If $X_n\overset{p}{\rightarrow}X$ and $\mathbb E(X_n)\rightarrow\mathbb E(X)$ then $\mathbb E|X_n−X|\rightarrow0$

Question

For $X_n>0$ almost surely, show that if $X_n\overset{p}{\rightarrow}X$ and $\mathbb E(X_n)\rightarrow\mathbb E(X)$, then $\mathbb E|X_n−X|\rightarrow0$

My try:

My idea was to consider the positive and negative parts of $|X_n−X|$ separately. We have

$$\begin{align*} \mathbb E|X_n-X| &=\mathbb E\left(\left(X_n-X\right)^{+}\right) + \mathbb E\left(\left(X_n-X\right)^{-}\right)\\\\ &=\mathbb E\left(\max\{X_n-X,0\}\right)+\mathbb E\left(\max\{X-X_n,0\}\right) \end{align*}$$

Since $\mathbb E(X_n)\rightarrow\mathbb E(X)$ then $\mathbb E(X_n-X)\rightarrow0$ and $\mathbb E(X-X_n)\rightarrow0$. Hence it seems it should be the case that $\mathbb E\left(\max\{X_n-X,0\}\right)$ converges to $0$ (and similarly for $\mathbb E\left(\max\{X-X_n,0\}\right)$). Hence $\mathbb E|X_n-X|\rightarrow0$.

Is my reasoning correct? I never made use of the fact that $X_n>0$ almost surely, nor the fact that $X_n\overset{p}{\rightarrow}X$ so I feel my approach might not be right.

Answer 1

First, note that this is false without the assumption $\mathbb{E}|X| < \infty$. Consider for instance $X(t) = 1/t$ on the space $(0,1]$ with respect to Lebesgue measure, and set $X_n(t) = \min(n, 1/t)$. Then we can check this satisfies all the assumptions (in fact, $X_n \to X$ surely), but $\mathbb{E}|X_n-X| = \infty$ for all $n$.

With that in mind, let's assume $X$ is integrable.

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Second, note that if you don't use either of the assumptions that $X_n\to X$ in probability and $X_n > 0$ a.s., then as I had mentioned in the comments earlier, you are effectively trying to prove $$\mathbb{E}(X_n-X) \to 0 \implies \mathbb{E}|X_n-X| \to 0$$ which is false, so the proof is flawed from the onset. In your proof, this error appears more precisely as $$\mathbb{E}(X_n-X) \to 0 \implies \mathbb{E}(\max\{X_n-X,0\}) \to 0$$

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Let's finally prove the modified statement.

Consider $(X_n-X)^-$. We have $0 \leq (X_n-X)^- \leq |X|$ a.s. and $(X_n-X)^- \to 0$ in probability. By the the Lebesgue dominated convergence theorem, $\mathbb{E}\left[(X_n-X)^-\right]\to 0$.

At this point, we're effectively finished. This is because we may write $$|X_n-X| = (X_n-X) + 2(X_n-X)^-$$ and the right-hand side is the sum of two sequences, each of whose expectations goes to zero.