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I read in some paper that convergence in probability implies the convergence in quadratic mean if all moments of higher order exists, but I don't know how to prove it. Can someone please provide the proof?

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Convergence in probability does not imply convergence in quadratic mean, did you accidentally write the reverse statement? Some good notes on convergence can be found here. The relevant parts to your question are reproduced below.

$X_{n}$ converges to $X$ in probability, $X_n \xrightarrow{P} X$, if for every $\epsilon$,

$$ P(|X_n - X| > \epsilon) \rightarrow 0 $$ as $n \rightarrow \infty$

$X_{n}$ converges to $X$ in quadratic mean, $X_n \xrightarrow{qm} X$, if

$$ E[(X_n - X)^2] \rightarrow 0 $$ as $n \rightarrow \infty$

As a counterexample to show convergence in probability does not imply convergence in quadratic mean, take $U \sim Unif(0, 1)$ and let $X_n = \sqrt{n} I_{[0, \frac{1}{n}]}(U)$. Then

$$ P(|X_n| > \epsilon) = P(|\sqrt{n} I_{[0, \frac{1}{n}]}(U)| > \epsilon) = P(0 \leq U \leq 1/n) = 1/n \rightarrow 0 $$ as $n \rightarrow \infty$ and therefore $X_n \xrightarrow{P} 0$

but

$$ E[X_{n}^2] = n\int_{0}^{\frac{1}{n}}dx = 1 $$

and therefore $X_n \not\xrightarrow{qm} 0$


With the additional constraint that $P(|X_n| < M) = 1$ for some constant $M > 0$, we do have that convergence in probability implies convergence in quadratic mean or L2. This can be seen as a restricted version of the following theorem from chapter 17 of Convergence of Random Variables by Jean Jacod and Philip Protter. The relevant chapter is available here.

Theorem 17.4 Suppose $X_n \xrightarrow{P} X$ and also that $|X_{n}| \leq Y$, all $n$, and $Y \in L^p$. Then $|X|$ is in $L^{p}$ and $X_{n} \xrightarrow{L^p} X$.

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    $\begingroup$ I'm sorry if my question was not precise enough. I'm familiar with the fact that convergence in moments implies convergence in probability but the reverse is not generally true. What I read in paper is that, under assumption of bounded variables, i.e P(|X_n|<M) = 1 for some constant M > 0, convergence in probability does imply convergence in quadratic mean, but I don't know how to prove it. $\endgroup$
    – Mare
    May 21 '17 at 11:36
  • $\begingroup$ @mare do you want to correct the question then $\endgroup$
    – seanv507
    May 21 '17 at 15:04
  • $\begingroup$ @seanv507 he's asking for the proof $\endgroup$
    – Taylor
    May 21 '17 at 15:15

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