For a Poisson process, what is the distribution of the ratio of one arrival time to another?

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The problem: Commuters arrive at a bus stop with the rate of $\lambda = 8$ per hour. Let $S_i$ denote the time when the $i^{th}$ arrival occurs.

Find the distribution of the ratio $W = \frac{S_1}{S_6}$.

My attempt: This looks like a standard transformation problem. My first line was:

$$\Pr(W <= s)=\Pr(\frac{S_1}{S_6}<=s)=\Pr(S_1<=S_6s)=\iint f_{S_1,S_6}(s_1, s_6)dS_1dS_6$$

The bounds on the integral being from $0$ to infinity on $S_6$ and $0$ to $s$ on $S_1$. When calculating, I got a nonsensical result from the integral (infinity$-0$), and I think that the culprit may be my joint probability density function. I was under the impression that the joint density function of the arrival times was equal to the product of interarrival times:

$$f_{S_1,S_6}(s_1, s_6) = f_{S_1}(s_1)f_{S_6}(t - s_6)$$

Now that I realize, though, this only applies to the conditional distribution, i.e. when $N(t) = n$ is given. I don't know where to start.

The key problem I have: How do I find the joint probability density function of $S_6$ and $S_1$?

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What you are thinking of is that: $f_{S_1,S_6}(s,t)~=~ f_{S_1}(s)~f_{S_6-S_1}(t-s)$, since $S_1$ and $S_6-S_1$ are independent interarrival times.   (Also recall, $S_6-S_1$ has identical distribution to $S_5$, and that $S_k\sim\mathcal {Erlang}(k,\lambda)$.)

Now, for $0\lt w\leq 1$ : $$\begin{align}\mathsf P(W\leq w)&=\mathsf P(S_1\leq wS_6) \\[1ex]&= \int_0^\infty \int_0^{wt} f_{S_1}(s)~f_{S_6-S_1}(t-s)~\mathsf d s~\mathsf d t\\[1ex]&= \int_0^\infty f_{S_1}(s)\int_{s/w}^\infty f_{S_6-S_1}(t-s)~\mathsf d t~\mathsf d s\\[1ex]&= \int_0^\infty f_{S_1}(s)\int_{s(1/w-1)}^\infty f_{S_6-S_1}(u)~\mathsf d u~\mathsf d s\\[1ex]&~~\vdots\end{align}$$