Difference between revisions of "ECE PhD QE CNSIP 2004 Problem1.3" - Rhea

Latest revision as of 00:37, 10 March 2015

3. (25 pts.)

Let $ \mathbf{X}_{1},\cdots,\mathbf{X}_{n} $ be independent identically distributed (i.i.d. ) random variables uniformaly distributed over the interval $ \left[0,1\right] $ .

(a)

Find the probability density function of $ \mathbf{Y}=\max\left\{ \mathbf{X}_{1},\cdots,\mathbf{X}_{n}\right\} $ .

ref.

This problem is almost identical to the example.

Solution

$ F_{\mathbf{Y}}(y)=P\left(\left\{ \mathbf{Y}\leq y\right\} \right)=P\left(\left\{ \max\left\{ \mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n}\right\} \leq y\right\} \right)=P\left(\left\{ \mathbf{X}_{1}\leq y\right\} \cap\left\{ \mathbf{X}_{2}\leq y\right\} \cap\cdots\cap\left\{ \mathbf{X}_{n}\leq y\right\} \right) $$ =P\left(\left\{ \mathbf{X}_{1}\leq y\right\} \right)P\left(\left\{ \mathbf{X}_{2}\leq y\right\} \right)\cdots P\left(\left\{ \mathbf{X}_{n}\leq y\right\} \right)=\left(F_{\mathbf{X}}\left(y\right)\right)^{n} $

where $ f_{\mathbf{X}}(x)=\mathbf{1}_{\left[0,1\right]}(x) $ and $ F_{\mathbf{X}}\left(x\right)=\left\{ \begin{array}{ll} 0 & \quad,\; x<0\\ x & \quad,\;0\leq x<1\\ 1 & \quad,\; x\geq1. \end{array}\right. $

$ f_{\mathbf{Y}}\left(y\right)=\frac{dF_{\mathbf{Y}}\left(y\right)}{dy}=n\left[F_{\mathbf{X}}\left(y\right)\right]^{n-1}\cdot f_{\mathbf{X}}\left(y\right)=n\cdot y^{n-1}\cdot\mathbf{1}_{\left[0,1\right]}(y). $

(b)

Find the probability density function of $ \mathbf{Z}=\min\left\{ \mathbf{X}_{1},\cdots,\mathbf{X}_{n}\right\} $ .

Solution

$ F_{\mathbf{Z}}(z)=P\left(\left\{ \mathbf{Z}\leq z\right\} \right)=1-P\left(\left\{ \mathbf{Z}>z\right\} \right)=1-P\left(\left\{ \min\left\{ \mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n}\right\} >z\right\} \right) $$ =1-P\left(\left\{ \mathbf{X}_{1}>z\right\} \cap\left\{ \mathbf{X}_{2}>z\right\} \cap\cdots\cap\left\{ \mathbf{X}_{n}>z\right\} \right)=1-\left(1-F_{\mathbf{X}}(z)\right)^{n}. $

$ f_{\mathbf{Z}}(z)=\frac{dF_{\mathbf{Z}}(z)}{dz}=n\left(1-F_{\mathbf{X}}(z)\right)^{n-1}\cdot f_{\mathbf{X}}(z)=n\left(1-z\right)^{n-1}\cdot\mathbf{1}_{\left[0,1\right]}\left(z\right). $