Difference between revisions of "ECE600 QE 2002 January" - Rhea

Latest revision as of 07:33, 27 June 2012

7.4 QE 2002 January

1. (20 pts)

Given two coins; the first coin is fair and the second coin has two heads. One coin is picked at random and tossed two times. It shows heads both times. What is the probability that the coin picked is fair?

• $ F=\left\{ \text{fair coin is selected}\right\} $

• $ S=\left\{ \text{the coin that has two heads is selected}\right\} $

• $ H2=\left\{ \text{heads are shown both times}\right\} $

• $ P\left(H2|F\right)=\frac{1}{4},\; P\left(H2|S\right)=1,\; P\left(F\right)=P\left(S\right)=\frac{1}{2}. $

• By using Bayes' theorem,$ P\left(F|H2\right)=\frac{P\left(H2|F\right)P\left(F\right)}{P\left(H2|F\right)P\left(F\right)+P\left(H2|S\right)P\left(S\right)}=\frac{P\left(H2|F\right)}{P\left(H2|F\right)+P\left(H2|S\right)}=\frac{\frac{1}{4}}{\frac{1}{4}+1}=\frac{1}{5}. $

2. (20 pts)

Let $ \mathbf{X}_{t} $ and $ \mathbf{Y}_{t} $ by jointly wide sense stationary continous parameter random processes with $ E\left[\left|\mathbf{X}\left(0\right)-\mathbf{Y}\left(0\right)\right|^{2}\right]=0 $ . Show that $ R_{\mathbf{X}}\left(\tau\right)=R_{\mathbf{Y}}\left(\tau\right)=R_{\mathbf{XY}}\left(\tau\right) $ .

$ E\left[\mathbf{X}\left(t\right)\left(\mathbf{X}^{\star}\left(t+\tau\right)-\mathbf{Y}^{\star}\left(t+\tau\right)\right)\right]=E\left[\mathbf{X}\left(t\right)\mathbf{X}^{\star}\left(t+\tau\right)\right]-E\left[\mathbf{X}\left(t\right)\mathbf{Y}^{\star}\left(t+\tau\right)\right]=R_{\mathbf{X}}\left(\tau\right)-R_{\mathbf{XY}}\left(\tau\right). $

$ E\left[\left|\mathbf{X}\left(t\right)\right|^{2}\right]=E\left[\mathbf{X}\left(t\right)\mathbf{X}^{\star}\left(t\right)\right]=R_{\mathbf{X}}\left(0\right). $

$ E\left[\left|\mathbf{X}\left(t+\tau\right)-\mathbf{Y}\left(t+\tau\right)\right|^{2}\right]=E\left[\left(\mathbf{X}\left(t+\tau\right)-\mathbf{Y}\left(t+\tau\right)\right)\left(\mathbf{X}^{\star}\left(t+\tau\right)-\mathbf{Y}^{\star}\left(t+\tau\right)\right)\right] $$ =R_{\mathbf{X}}\left(0\right)-R_{\mathbf{YX}}\left(0\right)-R_{\mathbf{XY}}\left(0\right)+R_{\mathbf{Y}}\left(0\right) $$ =E\left[\mathbf{X}\left(0\right)\mathbf{X}^{\star}\left(0\right)\right]-E\left[\mathbf{Y}\left(0\right)\mathbf{X}^{\star}\left(0\right)\right]-E\left[\mathbf{X}\left(0\right)\mathbf{Y}^{\star}\left(0\right)\right]+E\left[\mathbf{Y}\left(0\right)\mathbf{Y}^{\star}\left(0\right)\right] $$ =E\left[\mathbf{X}\left(0\right)\mathbf{X}^{\star}\left(0\right)-\mathbf{Y}\left(0\right)\mathbf{X}^{\star}\left(0\right)-\mathbf{X}\left(0\right)\mathbf{Y}^{\star}\left(0\right)+\mathbf{Y}\left(0\right)\mathbf{Y}^{\star}\left(0\right)\right] $$ =E\left[\left(\mathbf{X}\left(0\right)-\mathbf{Y}\left(0\right)\right)\left(\mathbf{X}\left(0\right)-\mathbf{Y}\left(0\right)\right)^{\star}\right]=E\left[\left|\mathbf{X}\left(0\right)-\mathbf{Y}\left(0\right)\right|^{2}\right]. $

By Cauchy-Schwarz inequality, $ \left|R_{\mathbf{X}}\left(\tau\right)-R_{\mathbf{XY}}\left(\tau\right)\right|^{2}\leq R_{\mathbf{X}}\left(0\right)E\left[\left|\mathbf{X}\left(0\right)-\mathbf{Y}\left(0\right)\right|^{2}\right]=0 $ .

$ \therefore\; R_{\mathbf{X}}\left(\tau\right)=R_{\mathbf{XY}}\left(\tau\right). $ Similarly,

$ E\left[\left(\mathbf{X}\left(t\right)-\mathbf{Y}\left(t\right)\right)\mathbf{Y}^{\star}\left(t+\tau\right)\right]^{2}\leq E\left[\left|\mathbf{X}\left(t\right)-\mathbf{Y}\left(t\right)\right|^{2}\right]E\left[\left|\mathbf{Y}\left(t+\tau\right)\right|^{2}\right] $$ \left|R_{\mathbf{XY}}\left(\tau\right)-R_{\mathbf{Y}}\left(\tau\right)\right|^{2}\leq E\left[\left|\mathbf{X}\left(0\right)-\mathbf{Y}\left(0\right)\right|^{2}\right]R_{\mathbf{Y}}\left(0\right)=0. $

$ \therefore\; R_{\mathbf{XY}}\left(\tau\right)=R_{\mathbf{Y}}\left(\tau\right). $

Thus, $ R_{\mathbf{X}}\left(\tau\right)=R_{\mathbf{Y}}\left(\tau\right)=R_{\mathbf{XY}}\left(\tau\right). $

3. (20 pts)

Let $ \mathbf{X}_{t} $ be a zero mean continuous parameter random process. Let $ g(t) $ and $ w\left(t\right) $ be measurable functions defined on the real numbers. Further, let $ w\left(t\right) $ be even. Let the autocorrelation function of $ \mathbf{X}_{t} $ be $ \frac{g\left(t_{1}\right)g\left(t_{2}\right)}{w\left(t_{1}-t_{2}\right)} $ . From the new random process $ \mathbf{Y}_{i}=\frac{\mathbf{X}\left(t\right)}{g\left(t\right)} $ . Is $ \mathbf{Y}_{t} $ w.s.s. ?

$ E\left[\mathbf{Y}\left(t\right)\right]=E\left[\frac{\mathbf{X}\left(t\right)}{g\left(t\right)}\right]=\frac{1}{g\left(x\right)}E\left[\mathbf{X}\left(t\right)\right]=0. $

$ E\left[\mathbf{Y}\left(t_{1}\right)\mathbf{Y}\left(t_{2}\right)\right]=E\left[\frac{\mathbf{X}\left(t_{1}\right)\mathbf{X}^{\star}\left(t_{2}\right)}{g\left(t_{1}\right)g\left(t_{2}\right)}\right]=\frac{1}{g\left(t_{1}\right)g\left(t_{2}\right)}E\left[\mathbf{X}\left(t_{1}\right)\mathbf{X}^{\star}\left(t_{2}\right)\right] $$ =\frac{1}{g\left(t_{1}\right)g\left(t_{2}\right)}\times\frac{g\left(t_{1}\right)g\left(t_{2}\right)}{w\left(t_{1}-t_{2}\right)}=\frac{1}{w\left(t_{1}-t_{2}\right)}, $

which depends on $ t_{1}-t_{2} $ .
$ \therefore\;\mathbf{Y}_{t}\text{ is wide-sense stationary.} $
4. (20 pts)

Let $ \mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n} $ be i.i.d. random variables with absolutely continuous probability distribution function $ F\left(x\right) $ . Let the random variable $ \mathbf{Y}_{j} $ be the $ j $ -th order statistic of the $ \mathbf{X}_{i} $ 's. that is: $ \mathbf{Y}_{j}=j\text{-th smallest of }\left\{ \mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n}\right\} $ .

(a)

What is another name for the first order statistic?

minimum

(b)

What is another name for the n/2 order statistic?

sample median

(c)

Find the probability density function of the first order statistic. (You may assume n is odd.)

$ F_{\mathbf{Y}_{1}}\left(y\right)=P\left(\left\{ \mathbf{Y}_{1}\leq y\right\} \right)=1-P\left(\left\{ \mathbf{Y}_{1}>y\right\} \right) $$ =1-P\left(\left\{ \mathbf{X}_{1}>y\right\} \cap\left\{ \mathbf{X}_{2}>y\right\} \cap\cdots\cap\left\{ \mathbf{X}_{n}>y\right\} \right) $$ =1-\prod_{i=1}^{n}P\left(\mathbf{X}_{i}>y\right)=1-\left(1-F_{\mathbf{X}}\left(y\right)\right)^{n}. $

$ f_{\mathbf{Y}_{1}}\left(y\right)=\frac{d}{dy}F_{\mathbf{Y}_{1}}\left(y\right)=n\left(1-F_{\mathbf{X}}\left(y\right)\right)^{n-1}f_{\mathbf{X}}\left(y\right). $

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) $ .

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}. $

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