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Suppose $X_1,X_2,...X_n$ are i. i. d. random variables with p. d. f. $$f(x)=xe^{-x}I_{(0,\infty)}\!(x)$$ and let $Y_1,...,Y_n$ be the order statistics for these variables.

a) Find the conditional p. d. f. of $Y_1$ given $Y_n=y_n$.

When working with order statistics, one should start with the distribution function: $$F_{Y_1|Y_n}(y_1\,|\,y_n) = \mathbb P(Y_1\le y_1\,|\,Y_n=y_n) = 1 - \mathbb P(Y_1>y_1\,|\,Y_n=y_n)\quad.$$ Because $Y_1 = \min_iX_i$, I have $$\begin{align} \mathbb P(Y_1>y_1\,|\,Y_n=y_n) &= \mathbb P(Y_1,...,Y_n>y_1\,|\,Y_n=y_n) =\\ &= \mathbb P(X_1,...,X_n>y_1\,|\,Y_n=y_n) \quad. \end{align}$$ By the independence of the $X_i$'s, setting $Y_n\triangleq X_j$, $$\begin{align} \mathbb P(X_1,...,X_n>y_1\,|\,Y_n=y_n) &= \prod_{i=1}^n\mathbb P(X_i>y_1\,|\,Y_n=y_n) =\\ &= \prod_{i=1}_{i\neq j}^n[1-\mathbb P(X_i\le y_1\,|\,Y_n=y_n)]\cdot I_{(-\infty,\,y_n)}\!(y_1)\\ &= [1-F_{|Y_n}(y_1\,|\,y_n)]^{n-1}I_{(-\infty,\,y_n)}\!(y_1)\quad. \end{align}$$

The conditional p. d. f. for each $X_i$ is $$f_{|Y_n}(x\,|\,y) = xe^{-x}I_{(0,y)}\!(x)\left[\int_0^y xe^{-x}\;dx\right]^{-1} = \frac{xe^{-x}}{1 - (y+1)e^{-y}}I_{(0,y)}\!(x)\quad,$$ thus the c. d. f. is $$\begin{align} F_{|Y_n}(x\,|\,y) &= \int_{-\infty}^xf_{|Y_n}(t\,|\,y)\;dt =\\ &= \frac1{1 - (y+1)e^{-y}}\left(\int_0^xte^{-t}\;dt\right)I_{(0,y)}\!(x) + I_{[y,\infty)}\!(x) =\\ &= \frac{1 - (x+1)e^{-x}}{1 - (y+1)e^{-y}}I_{(0,y)}\!(x) + I_{[y,\infty)}\!(x)\quad, \end{align}$$ and $$\begin{align} [1-F_{|Y_n}(y_1\,|\,y_n)]^{n-1} &= \left\{1 - \left[\frac{1 - (y_1+1)e^{-y_1}}{1 - (y_n+1)e^{-y_n}}I_{(0,y_n)}\!(y_1) + I_{[y_n,\infty)}\!(y_1)\right]\right\}^{n-1} =\\ &= I_{(-\infty,\,0]}(y_1) + \left[\frac{(y_1+1)e^{-y_1} - (y_n+1)e^{-y_n}}{1 - (y_n+1)e^{-y_n}}\right]^{n-1} \!\!\!I_{(0,y_n)}\!(y_1)\quad. \end{align}$$

