# How to show $\lim_{n\to\infty} \mathbb{E}|X_{n} - X| = 0$ given $\lim_{n\to\infty} X_{n} = X$?

Let $$X$$ be a random variable and let $$\{X_{n}\}$$ be a sequence of non-negative random variables such that $$E(X_{n}) = E(X) < \infty$$ for every $$n\in \mathbb{N}$$, and $$\lim_{n\to\infty} X_{n} = X$$ $$\mathbb{P}$$-a.s.

Prove that $$\lim_{n\to\infty} E|X_{n} - X| = 0$$.

I am sort of new to convergence in probability theory. But I am trying to learn on my own. I would really like your assistance with this problem. I have tried for long but cannot figure it out. I guess it might be because my understanding of convergence in probability theory is not so great.

I think I must use Markov's inequality. I tried it on $$E|X_{n} - X|$$ though with no help. Can you please help me solve this problem ?

• This is known as Scheffe's lemma, and only requires that $\mathbb E[X_n]\stackrel{n\to\infty}\longrightarrow \mathbb E[X]$, as I recall. Commented Nov 11, 2019 at 3:14
• Kudos to you @Math1000, I found a really simple & elegant proof here: theanalysisofdata.com/probability/8_4.html. Commented Nov 11, 2019 at 3:42
• $|c| = c + 2\max(-c,0)$ is such a strange identity; I doubt I would have come up with that on my own. Commented Nov 11, 2019 at 5:22

The trick is to bound the term by logically three parts, a part where $$X_n$$ varies from $$X$$ by no more than $$\epsilon$$, a part with $$X with diminishing measure, and finally a part of $$X>M$$ where the expectation diminishes.

Let $$\epsilon > 0,$$ $$A_n = \left\{ \vert X_n - X \vert \le \epsilon \right\}$$. Since $$\lim_{n\to\infty} X_{n} = X$$ a.s, we have $$\lim _{n\to\infty} P(A_n^c) = 0.$$

Let $$X^M=X \wedge M$$. By the dominated convergence theorem, $$\lim _{M\to\infty}E(X^M)=EX.$$ Hence $$\lim _{M\to\infty}E(X1_{(X>M)})=0.$$ Again by dominated convergence theorem, $$\lim _{n\to\infty}E(X_n1_{(X_n>M)})=E(X1_{(X>M)})$$, i.e. with any $$\delta > 0$$, for sufficiently large $$N$$, $$E(X_n1_{(X_n>M)})\le E(X1_{(X>M)}) + \delta\quad \text{for } n \ge N.$$

$$E|X_{n} - X| = \int | X_n - X |dP = \int_{A_n} | X_n - X |dP + \int_{A_n^C} | X_n - X |dP$$ $$\le \int_{A_n} | X_n - X |dP + \int_{A_n^C} X_n dP + \int_{A_n^C} X dP$$ $$= \int_{A_n} | X_n - X |dP + \int_{A_n^C \cap \{X_n\le M\}} X_n dP + \int_{A_n^C \cap \{X\le M\}} X dP + \int_{A_n^C \cap \{X_n> M\}} X_n dP + \int_{A_n^C \cap \{X> M\}} X dP$$ $$\le \int_{A_n} | X_n - X |dP + \int_{A_n^C \cap \{X_n\le M\}} X_n dP + \int_{A_n^C \cap \{X\le M\}} X dP + \int_{\{X_n> M\}} X_n dP + \int_{\{X> M\}} X dP$$ $$\le \epsilon P(A_n) + 2P(A_n^c)M + E(X_n1_{(X_n>M)}) + E(X1_{(X>M)})$$ $$\le \epsilon + 2P(A_n^c)M + E(X_n1_{(X_n>M)}) + E(X1_{(X>M)})$$ $$\le \epsilon + 2P(A_n^c)M + \delta + 2 E(X1_{(X>M)})$$

With the above inequality, we can start by choosing large enough $$M$$ so that the last term is small enough, next we can choose sufficiently small $$\epsilon$$ and sufficiently large $$n$$ so that the first three terms as well as the total are small enough.

Hence $$\lim_{n\to\infty} E|X_{n} - X| = 0$$.

• Thank you for your response. What does $X \wedge M$ mean?
– user709945
Commented Nov 11, 2019 at 3:21
• Value truncation, i.e. taking the minimum of the two. Commented Nov 11, 2019 at 3:22
• Thanks. That makes sense. Is there any way I can get an estimate on how large $M$/how small $\epsilon$ should be?
– user709945
Commented Nov 11, 2019 at 3:23
• $M$ value depends on how fast the limit goes to zero, i.e. $\lim _{M\to\infty}E(X_n1_{(X_n>M)})=0$. You don't need explicit formula for proving. All you need is the existence. Commented Nov 11, 2019 at 3:25