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==Question from [[ECE_PhD_QE_CNSIP_Jan_2001_Problem1|ECE QE January 2001]]== | ==Question from [[ECE_PhD_QE_CNSIP_Jan_2001_Problem1|ECE QE January 2001]]== | ||

− | + | 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.) | |

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==Share and discuss your solutions below.== | ==Share and discuss your solutions below.== | ||

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=Solution 1 (retrived from [[ECE600_QE_2000_August|here]])= | =Solution 1 (retrived from [[ECE600_QE_2000_August|here]])= | ||

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+ | 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.) | ||

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+ | We know that <math class="inline">E\left[\left|\mathbf{X}-\mathbf{X}_{n}\right|^{2}\right]\rightarrow0</math> as <math class="inline">n\rightarrow\infty</math> . | ||

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+ | By using [[ECE 600 Chebyshev Inequality|Chebyshev Inequality]], | ||

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+ | <math class="inline">\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.</math> | ||

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+ | <math class="inline">\therefore</math> A sequence of random variable that converge in mean square sense to the random variable <math class="inline">\mathbf{X}</math> , also converges in probability to <math class="inline">\mathbf{X}</math> . | ||

+ | ::<span style="color:green">Question: Should we prove Chebyshev Inequality to get full credit?</span> | ||

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==Solution 2== | ==Solution 2== |

## Revision as of 05:46, 17 July 2012

## Contents

## 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?

## Solution 2

Write it here.