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Let <math class="inline">\mathbf{X}</math> be a random variable with absolutely continuous probability distribution function. Show that for any <math class="inline">\alpha>0</math> and any real number <math class="inline">s</math> :<math class="inline">P\left(e^{s\mathbf{X}}\geq\alpha\right)\leq\frac{\phi\left(s\right)}{\alpha}</math> where <math class="inline">\phi\left(s\right)</math> is the moment generating function, <math class="inline">\phi\left(s\right)=E\left[e^{s\mathbf{X}}\right]</math> . Note: <math class="inline">\phi\left(s\right)</math> can be related to the Laplace Transform of <math class="inline">f_{\mathbf{X}}\left(x\right)</math> . | Let <math class="inline">\mathbf{X}</math> be a random variable with absolutely continuous probability distribution function. Show that for any <math class="inline">\alpha>0</math> and any real number <math class="inline">s</math> :<math class="inline">P\left(e^{s\mathbf{X}}\geq\alpha\right)\leq\frac{\phi\left(s\right)}{\alpha}</math> where <math class="inline">\phi\left(s\right)</math> is the moment generating function, <math class="inline">\phi\left(s\right)=E\left[e^{s\mathbf{X}}\right]</math> . Note: <math class="inline">\phi\left(s\right)</math> can be related to the Laplace Transform of <math class="inline">f_{\mathbf{X}}\left(x\right)</math> . | ||
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| + | ==Share and discuss your solutions below.== | ||
| + | ---- | ||
| + | ==Solution 1== | ||
Note | Note | ||
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<math class="inline">\therefore\; P\left(\left\{ e^{s\mathbf{X}}\geq\alpha\right\} \right)\leq\frac{\phi\left(s\right)}{\alpha}.</math> | <math class="inline">\therefore\; P\left(\left\{ e^{s\mathbf{X}}\geq\alpha\right\} \right)\leq\frac{\phi\left(s\right)}{\alpha}.</math> | ||
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Latest revision as of 16:39, 13 March 2015
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
January 2002
5. (20 pts)
Let $ \mathbf{X} $ be a random variable with absolutely continuous probability distribution function. Show that for any $ \alpha>0 $ and any real number $ s $ :$ P\left(e^{s\mathbf{X}}\geq\alpha\right)\leq\frac{\phi\left(s\right)}{\alpha} $ where $ \phi\left(s\right) $ is the moment generating function, $ \phi\left(s\right)=E\left[e^{s\mathbf{X}}\right] $ . Note: $ \phi\left(s\right) $ can be related to the Laplace Transform of $ f_{\mathbf{X}}\left(x\right) $ .
Solution 1
Note
This is similar to the proof of Chebyshev Inequality.
$ g_{1}\left(x\right)=1_{\left(x\right)_{\left\{ r:e^{sx}\geq\alpha\right\} }},\; g_{2}\left(x\right)=\frac{e^{sx}}{\alpha}. $
$ E\left[g_{2}\left(\mathbf{X}\right)-g_{1}\left(\mathbf{X}\right)\right]=E\left[g_{2}\left(\mathbf{X}\right)\right]-E\left[g_{1}\left(\mathbf{X}\right)\right]=\frac{\phi\left(s\right)}{\alpha}-P\left(\left\{ e^{s\mathbf{X}}\geq\alpha\right\} \right)\geq0. $
$ \therefore\; P\left(\left\{ e^{s\mathbf{X}}\geq\alpha\right\} \right)\leq\frac{\phi\left(s\right)}{\alpha}. $