Finally, using that $y_n$ has to be positive when it is given, $$\begin{align} F_{Y_1|Y_n}(y_1\,|\,y_n) &= 1 - \mathbb P(X_1,...,X_n>y_1\,|\,Y_n=y_n) =\\ &= 1 - [1-F_{|Y_n}(y_1\,|\,y_n)]^{n-1}I_{(-\infty,\,y_n)}\!(y_1) =\\ &= 1 - \left\{I_{(-\infty,\,0]}(y_1) + \left[\frac{(y_1+1)e^{-y_1} - (y_n+1)e^{-y_n}}{1 - (y_n+1)e^{-y_n}}\right]^{n-1} \!\!\!I_{(0,\,y_n)}\!(y_1)\right\} =\\ &= \left\{1 - \left[\frac{(y_1+1)e^{-y_1} - (y_n+1)e^{-y_n}}{1 - (y_n+1)e^{-y_n}}\right]^{n-1}\right\}I_{(0,\,y_n)}\!(y_1) + I_{[y_n,\infty)}\!(y_1)\quad, \end{align}$$ which implies that its derivative is $$\begin{align} f_{Y_1|Y_n}(y_1\,|\,y_n) &= -\frac\partial{\partial y_1}\! \left\{\left[\frac{(y_1+1)e^{-y_1} - (y_n+1)e^{-y_n}} {1 - (y_n+1)e^{-y_n}}\right]^{n-1}\right\}I_{(0,\,y_n)}\!(y_1) =\\ &= -\frac{(n-1)[(y_1+1)e^{-y_1} - (y_n+1)e^{-y_n}]^{n-2}}{1 - (y_n+1)e^{-y_n}}\frac\partial{\partial y_1}\![\ldots]\,I_{(0,\,y_n)}\!(y_1) =\\ &= \frac{n-1}{1 - (y_n+1)e^{-y_n}}[(y_1+1)e^{-y_1} - (y_n+1)e^{-y_n}]^{n-2}y_1e^{-y_1} I_{(0,\,y_n)}\!(y_1) \quad. \end{align}$$

b) Find the p. d. f. of the amplitude $W\triangleq Y_n\,–Y_1$.

A variable transformation shows that $$\begin{align} f_{W,Y_1}(w,y_1) &= f_{Y_1,Y_n}(y_1,y_n)\left|\frac{\partial(w,y_1)}{\partial(y_1,y_n)}\right|^{-1} =\\ &= \frac{f_{Y_1,Y_n}(y_1,y_n)}{|-1\cdot0 - 1\cdot1|} =\\ &= f_{Y_1,Y_n}(y_1,y_1+w) =\\ &= f_{Y_1|Y_n}(y_1\,|\,y_1+w)f_{Y_n}(y_1+w)\quad. \end{align}$$

To obtain the marginal density of $Y_n$, I note that its c. d. f. is $$\begin{align} F_{Y_n}(y) &= \mathbb P(Y_n\le y) =\\ &= \mathbb P(Y_1,...,Y_n\le y) =\\ &= \mathbb P(X_1,...,X_n\le y) =\\ &= \prod_{i=1}^n\mathbb P(X_i\le y) =\\ &= F(y)^n =\\ &= \left(\int_0^y te^{-t}\;dt\cdot I_{(0,\infty)}(y)\right)^n =\\ &= [1 - (y+1)e^{-y}]^n\,I_{(0,\infty)}\!(y)\quad, \end{align}$$ therefore, $$f_{Y_n}(y) = n[1 - (y+1)e^{-y}]^{n-1}ye^{-y}I_{(0,\infty)}\!(y)\quad,$$ and $$\begin{align} f_{W,Y_1}(w,y) &= (n-1) \frac{[(y+1)e^{-y} - (y+w+1)e^{-(y+w)}]^{n-2}ye^{-y}}{1 - (y+w+1)e^{-(y+w)}} I_{(0,\,y+w)}\!(y) \cdot\\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\cdot\;n[1 - (y+w+1)e^{-(y+w)}]^{n-1}(y+w)e^{-(y+w)} I_{(0,\infty)}\!(y+w) =\\ &= n(n-1)\frac{[g(y)-g(y+w)]^{n-2}[1 - g(y+w)]^{n-1}g(y+w)ye^{-y}}{1 - g(y+w)}\cdot\\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\cdot\;I_{(0,\infty)}\!(w)I_{(-w,\infty)}\!(y)\quad, \end{align}$$ where $g(t)\triangleq (t+1)e^{-t}$. Now I'd have to integrate this function with respect to the second variable over $\mathbb R$, which seems insane. There is another way to do it, right?

