Poisson Process and Exponential Random Variables

Question

Suppose we have a planetary system with $m$ planets, orbiting a star. Suppose the distance from the star to each planet is independent exponential random variable with rate $\mu$. I want to find the expected value of the distance between the star and the furthest planet. Now, I was given a hint to use the fact this Poisson process was memoryless. Now, my thinking is that since the distance is exponential random variable, then the mean distance from any planet to another planet should be $\frac{1}{\mu}$. Then we can use the memoryless property to conclude that the expected value of the distance between the star and the furthers planet is $\frac{1}{\mu} +\frac{1}{\mu}+...+\frac{1}{\mu}$ $m$ times. So expected distance would be $\frac{m}{\mu}$. Does this thinking seem right?

Answer 1

For each $1\leq j\leq m$ let $D_j\sim \text{Exponential}(\mu)$ denote the distance between the $j^{\text{th}}$ planet and the star.

Take $D_{(1)},\dots,D_{(m)}$ as the corresponding order statistics. It follows $D_{(1)}\sim\text{Exponential}(m\mu)$ and $D_{(j+1)}-D_{(j)}\sim \text{Exponential}\big(\mu (m-j)\big)$ for each $1\leq j \leq m-1$. You can justify this last fact by using the memoryless property of the exponential distribution and the total law of probability. (If you want addition details on this calculation let me know.)

You want to compute $\mathbb{E}\big(D_{(m)}\big)$. We get $$\begin{eqnarray*} \mathbb{E}(D_{(m)}) & = & \mathbb{E}(D_{(m)}-D_{(m-1)})+\dots +\mathbb{E}(D_{(2)}-D_{(1)}) +\mathbb{E}(D_{(1)})\\ & = & \frac{1}{\mu\big[m-(m-1)\big]}+\dots +\frac{1}{\mu\big[m-1\big]}+\frac{1}{\mu m} \\ & = & \frac{1}{\mu} \sum_{k=1}^{m}\frac{1}{k} \end{eqnarray*}$$

Another way to proceed is by explicitly calculating the pdf of $D_{(m)}$ as I mentioned in the comments: $$f_{D_{(m)}}(x)=m \mu e^{-\mu x}\Big(1-e^{-\mu x}\Big)^{m-1}$$ We get that $$\mathbb{E}(D_{(m)})=\int_0^{\infty}xf_{D_{(m)}}(x)\mathrm{d}x$$ You can expand $\Big(1-e^{-\mu x}\Big)^{m-1}$ via the binomial theorem and calculate the resulting integrals to get the same result.

Edit: Let me elaborate on the details for the distribution of $X=D_{(j+1)}-D_{(j)}$ where $j\in \{1,\dots,m-1\}$ is fixed. From the total law of probability, we have for $x\geq 0$ that$$F_X(x)=\int_0^{\infty}P(D_{(j+1)}\leq x+t|D_{(j)}=t)f_{D_{(j)}}(t)\mathrm{d}t$$ The event $\{D_{(j)}=t\}$ reveals that the furthest $m-j$ planets are at least $t$ units away from the star. The pdf of each of these $m-j$ distances conditioned on this information is given by $$x \longrightarrow \frac{\mu e^{-\mu x}}{\int_t^{\infty}\mu e^{-\mu s}\mathrm{d}s}$$ for $x\in [t,\infty)$. Therefore $$P(D_{(j+1)}\leq x+t|D_{(j)}=t)=1-P(D_{(j+1)}> x+t|D_{(j)}=t)=1-\Bigg(\frac{\int_{x+t}^{\infty}\mu e^{-\mu s}\mathrm{d}s}{\int_t^{\infty}\mu e^{-\mu s}\mathrm{d}s}\Bigg)^{m-j}$$ The last expression on the right hand side simplifies to $1-e^{-\mu(m-j)x}$ which doesn't depend on $t$ (since exponential distribution is memoryless). Therefore $$F_X(x)=\Big[1-e^{-\mu(m-j)x}\Big]\int_0^{\infty}f_{D_{(j)}}(t)\mathrm{d}t=1-e^{-\mu(m-j)x}$$ which is the cdf for an $\text{Exponential}\big(\mu(m-j)\big)$ distribution.