Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$X$ 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.

share|cite|improve this question
up vote 1 down vote accepted

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 $$ \mathbb{E}\left(Z\right) = \int_{0}^\infty \mathbb{P}\left(Z>t\right) \mathrm{d}t - \int_{0}^\infty \mathbb{P}\left(Z<-t\right) \mathrm{d}t = \int_{0}^\infty \left(\mathbb{P}\left(Z>t\right)-\mathbb{P}\left(Z\leqslant-t\right)\right)\mathbb{t} = \mathbb{E}\left(X \int_0^\infty \left(1_{\ln X >t } - 1_{\ln X < -t}\right) \mathrm{d}t\right) = \mathbb{E}\left(X \left( 1_{X>1} \cdot \ln X - 1_{X<1} \left(-\ln X\right) \right) \mathrm{d}t\right) = \mathbb{E}\left(X \ln X\right) $$

share|cite|improve this answer
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 '14 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 '14 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 $E|Z|<\infty$.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.