# 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

$$\rm{P} (N = n) = \frac{1}{n! (e − 1)} \text{for n = 1, 2, 3, . . . .}$$

(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 ?

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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 $$f_Y (y) = \frac{1}{{({\rm e} - 1)\theta }}{\rm e}^{ 1 - y/\theta}, \;\; 0 < y < \theta.$$

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Neat explanation. Understood Clearly. Thanks. –  Sunil Feb 8 '11 at 23:01
@Sunil: Could you explain what in Shai's answer left you more satisfied than the previous answer? Your explanation could include what was not clear in the explanations of the previous answer. Also, what happened to part (2) of your question? –  Did Feb 9 '11 at 7:39
@Sunil: You may skip assurances that my solution is very correct or not wrong, they are not needed. Two more points. First, you do not specify what minor steps I skipped (that is, other than steps already explicit in your question such as formulas for $F_{X(1)}$ and $F_{X(n)}$). So what you mean is still rather mysterious. Second, part (2) of your post seems to ask for results valid for distributions of $X$ more general than uniform, and my answer is phrased so as to be valid (as late as possible) in the general case, but you say you didn't get any answers to your part (2)... Funny. –  Did Feb 9 '11 at 14:25
For the first part the mistake I was making in my trials was I didn't know how to calculate the value of $[F_x(z)]^n$. I didn't know that I should use pdf of uniform at that point and it was clear when Shai pointed it out to be equal to $(z/\theta)^n$. That is the single point where I understood how everything was working. For the second part consider this example, the fn. ($x1^2$ + $x2^2$) having two rv's x1, x2 that are iid's from N(0,1) follow chi-square distribution. So I was wondering if the densities obtained here from the given function should follow any known distribution ? –  Sunil Feb 9 '11 at 15:20
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 $$f_Y(y)=(c\mathrm{e}/\theta)\mathrm{e}^{-y/\theta}, \quad f_Z(z)=(c/\theta)\mathrm{e}^{z/\theta}.$$
@Sunil: Together with the term $P(...)^n$ it yields the series expansion of the exponential. –  Did Feb 8 '11 at 20:43
@Sunil: They are certainly not Gamma distributed (they are bounded from above by $\theta$). –  Shai Covo Feb 8 '11 at 21:50