# Convergence in probability implies mean squared convergence

Let $$(\Omega, \mathcal{F}, \mathbb{P})$$ be a probability space. Let $$(X_n)_{n \in \mathbb{N}}$$ be a sequence of $$\mathcal{F}$$ measurable random variables. Let $$X$$ be another $$\mathcal{F}$$ measurable random variable. I have $$X_n \rightarrow X$$ in probability. Additionally, $$\mathbb{P}(|X_n|, where $$L$$ is a constant independent of $$n$$. I have to show that $$X_n \rightarrow X$$ in mean squared sense, i.e. as $$n \rightarrow \infty$$, $$\mathbb{E}(X_n - X)^2 \rightarrow 0$$. How do I go about this? Thanks.

• @OliverDiaz I'm sorry I guess the sequence in the answer does converge to zero in mean squared sense Jul 1, 2020 at 17:28

Convergence in probability: For any $$\delta>0$$, $$\lim_{n\to\infty}\mathbb{P}(|X_n-X|>\delta)=0$$.
Also, since $$\mathbb{P}(|X_n|, we have that $$|X_n| almost surely for all $$X_n$$. Since convergence in probability implies almost-everywhere convergence of a subsequence, we also have that $$\mathbb{P}(|X|, i.e. $$|X| almost surely. Now let $$\delta>0$$. We have $$\mathbb{E}[X_n-X]^2=\int|X_n-X|^2=\int_{\{|X_n-X|>\delta\}}|X_n-X|^2+\int_{\{|X_n-X|<\delta\}}|X_n-X|^2\leq$$ $$\leq\int_{\{|X_n-X|>\delta\}}|X_n-X|^2+\delta^2\leq\mathbb{P}(|X_n-X|>\delta)\cdot (4L^2)+\delta^2\to\delta^2$$
Since $$\delta>0$$ was arbitrary and we have that $$\limsup_{n\to\infty}\mathbb{E}|X_n-X|^2\leq\delta^2$$, we conclude that $$\mathbb{E}|X_n-X|^2\to0$$.