# Convergence of Random Variables in L1

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Let ##\{X_n\}## be a sequence of integrable, real random variables on a probability space ##(\Omega, \mathscr{F}, \mathbb{P})## that converges in probability to an integrable random variable ##X## on ##\Omega##. Suppose ##\mathbb{E}(\sqrt{1 + X_n^2}) \to \mathbb{E}(\sqrt{1 + X^2})## as ##n\to \infty##. Show that ##X_n\xrightarrow{L^1} X##.

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I have no solution attempt, but thought I would write some random stuff to get the conversation going:
Converges in probability means ##P(|X_n-X|>\epsilon)\to 0## for all ##\epsilon>0##.

Converges in ##L_1## means ##E(|X_n-X|)\to 0##. One example where these aren't the same is: ##X## is identically zero, ##X_n## is ##n## with probability ##1/n## and 0 otherwise. ##P(|X_n-X|>\epsilon)\leq 1/n\to 0##, but ##E(|X_n-X|)=1## for all ##n##.

The expected value condition is interesting, I wonder if the ##1+## piece is necessary.

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Here is a hint: Convergence in probability implies convergence almost surely.

topsquark
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Let ##\{X_{n_k}\}## be a subsequence of ##\{X_n\}##. Since ##X_n\to X## in probability, there is a further subsequence ##\{X_{n_{k_j}}\}## of ##\{X_{n_k}\}## that converges to ##X## almost surely. Now ##|X_{n_{k_j}}| \le \sqrt{1 + X_{n_{k_j}}^2}## and ##\mathbb{E}(\sqrt{1+X_{n_{k_j}}^2}) \to \mathbb{E}(\sqrt{1+X^2}) < \infty##, so by the generalized dominated convergence theorem ##X_{n_{k_j}} \xrightarrow{L^1} X##. Since ##\{X_{n_k}\}## is an arbitrary subsequence of ##\{X_n\}##, the result follows.

topsquark