If $X, X_1, X_2, \ldots $ are positive and $X_n\stackrel{P}\to X $ and $E(X_n) \to E(X)$, then $X_n \stackrel{L_1}\to X$

Question

Let $X, X_1, X_2, \ldots $ be positive random variables. Prove that if $X_n\stackrel{P}\to X $ and $E(X_n) \to E(X)$, then $X_n \stackrel{L_1}\to X$

My attempt: I tried to truncate $E(|X_n-X|)$ in two parts like $$E(|X_n-X|)=E(|X_n-X|\mathbb{1}_{|X_n-X|\geq\epsilon} ) + E(|X_n-X|\mathbb{1}_{|X_n-X|\leq\epsilon} )$$

I could show the second part is less than $\epsilon$ but I couldn't bound the first part.

Is it possible to bound the first part? Am I even going in the right direction?

Answer 1

Hint: if the $X_n$ were bounded say by $M$, everything would be great, because then as $P(\{ \omega : |X_n(\omega)-X(\omega)| \geq \varepsilon \})$ becomes less than $\delta$, it would only contribute at most $M\delta$ to the error. In general they are not bounded, but you can approximate by a bounded function and then send the bound to $\infty$. That is, you can write

$$Y_M = \min \{ X,M \} \\ |X_n - X| \leq |X_n - Y_M| + |Y_M - X|$$

By choosing $M$ large enough you can make the expectation of the second term small using the bounded convergence theorem. Now try to play with the first term.

Answer 2

The problem gives a criterion for the convergence in $\mathbb L^1$ of a sequence of non-negative random variables.

If $(X_{n_k})$ is an almost everywhere convergent subsequence, define $Y_k:=X_{n_k}+X-|X_{n_k}-X|$. Using Fatou's lemma and the assumption $\mathbb E[X_n]\to\mathbb E[X]$, we obtain that $\mathbb E|X_{n_k}-X| \to 0$.

Using the fact that if $X_n\to X$ in probability, then a subsequence converges almost surely, we can prove that if $(X_{m_j})$ is a subsequence of $(X_j)$, then we can extract a further subsequence $(X_{m_{j'} })$ such that $\mathbb E|X_{m_{j'} }-X| \to 0$.