Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2003

3. (20% of Total)

Consider three independent random variables, $\mathbf{X}$ , $\mathbf{Y}$ , and $\mathbf{Z}$ . Assume that each one is uniformly distributed over the interval $\left(0,1\right)$ . Call “Bin #1” the interval $\left(0,\mathbf{X}\right)$ , and “Bin #2” the interval $\left(\mathbf{X},1\right)$ .

a. (10%)

Find the probability that $\mathbf{Y}$ falls into Bin #1 (that is, $\mathbf{Y}<\mathbf{X}$ ). Show your work.

Solution 1 - Wrong

$P\left(\left\{ \mathbf{Y}<\mathbf{X}\right\} \right)=\int_{0}^{1}P\left(\left\{ \mathbf{Y}<k\right\} \cap\left\{ \mathbf{X}\geq k\right\} \right)dk=\int_{0}^{1}P\left(\left\{ \mathbf{Y}<k\right\} \right)\cdot P\left(\left\{ \mathbf{X}\geq k\right\} \right)dk$$=\int_{0}^{1}k\left(1-k\right)dk=\int_{0}^{1}k-k^{2}dk=\frac{1}{2}k^{2}-\frac{1}{3}k^{3}\biggl|_{0}^{1}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}.$

Solution 2

 $\int_{0}^{1}\int_{0}^{1}cdxdy=1$ $\int_{0}^{1}cdy=1$ $c=1$

$P\left(\left\{ \mathbf{Y}<\mathbf{X}\right\} \right)=\int_{0}^{1}\int_{0}^{x}1dydx=\int_{0}^{1}xdx=\frac{x^{2}}{2}\biggl|_{0}^{1}=\frac{1}{2}.$

b. (10%)

Find the probability that both $\mathbf{Y}$ and $\mathbf{Z}$ fall into Bin #1. Show your work.

$P\left(\left\{ \mathbf{Y}<\mathbf{X}\right\} \cap\left\{ \mathbf{Z}<\mathbf{X}\right\} \right)=\int_{0}^{1}\int_{0}^{x}\int_{0}^{x}1dzdydx=\int_{0}^{1}\int_{0}^{x}xdydx=\int_{0}^{1}x^{2}dx=\frac{1}{3}.$

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009