Difference between revisions of "ECE PhD QE CNSIP 2003 Problem1.3" - Rhea

Latest revision as of 23:54, 9 March 2015

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

$ 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}. $