Having already worked out the distributions of $\Delta_{(2)}X=X_{(2)}-X_{(1)}\sim\text{Exp}(\lambda)$ and of $\Delta_{(1)}X=X_{(1)}\sim\text{Exp}(2\lambda)$ where $X_{(i)}$ are the $i$th order variables from the 2-sample of independent exponentially distributed random variables $X_1,X_2\sim \text{Exp}(\lambda)$ i.e $\text{P}(X_i\ge r)=\exp(-\lambda r)$. We are asked to compute:

$\text{E}(X_{(2)}\mid X_{(1)}=r_1)$

The solution uses the fact that $\Delta_{(2)}X+\Delta_{(1)}X=X_{(2)}$ (and $\Delta_{(1)}X\equiv X_{(1)}$) so that the expectation is equivalent to:

$\text{E}(\Delta_{(2)}X+\Delta_{(1)}X \mid\Delta_{(1)}X=r_1)=\text{E}(\Delta_{(2)}X\mid\Delta_{(1)}X=r_1)+\text{E}(\Delta_{(1)}X\mid\Delta_{(1)}X=r_1)$

Since I have already worked out that $\Delta_{(2)}X,\Delta_{(1)}X$ are independent $\text{E}(\Delta_{(2)}X\mid\Delta_{(1)}=r_1)=\text{E}(\Delta_{(2)}X)$

So that $\text{E}(X_{(2)}|X_{(1)}=r_1)=\frac{1}{\lambda}+r_1$

Is there another way to get this result? I don't want to rely on the relation, $\Delta_{(2)}X+\Delta_{(1)}X=X_{(2)}$, even though it seems to maintain its convenience for larger $n$-samples, i.e. for $X_{(n)}=\sum_{i=1}^nX_{(i)}-X_{(i-1)}=\sum_{i=1}^n\Delta_{(i)}X$.

I was thinking of the integral $\text{E}[X_{(2)}\mid X_{(1)}=r_1]=\int_0^\infty \xi f_{X_{(2)}\mid X_{(1)}}(\xi\mid r_1)d\xi$, but I cannot get the correct answer so I think that my densities are incorrect.

The density functions I have used in the integral are $f_{X_{(1)}}(x)=\lambda\exp(-\lambda x)$ and $f_{X_{(1)},X_{(2)}}(x,\xi)=\lambda^2 \exp(-\lambda(\xi+x))$ with $$f_{X_{(2)}|X_{(1)}}(\xi,x)=\frac{f_{X_{(1)},X_{(2)}}(x,\xi)}{f_{X_{(2)}}(x)}$$

If I integrate this I get $\frac{1}{\lambda}$

  • $\begingroup$ If $X_1,...,X_n$ are independent exponentially distributed random variables, then $X_{(1)}$ can not have the same distribution as $X_1$. So it is false that $X_{(1)} \sim \mathcal E(\lambda)$. $\endgroup$ – roger Apr 27 '13 at 9:09
  • $\begingroup$ There are only two, $X_1,X_2 \sim \text{Exp}(\lambda)$. I have edited my post, I think that that was unclear to begin with. $\endgroup$ – shilov Apr 27 '13 at 9:23
  • $\begingroup$ Also, I made a typo in the original; $X_{(1)}\sim \text{Exp}(2\lambda)$. Could you also briefly explain the reason for $X_{(1)}$ not being able to have the same distribution as $X_1$? $\endgroup$ – shilov Apr 27 '13 at 9:28

Is there another way to get this result


  • find the joint pdf of the $1^{\text{st}}$ and $2^{\text{nd}}$ order statistics, with a sample size of $n=2$.

  • Now that you have the joint pdf of $(X_{(1)}, X_{(2)})$, find the conditional pdf of $X_{(2)}$ given $X_{(1)}=r_1$ . You now have a univariate distribution on $X_{(2)}$, say pdf $h(x_{(2)})$, with domain of support $(r_1, \infty)$.

  • Find $E[X_{(2)}]$ wrt pdf $h$. Doing so yields the same solution: $\frac{1}{\lambda}+r_1$

In your above workings, you note:

$$f_{X_{(1)},X_{(2)}}(x,\xi)=\lambda^2 \exp(-\lambda(\xi+x))$$

This result is incorrect in two respects. The first is that you are missing a 2 in the numerator (i.e. multiply your answer by 2), and the second is that the domain of support is $0 < x_1 < x_2$.

Here are the workings in mathStatica/Mathematica to derive the joint pdf $(X_{(1)}, X_{(2)})$ result:

enter image description here

So, you were pretty close :)

  • $\begingroup$ Is there a way for me to get OrderStat in Mathematica, or is it just mathStatica? $\endgroup$ – shilov Apr 27 '13 at 16:48
  • $\begingroup$ @shilov OrderStat is a mathStatica function for Mathematica: it is a Mathematica package, so you need Mathematica to use it. $\endgroup$ – wolfies Apr 27 '13 at 18:29

Given two random variables $X$ and $Y$, the joint density of $W=\min(X,Y)$ and $Z=\max(X,Y)$ is $$f_{W,Z}(x,y) = \begin{cases}f_{X,Y}(x,y)+f_{X,Y}(y,x),&y > x,\\0, & y<x.\end{cases}$$ See, for example, this answer for details. Thus, for independent exponential random variables $X$ and $Y$, the joint density is $$f_{W,Z}(x,y) = \begin{cases}2\lambda^2\exp(-\lambda(x+y)),&y > x > 0,\\0, & \mathrm{otherwise}.\end{cases}$$ So, $f_{Z\mid W}(y\mid W=x)$, the conditional density of $Z$ given $W=x > 0$ is proportional to $[2\lambda^2\exp(-\lambda x)]\exp(-\lambda y)\mathbf 1_{y > x > 0} = c\cdot \exp(-\lambda y)\mathbf 1_{y > x > 0}$, and so must be an exponential density with parameter $\lambda$ displaced $x$ to the right. Note that we do not even need to compute $f_W(x)$ in order to reach this conclusion, though we can make assurance doubly sure by working out the details. The conditional mean is thus $E[Z\mid W=x] = x+\frac{1}{\lambda}$.


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