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  • $\begingroup$ What is your $I_{(0,\infty)}$ function? $\endgroup$ – Hong Ooi Jan 22 '14 at 16:25
  • $\begingroup$ @Hong Because the title calls these Gamma variables, and Gamma distributions have non-negative support, we deduce that $I_{(0,\infty)}$ must be an indicator function: $I_{(0,\infty)}(x)$ is equal to $1$ when $x\in (0,\infty)$ and is equal to $0$ otherwise. $\endgroup$ – whuber Jan 22 '14 at 20:15
  • $\begingroup$ @whuber Exactly. $\endgroup$ – Luke Jan 23 '14 at 1:29
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Given: $X$ has pdf $f(x)$:


(source: tri.org.au)

Then, the joint pdf of the 1st and $n$-th order statistics, in a sample of size $n$, is say $g(x_1,x_n)$:


(source: tri.org.au)

with domain of support:


(source: tri.org.au)

... where I am using the OrderStat function from the mathStatica package for Mathematica to automate the nitty-gritties.


Part (a): the pdf of $X_1|x_n$

Then, the pdf of $X_1|x_n$ can be obtained simply with:


(source: tri.org.au)

[ By the way, the easy way to see that your solution must be incorrect is to note that the conditional pdf of the first order statistic [GIVEN the sample maximum] should depend on the sample maximum (but your $y_n$ term has effectively disappeared from your solution).]


Part (b): the pdf of the sample range $W = X_n - X_1$

The cdf of the sample range $W = X_n - X_1$ is $P(X_n - X_1 < w)$, calculated wrt to the joint pdf $g(x_1,x_n)$:


(source: tri.org.au)

where mathStatica's Prob function calculates the required probability, and where ExpIntegralE is the exponential integral function described here: http://reference.wolfram.com/mathematica/ref/ExpIntegralE.html

Finally, the pdf of $W$ is just the derivative of the cdf wrt $w$:


(source: tri.org.au)

Here is a plot of the solution: i.e. the pdf of the sample range, as the sample size $n$ varies:


(source: tri.org.au)


Notes:

  1. I should perhaps add that I am one of the authors of the mathStatica software used above.

  2. When using computer algebra systems or any sort of automated tool, it is always a good idea to check one's work using Monte Carlo methods. Here is a quick check that compares a Monte Carlo simulation of the pdf of the sample range when $n = 5$ (blue squiggly curve) to the theoretical solution derived above (red dashed) ...


(source: tri.org.au)

... Looks good.

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    $\begingroup$ +1 This is a wonderful answer in how it combines deft use of appropriate software, a clear explanation of the steps followed, and references to details of the question itself, together with a final reality check. $\endgroup$ – whuber Jan 22 '14 at 20:13
  • $\begingroup$ Very nice computational solution, but this question was on a test, so I'll keep looking for a manual solution. $\endgroup$ – Luke Jan 27 '14 at 4:21
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I think you are going wrong very early, when you state that $$\begin{align} \mathbb P(Y_1>y_1\,|\,Y_n=y_n) &= \mathbb P(Y_1,...,Y_n>y_1\,|\,Y_n=y_n) =\\ &= \mathbb P(Y_1,...,Y_{n-1}>y_1)I_{(y_1,\infty)}\!(y_n). \end{align}$$ You need to keep the information that all the variables with the exception of $Y_n$ are less than or equal to $y_n$. So (wlog assume $X_n = Y_n$) $$\begin{align} \mathbb P(Y_1>y_1\,|\,Y_n=y_n) &= \mathbb P(Y_1,...,Y_n>y_1\,|\,Y_n=y_n) =\\ &= \mathbb P(y_1 <X_1,...,X_{n-1}\leq y_n)I_{(y_1,\infty)}\!(y_n) = \\ & = [F(y_n) - F(y_1)]^{n-1} I_{(y_1,\infty)}\!(y_n). \end{align}$$

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