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Let X and $X_1, X_2, ...$ be random variables on $(\Omega, F, P)$Show that if $X_n \rightarrow X$ in $L^2$ as $n$ $ \rightarrow \infty$ then $X_n \rightarrow X$ in $L^1$ as well.

I'm trying to prove this without using the Cauchy-Schwarz inequality, this is what I've got so far.

Convergence in $L^2$ implies that $\mathbb{E}[|X_n - X|^2] \rightarrow 0$ as $n \rightarrow \infty$.

I know that $|\mathbb{E}[X_n] - \mathbb{E}[X]]| = |\mathbb{E}[(X_n - X)]| \leq \mathbb{E}[|X_n - X|] \leq \mathbb{E}[|X_n - X|^2]$

Is it enough to say that as the $\lim_{n\to\infty} \mathbb{E}[|X_n - X|^2] = 0$then this would imply that $\lim_{n\to\infty} \mathbb{E}[|X_n - X|] = 0$, and then convergence in mean square implies convergence in mean?

Also, how could I use the fact that $ 0 \leq Var(X) = \mathbb{E}[X^2]-\mathbb{E}[X]^2 = \mathbb{E}[(X - \mathbb{E}[X])^2]$ to prove it?

Thanks a lot!

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1 Answer

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We know that $\mathbb{E}[|X_n - X|^2] \rightarrow 0$ as $n \rightarrow \infty$ and we want to show that $\mathbb{E}[|X_n - X|] \rightarrow 0$.

For any random variable X we know that:$$ 0 \le Var(X) = \mathbb{E}[X^2] - E[X]^2$$

By this, we can say that $$\mathbb{E}[|X_n-X|] \le \sqrt{\mathbb{E}[|X_n-X|^2]}$$

Since $\mathbb{E}[|X_n-X|] \geq 0$, and $\mathbb{E}[|X_n-X|^2]$ goes to zero as $n \rightarrow \infty$. This is enough to prove $L^1$ convergence.

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