0
$\begingroup$

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 $\frac{1}{\lambda}$

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

Is there another way to get this result

Yes:

  • 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 :)

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $E[Z\mid W=x] = x+\frac{1}{\lambda}$.

$\endgroup$

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.