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For a random variable $X \colon \Omega \to \mathbb{R}$ and a sequence of random variables $X_n$ with

$$ \lim_{n \to \infty} \mathbb{E} [|X_n -X|] = 0,$$

I have found that

$$ \lim_{n\to \infty} \mathbb{E} [f \circ X_n] = \mathbb{E}[f \circ X] \quad (1)$$

holds for any $f \in C_b(\mathbb{R})$.

First Question: I would like to use $(1)$ with $f(x)=e^x \notin C_b(\mathbb{R})$, and I know that $\mathbb{E}[e^X]$ exists. Does this already imply $(1)$? $X$ has negative and positive values.

Second Question: Does $$\mathbb{E} [|X_n -X|] \in O\left(\sqrt {\frac{1}{n}}\right)$$ imply $$ \mathbb{E}[f \circ X_n] - \mathbb{E}[f \circ X] \in O\left(\sqrt {\frac{1}{n}}\right) \quad ?$$ If so, is this also true for $f(x)=e^x$?

Adam
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1 Answers1

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To answer both questions (negatively), consider $X_n$ with distribution $x_n\delta_n+(1-x_n)\delta_0$ with $x_n\to0$. Then (1) holds (with $X=0$ almost surely) as soon as $x_n\ll1/n$ but the convergence for $f:x\mapsto\mathrm e^x$ requires that $x_n\ll1/\mathrm e^n$.

Did
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  • What is the probability space in your example? In particular, what is $\mathbb{E}[\exp(0)]=\mathbb{E}[1]$? – Adam Jun 17 '14 at 13:48
  • ?? Surely you know that the probability space is never relevant in these situations... If one insists, $\Omega=[0,1]$ endowed with the Borel sigma-algebra and the Lebesgue measure would do. – Did Jun 17 '14 at 16:27