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Let $Y_n, n \in \mathbb{N}$ be a sequence of non-negative random variables on $L_1(\Omega, \mathcal{F}, P)$ such that

  1. $\mathbb{E}Y_n = 1$ and $\mathbb{E}(Y_n \,\ln Y_n) \leq 1$, $\forall n\in \mathbb N$
  2. $\mathbb{E}(Y_n \,Z) \rightarrow \mathbb{E}(Y Z) $ for all bounded random variables $Z$ on $(\Omega, \mathcal{F}, P)$

How do I prove that $Y\geq 0$ a.s., $\mathbb{E}Y = 1 $ and $\mathbb{E}(Y \ln Y) \leq 1$?

I was thinking that maybe I should first prove that $Y_n \rightarrow Y$ a.s. which should somehow follow from (2.) and than apply convergence theorems.

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    $\begingroup$ Don't you need Y_n,Y strictly greater than zero to take the ln? $\endgroup$
    – Jimmy R.
    Oct 30, 2016 at 8:32
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    $\begingroup$ Maybe we can assume that $Y_n, Y$ take values in the extended real line $\endgroup$
    – Jules
    Oct 30, 2016 at 11:31
  • $\begingroup$ Yes, if course. By the way, I am not happy with some implications in my answer, so I am reconsidering it. $\endgroup$
    – Jimmy R.
    Oct 30, 2016 at 11:55

1 Answer 1

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The random variable $Z(ω)=\mathbf 1_{\{ω\in Ω\}}(ω)$ or equivalentlty $Z\equiv 1$, is bounded. Hence, by (2.) $$\mathbb E(Y)=\lim_{n\to\infty}\mathbb E(Y_n)\overset{(1.)}=\lim_{n\to\infty}1=1$$ Moreover, since for any $A \in \mathcal F$ the function $Z_A=\mathbf 1_{\{ω\in A\}}$ is bounded, you have by (2.) and the Portmanteu lemma that $Y_n\to Y$ in distribution. The continuous mapping theorem, implies that also $g(Y_n):=Y_n\ln Y_n\to g(Y):=Y\ln Y$ in distribution. Since, $P(Y_n<0)=0$ for any $n\in\mathbb N$, convergence in distribution implies that $$0=\lim_{n\to\infty}P(Y_n<0)=P(Y<0)$$ hence $Y\ge 0$ a.s. It remains to show that $\mathbb E(Y\ln{Y})\le 1$.

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  • $\begingroup$ $x < \ln(x)$ for any $x > 0$? What about $x =1$? $\endgroup$
    – user159517
    Oct 29, 2016 at 13:16
  • $\begingroup$ "Since $\mathbb{E}(Y_n) = 1$, you know that the $Y_n's$ are a.s. bounded". What about letting $X_n= U(0,1)$ (uniform distribution on (0,1)) and taking $Y_n = -\ln(X_n)$? $\endgroup$
    – user159517
    Oct 29, 2016 at 13:19
  • $\begingroup$ @user159517 If I am correct, in your example $Y_n \sim \exp(1)$ and hence $\mathbb E(Y_n)=1$ but the $Y_n$'s are unbounded, right? $\endgroup$
    – Jimmy R.
    Oct 29, 2016 at 13:23
  • $\begingroup$ That's what I'm saying, yes. $\endgroup$
    – user159517
    Oct 29, 2016 at 13:26
  • $\begingroup$ @user159517 Ok, I see the problem in my reasoning. Hmmm, let me think how I can correct it. Thanks a lot for the feedback. $\endgroup$
    – Jimmy R.
    Oct 29, 2016 at 13:27

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