Part 3

Let the $ \mathbf{X}_{1},\mathbf{X}_{2},\cdots $ be a sequence of random variables that converge in mean square to the random variable $ \mathbf{X} $ . Does the sequence also converge to $ \mathbf{X} $ in probability? (A simple yes or no answer is not acceptable, you must derive the result.)

Share and discuss your solutions below.

Solution 1 (retrived from here)

Let the $ \mathbf{X}_{1},\mathbf{X}_{2},\cdots $ be a sequence of random variables that converge in mean square to the random variable $ \mathbf{X} $ . Does the sequence also converge to $ \mathbf{X} $ in probability? (A simple yes or no answer is not acceptable, you must derive the result.)

We know that $ E\left[\left|\mathbf{X}-\mathbf{X}_{n}\right|^{2}\right]\rightarrow0 $ as $ n\rightarrow\infty $ .

By using Chebyshev Inequality,

$ \lim_{n\rightarrow\infty}P\left(\left\{ \mathbf{X}-\mathbf{X}_{n}\right\} \geq\epsilon\right)\leq\lim_{n\rightarrow\infty}\left(\frac{E\left[\left|\mathbf{X}-\mathbf{X}_{n}\right|^{2}\right]}{\epsilon^{2}}\right)=\frac{\lim_{n\rightarrow\infty}E\left[\left|\mathbf{X}-\mathbf{X}_{n}\right|^{2}\right]}{\epsilon^{2}}=0. $

$ \therefore $ A sequence of random variable that converge in mean square sense to the random variable $ \mathbf{X} $ , also converges in probability to $ \mathbf{X} $ .

Question: Should we prove Chebyshev Inequality to get full credit?

Solution 2

Write it here.

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