I want to do exercise 3.2.4 from Rick Durett, Probability: Theory and Examples page 86.

$$\text{Let } g \geq 0 \text{ be continuous. If }X_n \Rightarrow X_{\infty} \text{ then } \liminf_{n\rightarrow \infty} \mathbb E(g(X_n))\geq \mathbb E(g(X_{\infty}))$$

My attempt:

Because $X_n\Rightarrow X_{\infty}$ there exists a random variabel $Y_n$ (with the same distribution as $X_n$) which converges to another random variable $Y_{\infty}$ almost surely.

So, we have $\liminf_{n \rightarrow \infty}\mathbb E(g(X_n))=\liminf_{n \rightarrow \infty}\mathbb E(g(Y_n))\geq \mathbb E(g(Y_{\infty}))$ by fatou and the continuity of g.

But can we say that $\mathbb E(g(Y_{\infty}))=\mathbb E(g(X_{\infty})$, if yes, then the prove would be finish.


1 Answer 1


It follows from the very definition that $X_n \to X_{\infty}$ in distribution is equivalent to $Y_n \to X_{\infty}$ in distribution for any sequence $(Y_n)_{n \in \mathbb{N}}$ such that $Y_n \sim X_n$.

Now in your case, as $Y_n \to Y_{\infty}$ almost surely, we have in particular $Y_n \to Y_{\infty}$ in distribution and therefore $X_{\infty} = Y_{\infty}$ in distribution.

  • $\begingroup$ Hi. From your answer I guess that the condition $Y_{\infty} \sim X_{\infty}$ is needed, but this is not part of the assumption. (2nd line) $\endgroup$ Feb 1, 2015 at 18:16
  • $\begingroup$ @Mathlearner Hopefully it's clearer now. $\endgroup$
    – saz
    Feb 1, 2015 at 18:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .