## Question from ECE QE January 2001

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.)

# 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?

Write it here.

## Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva