Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
January 2006
2 (33 points)
Suppose that $ \mathbf{X} $ and $ \mathbf{N} $ are two jointly distributed random variables, with $ \mathbf{X} $ being a continuous random variable that is uniformly distributed on the interval $ \left(0,1\right) $ and $ \mathbf{N} $ being a discrete random variable taking on values $ 0,1,2,\cdots $ and having conditional probability mass function $ p_{\mathbf{N}}\left(n|\left\{ \mathbf{X}=x\right\} \right)=x^{n}\left(1-x\right),\quad n=0,1,2,\cdots $ .
(a)
Find the probability that \mathbf{N}=n .
$ f_{\mathbf{X}}\left(x\right)=\left\{ \begin{array}{lll} 1 & & ,0\leq x\leq1\\ 0 & & ,\text{otherwise.} \end{array}\right. $
$ P\left(\left\{ \mathbf{N}=n\right\} \right)=\int_{-\infty}^{\infty}p_{\mathbf{N}}\left(n|\left\{ \mathbf{X}=x\right\} \right)f_{\mathbf{X}}\left(x\right)dx=\int_{0}^{1}x^{n}\left(1-x\right)dx $
$ =\frac{1}{n+1}x^{n+1}-\frac{1}{n+2}x^{n+2}\Bigl|_{0}^{1}=\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{\left(n+1\right)\left(n+2\right)}. $
(b)
Find the conditional density of $ \mathbf{X} $ given $ \left\{ \mathbf{N}=n\right\} $ .
By using Bayes' theorem,
$ f_{\mathbf{X}}\left(x|\left\{ \mathbf{N}=n\right\} \right)=\frac{p_{\mathbf{N}}\left(n|\left\{ \mathbf{X}=x\right\} \right)f_{\mathbf{X}}\left(x\right)}{p_{\mathbf{N}}\left(n\right)}=\left\{ \begin{array}{lll} \left(n+1\right)\left(n+2\right)x^{n}\left(1-x\right) & & ,0\leq x\leq1\\ 0 & & ,\text{otherwise.} \end{array}\right. $
(c)
Find the minimum mean-square error estimator of $ \mathbf{X} $ given $ \left\{ \mathbf{N}=n\right\} $ .
$ MMSE=E\left[\mathbf{X}|\left\{ \mathbf{N}=n\right\} \right]=\int_{-\infty}^{\infty}x\cdot f_{\mathbf{X}}\left(x|\left\{ \mathbf{N}=n\right\} \right)dx=\int_{0}^{1}\left(n+1\right)\left(n+2\right)x^{n+1}\left(1-x\right)dx $$ =\left(n+1\right)\left(n+2\right)\left(\frac{1}{n+2}x^{n+2}-\frac{1}{n+3}x^{n+3}\right)\biggl|_{0}^{1}=\left(n+1\right)\left(n+2\right)\left(\frac{1}{n+2}-\frac{1}{n+3}\right) $$ =\frac{\left(n+1\right)\left(n+2\right)}{\left(n+2\right)\left(n+3\right)}=\frac{n+1}{n+3}. $
