Example. Sequence of exponentially distributed random variables
Let $ \mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n} $ be a collection of i.i.d. exponentially distributed random variables, each having mean $ \mu $ . Define
$ \mathbf{Y}=\max\left\{ \mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n}\right\} $
and
$ \mathbf{Z}=\min\left\{ \mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n}\right\} $ .
(a) Find the pdf of $ \mathbf{Y} $ .
$ F_{\mathbf{Y}}\left(y\right)=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) $$ =P\left(\left\{ \mathbf{X}_{1}\leq y\right\} \right)P\left(\left\{ \mathbf{X}_{2}\leq y\right\} \right)\cdots P\left(\left\{ \mathbf{X}_{n}\leq y\right\} \right)=\left(F_{\mathbf{X}}\left(y\right)\right)^{n}=\left(1-e^{-y/\mu}\right)^{n}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(y\right) $
$ f_{\mathbf{Y}}(y)=\frac{dF_{\mathbf{Y}}(y)}{dy}=n\left(1-e^{-y/\mu}\right)^{n-1}\cdot\frac{1}{\mu}e^{-y/\mu}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(y\right)=\frac{n}{\mu}e^{-y/\mu}\left(1-e^{-y/\mu}\right)^{n-1}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(y\right) $
(b) Find the pdf of $ \mathbf{Z} $
$ F_{\mathbf{Z}}(z)=P\left(\left\{ \mathbf{Z}\leq z\right\} \right)=1-P\left(\left\{ \mathbf{Z}>z\right\} \right)=1-P\left(\left\{ \min\left\{ \mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n}\right\} >z\right\} \right) $$ =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} $$ =\left[1-\left(1-\left(1-e^{-z/\mu}\right)\right)^{n}\right]\cdot\mathbf{1}_{\left[0,\infty\right)}\left(y\right) $$ =\left[1-\left(e^{-z/\mu}\right)^{n}\right]\cdot\mathbf{1}_{\left[0,\infty\right)}\left(y\right)=\left(1-e^{-nz/\mu}\right)\cdot\mathbf{1}_{\left[0,\infty\right)}\left(y\right) $
$ f_{Z}(z)=\frac{dF_{\mathbf{Z}}(z)}{dz}=\frac{n}{\mu}e^{-nz/\mu}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(y\right) $
(c) In words, give as complete a description of the random variable $ \mathbf{Z} $ as you can.
$ \mathbf{Z} $ is an exponetially distributed random variable with mean $ \frac{\mu}{n} $ .
