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Let $X_i \sim \mathrm{Exp}(\lambda_i)$, $i = 1,2,3$ be independent, find $\mathsf E(\max(X_i) \mid X_1<X_2<X_3)$


I have found out two solutions as follow:

Solution 1 Solution 1

I am wondering if someone could explain this solution 1 intuitively while referring to the memoryless property of the exponential distribution. I am not able to connect them with M.L.P.


Solution 2 Solution 2

For solution 2, can anyone explain why $\mathsf E(X_2 - X_1 \mid X_1 \lt X_2 \lt X_3)$ implies $\mathsf E(X_2 \mid X_2 \lt X_3)$, why we could drop the $X_1$. Also, why $\mathsf E(X_3 - X_2 \mid X_1 \lt X_2 \lt X3)$ implies $\mathsf E(X_3)$ and why we could drop the $X_1 \lt X_2 \lt X_3$


I know those two solutions are both connected to the memoryless property of the exponential distribution, but I couldn't get through them. Hope someone could make it clear for me. Any help will be appreciated, Thanks in advance :)

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The quantity $\lambda_1+\lambda_2+\lambda_3$ is the "rate", i.e. the average number of arrivals per unit of time, when three processes are running, with rates $\lambda_1$, $\lambda_2$, and $\lambda_3$. After the first arrival, only two more arrivals can happen, and those have rates $\lambda_2$ and $\lambda_3$. After the second arrival, only one more can happen, with rate $\lambda_3$.

The expectation $\operatorname{E}(X_2-X_1\mid X_1<X_2<X_3) = \operatorname{E}(X_2-X_1\mid X_1<X_2<X_3)$ is the same as $\operatorname{E}(X_2\mid X_2<X_3)$ because once the first arrival has happened, memorylessness says the probability distribution of the remaining time until the second arrival does not depend on how long one has waited for it so far. So it's the same as if you had started the whole thing running only the second and third processes.

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  • $\begingroup$ Thank you for the answer, @Michael Hardy. In the memoryless property $P(X \gt t+s \mid X \gt t) = P(X \gt s)$, where both $t$ and $s$ are constant numbers. But in $\operatorname{E}(X_2-X_1\mid X_1<X_2<X_3)$, $X_1$ is a random variable, can we still drop it? Or you just treat it as a constant after the first arrival has happened. If so, Can I write $\operatorname{E}(X_2-X_1\mid X_1<X_2<X_3) = \operatorname{E}(X_2-X_1\mid X_1<X_2<X_3, X_1 = c)$, where $c$ is a constant ? $\endgroup$ – SundayCat Apr 1 '16 at 3:32
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This is to supplement the explanation given above.

Let's look at the joint densities. When we condition on $\{X_1<X_2<X_3\}$ we have joint density equal to the original density (up to multiplicative constant):

$$ e^{-\lambda_1 x_1} e^{-\lambda_2 x_2} e^{-\lambda_3 x_3},$$ expect two changes:

  1. This formula is valid only for $x_1<x_2<x_3$, and otherwise the density is zero.
  2. We need a normalizing constant $c$ (equal to $\lambda_1\lambda_2 \lambda_3$ for the IID.

To make this more manageable, we choose a natural change of variables: $y_1 =x_1$, $y_2=x_2-x_1$, $y_3=x_3-x_2$.

The constraints in 1. become simpler: $y_1 \ge 0,y_2\ge 0,y_3\ge 0$, no interdependence ! What about the density?

$$ c e^{-\lambda_1 y_1} e^{-\lambda_2 (y_2+y_1)}e^{-\lambda_3(y_3+y_2+y_1)}=c e^{-(\lambda_1+ \lambda_2 + \lambda_3)y_1} e^{-(\lambda_2+\lambda_3)y_2}e^{-\lambda_3 y_3}.$$

Now the change of variables has Jacobian $1$ (the matrix is triangular with $1$ on the diagonal. As a result,

  1. We have in immediate expression for the normalizing constant: $(\lambda_1+\lambda_2+\lambda_3)(\lambda_2+\lambda_3)\lambda_3$.
  2. This is the joint density of independent random variables. From the very definition, these are $X_1,X_2-X_1,X_3-X_2$.

Summarizing, conditioned on $\{X_1<X_2<X_3\}$, $X_1,X_2-X_1,X_3-X_2$ are independent exponentials with rates $\lambda_1+\lambda_2+\lambda_3$, $\lambda_2+\lambda_3$, and $\lambda_3$, respectively.

I think this explains all you asked about. For example: $E[X_2-X_1 \mid X_1<X_2<X_3 ] $ is the expectation of an exponential $\lambda_2+\lambda_3$. But for similar reasons so is $E[ X_2 \mid X_2<X_3]$, etc.

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