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 ?

  • $\begingroup$ 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. $\endgroup$
    – Math1000
    Nov 11 '19 at 3:14
  • $\begingroup$ Kudos to you @Math1000, I found a really simple & elegant proof here: theanalysisofdata.com/probability/8_4.html. $\endgroup$ Nov 11 '19 at 3:42
  • $\begingroup$ $|c| = c + 2\max(-c,0)$ is such a strange identity; I doubt I would have come up with that on my own. $\endgroup$
    – Math1000
    Nov 11 '19 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 $X<M$ 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 $\lim_{n\to\infty} E|X_{n} - X| = 0$.

  • $\begingroup$ Thank you for your response. What does $X \wedge M$ mean? $\endgroup$
    – user709945
    Nov 11 '19 at 3:21
  • $\begingroup$ Value truncation, i.e. taking the minimum of the two. $\endgroup$ Nov 11 '19 at 3:22
  • $\begingroup$ Thanks. That makes sense. Is there any way I can get an estimate on how large $M$/how small $\epsilon$ should be? $\endgroup$
    – user709945
    Nov 11 '19 at 3:23
  • $\begingroup$ $M$ 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. $\endgroup$ Nov 11 '19 at 3:25

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