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?

  • $\begingroup$ 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

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)


  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.

  • 2
    $\begingroup$ +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

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 $X_n = Y_n$) $$\begin{align} \mathbb P(Y_1>y_1\,|\,Y_n=y_n) &= \mathbb P(Y_1,...,Y_n>y_1\,|\,Y_n=y_n) =\\ &= \mathbb P(y_1 <X_1,...,X_{n-1}\leq y_n)I_{(y_1,\infty)}\!(y_n) = \\ & = [F(y_n) - F(y_1)]^{n-1} I_{(y_1,\infty)}\!(y_n). \end{align}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.