(New page: Category:ECE Category:QE Category:CNSIP Category:problem solving Category:random variables ==Question from ECE QE January 2001==...) |
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[[Category:random variables]] | [[Category:random variables]] | ||
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| + | [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]] | ||
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| + | Communication, Networking, Signal and Image Processing (CS) | ||
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| + | Question 1: Probability and Random Processes | ||
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| + | 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.) | ||
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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== | ||
Latest revision as of 09:36, 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.
