this problem? I'd be grateful for any hints for this problem.

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2 Answers 2

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Using $\vert X \ln X \vert \leqslant \max\left(\mathrm{e}^{-1}, X^2\right)$, valid for $X > 0$, we have $$ \mathbb{E}\left(\vert X \ln X \vert \right) \leqslant \mathrm{E}\left(X^2 1_{X> \mathrm{e}^{-1/2}}\right) + \mathbb{E}\left(\mathrm{e}^{-1} 1_{X^2 < \mathrm{e}^{-1}}\right) < \mathbb{E}\left(X^2\right) < \infty $$ For the second part, note that $\Pr(Z<0) > 0$, hence $$ \mathbb{E}\left(Z\right) = \int_{0}^\infty \mathbb{P}\left(Z>t\right) \mathrm{d}t - \int_{0}^\infty \mathbb{P}\left(Z<-t\right) \mathrm{d}t = \int_{0}^\infty \left(\mathbb{P}\left(Z>t\right)-\mathbb{P}\left(Z\leqslant-t\right)\right)\mathbb{t} = \mathbb{E}\left(X \int_0^\infty \left(1_{\ln X >t } - 1_{\ln X < -t}\right) \mathrm{d}t\right) = \mathbb{E}\left(X \left( 1_{X>1} \cdot \ln X - 1_{X<1} \left(-\ln X\right) \right) \mathrm{d}t\right) = \mathbb{E}\left(X \ln X\right) $$

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The inequality $|x\ln x|\leqslant C\cdot x^2$ does not hold for every $x\gt0$, for any finite $C$, hence the first part of this post needs to be revised. –  Did Sep 1 at 13:13
    
@Did My bad, the inequality should have been $\vert X \ln X \vert \leqslant \max\left(\mathrm{e}^{-1},X^2\right)$. –  Sasha Sep 4 at 20:53

The first claim is clear because $x\ln x$ is bounded near $x=0$ and $\lesssim x^2$ for large $x$.

Let $d\mu(x)$ be the distribution of $X$; this can be viewed as a measure on $(0,\infty)$. Let $\nu$ be the image measure under the transformation $s=\ln x$. Then $$ P(Z\le t) = \int_{(0,e^t]} x\, d\mu(x) = \int_{(-\infty,t]} e^s\, d\nu(s) . $$ Thus the distribution of $Z$ is $d\mu_Z(t) = e^t\, d\nu(t)$. Now the claim follows by working out $EZ=\int t\, d\mu_Z(t)$ and undoing the above substitution. (This last part also establishes that $E|Z|<\infty$.)

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