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}.$

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