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Suppose that $Y_N \overset{p}{\to} Y$ as $N \rightarrow \infty$, i.e., $Y_N$ converges to $Y$ in probability. How do I prove that $$ E(Y_N) \rightarrow E(Y) $$ as $N \rightarrow \infty$? This is easy when $Y_N \overset{a.s.}{\to} Y$ as one can apply dominated convergence theorem (since integration over a set of measure zero yields zero). My definition for the expected value is $$ E(X)=\int_{\Omega} X dP, $$ where $\Omega$ is the space and $P$ probability measure.

mathslover
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    To add to the answer below, note that convergence in probability implies convergence in distribution. But convergence in distribution does not imply convergence of expectation; see https://math.stackexchange.com/questions/153293/does-convergence-in-distribution-implies-convergence-of-expectation. However, when your sequence of random variables is uniformly integrable, then you do indeed get convergence in expected value. – Satana Feb 15 '23 at 21:11
  • This can work if the $Y_n$ process is bounded, say, $|Y_n|\leq b$ always for some finite constant $b$. – Michael Feb 15 '23 at 23:01
  • Does this answer your question? Does convergence in distribution implies convergence of expectation? (not an exact duplicate, but it the counterexample given there works here). – Mike Earnest Feb 16 '23 at 21:26

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This is not true. Even if $Y_n\to Y$ almost-surely we do not in general have $\mathbb EY_n\to\mathbb E Y$.

As a typical example of this, consider the random variables $Y_n$ defined by $$\mathbb P[Y_n=n]=1/n, \mathbb P[Y_n=0]=1-1/n.$$ Then clearly $Y_n\to 0$ but $$\mathbb EY_n = 1, \hspace{0.5cm} \text{ for all } n \in\mathbb N.$$

In fact, knowing $\textbf{when}$ we can deduce convergence of expected values is entirely the point of the convergence theorems (dominated convergence, monotone convergence...).

Small Deviation
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  • I see. What kind of additional assumptions on $Y_N$:s do I need for this to hold? – mathslover Feb 15 '23 at 21:09
  • Well, any assumption that would allow you to apply some convergence theorem. For example (1) $0\leq Y_n\leq Y_{n+1}$ for all $n$ or (2) there exists an integrable $Z$ with $|Y_n|\leq Z$ or (3) $(Y_n)_{n\in\mathbb N}$ is uniformly integrable. – Small Deviation Feb 15 '23 at 21:15