(One intermediate revision by the same user not shown) | |||

Line 4: | Line 4: | ||

[[Category:problem solving]] | [[Category:problem solving]] | ||

[[Category:random variables]] | [[Category:random variables]] | ||

+ | [[Category:probability]] | ||

− | = | + | <center> |

+ | <font size= 4> | ||

+ | [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]] | ||

+ | </font size> | ||

+ | |||

+ | <font size= 4> | ||

+ | Communication, Networking, Signal and Image Processing (CS) | ||

+ | |||

+ | Question 1: Probability and Random Processes | ||

+ | </font size> | ||

+ | |||

+ | January 2001 | ||

+ | </center> | ||

+ | ---- | ||

+ | ---- | ||

+ | =Part 3= | ||

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

---- | ---- |

## Latest revision as of 10:36, 13 September 2013

Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

January 2001

## Contents

# 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.)

# 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.