Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2008

3

Consider a random process $\mathbf{X}\left(t\right)$ that assumes values $\pm1$ . Suppose that $\mathbf{X}\left(0\right)=\pm1$ with probability $1/2$ , and suppose that $\mathbf{X}\left(t\right)$ then changes polarity with each occurrence of an event in a Poisson process of rate $\lambda$ .

Note:

You might find the equations $\frac{1}{2}\left(e^{x}+e^{-x}\right)=\sum_{j=0}^{\infty}\frac{x^{2j}}{\left(2j\right)!}$ and $\frac{1}{2}\left(e^{x}-e^{-x}\right)=\sum_{j=0}^{\infty}\frac{x^{2j+1}}{\left(2j+1\right)!}$ helpful.

a. (15 points)

Find the probability mass function of $\mathbf{X}\left(t\right)$ .

$P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} |\left\{ \mathbf{X}\left(0\right)=1\right\} \right)=P\left(\left\{ \mathbf{N}\left(t\right)=\text{even}\right\} \right)=\sum_{n=0}^{\infty}e^{-\lambda t}\frac{\left(\lambda t\right)^{2n}}{\left(2n\right)!}=e^{-\lambda t}\sum_{n=0}^{\infty}\frac{\left(\lambda t\right)^{2n}}{\left(2n\right)!} $$=e^{-\lambda t}\cdot\frac{1}{2}\left(e^{\lambda t}+e^{-\lambda t}\right)=\frac{1}{2}\left(1+e^{-2\lambda t}\right). P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} |\left\{ \mathbf{X}\left(0\right)=-1\right\} \right)=P\left(\left\{ \mathbf{N}\left(t\right)=\text{odd}\right\} \right)=\sum_{n=0}^{\infty}e^{-\lambda t}\frac{\left(\lambda t\right)^{2n+1}}{\left(2n+1\right)!}=e^{-\lambda t}\sum_{n=0}^{\infty}\frac{\left(\lambda t\right)^{2n+1}}{\left(2n+1\right)!}$$ =e^{-\lambda t}\cdot\frac{1}{2}\left(e^{\lambda t}-e^{-\lambda t}\right)=\frac{1}{2}\left(1-e^{-2\lambda t}\right).$

$P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} \right)=P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} |\left\{ \mathbf{X}\left(0\right)=1\right\} \right)P\left(\left\{ \mathbf{X}\left(0\right)=1\right\} \right)+P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} |\left\{ \mathbf{X}\left(0\right)=-1\right\} \right)P\left(\left\{ \mathbf{X}\left(0\right)=-1\right\} \right) $$=\frac{1}{2}\left(1+e^{-2\lambda t}\right)\cdot\frac{1}{2}+\frac{1}{2}\left(1-e^{-2\lambda t}\right)\cdot\frac{1}{2}=\frac{1}{2}. Thus, P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} \right)=P\left(\left\{ \mathbf{X}\left(t\right)=-1\right\} \right)=\frac{1}{2} . b. (15 points) Find the autocovariance function of the random process \mathbf{X}\left(t\right) . Note The explanation about the autocovariance function is here. E\left[\mathbf{X}\left(t\right)\right]=1\cdot\frac{1}{2}+\left(-1\right)\cdot\frac{1}{2}=-\frac{1}{2}+\frac{1}{2}=0. C_{\mathbf{XX}}\left(t_{1},t_{2}\right)=E\left[\left(\mathbf{X}\left(t_{1}\right)-\mu_{\mathbf{X}}\left(t_{1}\right)\right)\left(\mathbf{X}\left(t_{2}\right)-\mu_{\mathbf{X}}\left(t_{2}\right)\right)^{*}\right]=R_{\mathbf{XX}}\left(t_{1},t_{2}\right)-\mu_{\mathbf{X}}\left(t_{1}\right)\mu_{\mathbf{X}}\left(t_{2}\right)^{*}$$ =R_{\mathbf{XX}}\left(t_{1},t_{2}\right)=1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}=1\right)\cap\mathbf{X}\left(t_{2}=1\right)\right\} \right)-1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}=-1\right)\cap\mathbf{X}\left(t_{2}=1\right)\right\} \right) $$-1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}=1\right)\cap\mathbf{X}\left(t_{2}=-1\right)\right\} \right)+1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}=-1\right)\cap\mathbf{X}\left(t_{2}=-1\right)\right\} \right)$$ =1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}\right)=\mathbf{X}\left(t_{2}\right)\right\} \right)-1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}\right)\neq\mathbf{X}\left(t_{2}\right)\right\} \right).$

$P\left(\left\{ \mathbf{X}\left(t_{1}\right)=\mathbf{X}\left(t_{2}\right)\right\} \right)=P\left(\left\{ \text{number of events occuring within }\left(t_{1},t_{2}\right)\text{is even}\right\} \right)$

$=\sum_{n=0}^{\infty}e^{-\lambda\left(t_{2}-t_{1}\right)}\frac{\left[\lambda\left(t_{2}-t_{1}\right)\right]^{2n}}{\left(2n\right)!}=\frac{1}{2}e^{-\lambda\left(t_{2}-t_{1}\right)}\left(e^{\lambda\left(t_{2}-t_{1}\right)}+e^{-\lambda\left(t_{2}-t_{1}\right)}\right)$$=\frac{1}{2}\left(1+e^{-2\lambda\left(t_{2}-t_{1}\right)}\right).$

By using the same method above, we can get $P\left(\left\{ \mathbf{X}\left(t_{1}\right)\neq\mathbf{X}\left(t_{2}\right)\right\} \right)=\frac{1}{2}\left(1-e^{-2\lambda\left(t_{2}-t_{1}\right)}\right)$ . Thus,

$C_{\mathbf{XX}}\left(t_{1},t_{2}\right)=\frac{1}{2}\left(1+e^{-2\lambda\left(t_{2}-t_{1}\right)}\right)-\frac{1}{2}\left(1-e^{-2\lambda\left(t_{2}-t_{1}\right)}\right)=e^{-2\lambda\left(t_{2}-t_{1}\right)}.$

## Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.