Order statistic I've two questions
(1) How do I determine the distribution of the first order and the highest order statistic for sample of random size N is taken from the continuous uniform(0, $\theta$) and  
\begin{equation} \rm{P} (N = n) = \frac{1}{n! (e − 1)} \text{for n = 1, 2, 3, . . . .} \end{equation}
(2) In general without any distribution of any random variables given, the distribution of highest order statistic is $F_{X(n)}(a) = [F_{X}(a)]^n$ and the distribution of the lowest order statistic is $F_{X(1)}(a) = 1 - [1 - F_X(a)]^n$. I understand this derivation. But what does these values say about the distribution ?
 A: The idea is to consider simultaneously the minimum $Y$ and the maximum $Z$ of the sample. For $y\le z$, the event $[y\le Y,Z\le z]$ is equivalent to the whole sample being between $y$ and $z$, hence, for a sample of size $n$, $P(y\le Y,Z\le z)$ would be $P(y\le X\le z)^n$. Here one considers a sample of random size $N$,
hence
$$
P(y\le Y,Z\le z)=\sum_{n\ge1}P(N=n)P(y\le X\le z)^n=c(\mathrm{e}^{P(y\le X\le z)}-1),
$$
with
$$
c=1/(\mathrm{e}-1).
$$
To get the joint distribution of $(Y,Z)$, one should differentiate $P(y\le Y,Z\le z)$ twice, with respect to $y$ and $z$, yielding
$$
P(Y\in\mathrm{d}y,Z\in\mathrm{d}z)=c\mathrm{e}^{P(y\le X\le z)}P(X\in\mathrm{d}y)P(X\in\mathrm{d}z).
$$
To get the distribution of $Y$ alone is even simpler, one differentiates $P(y\le Y,Z\le z)$ once with respect to $y$ and one lets $z\to+\infty$, hence
$$
P(Y\in\mathrm{d}y)=c\mathrm{e}^{P(X\ge y)}P(X\in\mathrm{d}y).
$$
Likewise, to get the distribution of $Z$ alone, one differentiates $P(y\le Y,Z\le z)$ once with respect to $z$ and one lets $y\to-\infty$, hence
$$
P(Z\in\mathrm{d}z)=c\mathrm{e}^{P(X\le z)}P(X\in\mathrm{d}z).
$$
In the special case where the sample is uniform on $(0,\theta)$, for $0\le y\le z\le\theta$, the density of the distribution of $(Y,Z)$ at $(y,z)$ is
$$
f_{Y,Z}(y,z)=(c/\theta^{2})\mathrm{e}^{(z-y)/\theta}.
$$
Finally, the densities of the distributions of $Y$ and $Z$ on $(0,\theta)$ are
$$
f_Y(y)=(c\mathrm{e}/\theta)\mathrm{e}^{-y/\theta},
\quad
f_Z(z)=(c/\theta)\mathrm{e}^{z/\theta}.
$$
A: Didier gave much more than the OP requested. My solution below addresses the original question (and is thus somewhat simpler, in my opinion).
If $Z$ denotes the maximum, then
$$
{\rm P}(Z \le z) = \sum\limits_{n = 1}^\infty  {{\rm P}(Z \le z|N = n){\rm P}(N = n)}  = \sum\limits_{n = 1}^\infty  {F_{X_{(n)} } (z)\frac{1}{{n!({\rm e} - 1)}}}. 
$$
Since $F_{X_{(n)} } (z) = [F_X (z)]^n = (\frac{z}{\theta })^n$, $0 \leq z \leq \theta$, we get 
$$
{\rm P}(Z \le z) = \frac{1}{{{\rm e} - 1}}\sum\limits_{n = 1}^\infty  {\frac{{( z/\theta)^n }}{{n!}}}  = \frac{1}{{{\rm e} - 1}}({\rm e}^{z/\theta}  - 1).
$$
Hence $Z$ has density $f_Z$ given by
$$
f_Z (z) = \frac{1}{{({\rm e} - 1)\theta }}{\rm e}^{ z/\theta}, \;\; 0 < z < \theta. 
$$
Similarly, if $Y$ denotes the minimum, then
$$
{\rm P}(Y \le y) = \sum\limits_{n = 1}^\infty  {{\rm P}(Y \le y|N = n){\rm P}(N = n)}  = \sum\limits_{n = 1}^\infty  {F_{X_{(1)} } (y)\frac{1}{{n!({\rm e} - 1)}}}. 
$$
Since $F_{X_{(1)} } (y) = 1 - [1-F_X (y)]^n = 1 - [1 - \frac{y}{\theta }]^n$, $ 0 \leq y \leq \theta$, we get
$$
{\rm P}(Y \le y) = \sum\limits_{n = 1}^\infty  {\frac{{1 - [1 - y/\theta ]^n }}{{n!({\rm e}-1)}}}  = 1 -  \sum\limits_{n = 1}^\infty  {\frac{{(1 - y/\theta )^n }}{{n!({\rm e}-1)}}}  = 1 - \frac{{{\rm e}^{1 - y/\theta }  - 1}}{{{\rm e} - 1}}.
$$
Hence $Y$ has density $f_Y$ given by
$$
f_Y (y) = \frac{1}{{({\rm e} - 1)\theta }}{\rm e}^{ 1 - y/\theta}, \;\; 0 < y < \theta. 
$$
