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Sheldon Ross's Introduction to Probability Models, exercise 5.44.b: cars pass a certain street location according to a Poisson process with rate $\lambda$. A woman who wants to cross the street at that location waits until she can see that no cars will come by in the next $\tilde t$ time units. I'm trying to find her expected waiting time.

I am given a hint to condition on the time of the first car, which I'll call $T_1$. The process is noted $\{N(t) \;|\; t \geq 0\}$. Her waiting time can be defined as $$S := \min_{t\geq0}\{N(t+\tilde t) - N(t) = 0\} \geq 0\quad.$$

Clearly, $$\mathbb E[S\;|\;T_1 \geq \tilde t] = \mathbb E[0] = 0\quad.$$ If this approach is good, how should I proceed for $T_1 < \tilde t$?

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Why the unaccept decision? Trying to establish a new obnoxiousness record? –  Did Aug 23 '13 at 17:25

1 Answer 1

up vote 5 down vote accepted

The exercise seems to ask for $E(S)$ where $S=\inf\{t\ge0;N(t+\tilde t)=N(t)\}$ hence $S\ge0$, in any case this is the convention used below.

Using the random time $T_1$ when the first car passes, as you suggested, one gets $S=0$ on $T_1> \tilde t$. On $T_1\le\tilde t$, the pedestrian must wait until time $T_1$ and then wait for the first suitable time to cross the street. The lack of memory of the Poisson process at time $T_1$ implies that this residual time is distributed like $S$ and is independent on $T_1$. Hence, $$ E(S)=E(E(S|T_1);T_1\le\bar t)=E(T_1+E(S);T_1\le\bar t)=E(T_1;T_1\le\bar t)+E(S)P(T_1\le\bar t). $$ Solving for $E(S)$ and using the fact that $T_1$ is exponentially distributed with parameter $\lambda$, one gets $$ E(S)=\frac{E(T_1;T_1\le\bar t)}{P(T_1\ge\bar t)}=\mathrm{e}^{\lambda\bar t}\int_0^{\bar t}\lambda s\mathrm{e}^{-\lambda s}\mathrm{d}s, $$ hence $$ E(S)=\frac1\lambda(\mathrm{e}^{\lambda\bar t}-1-\lambda\bar t). $$ Post hoc checks: The distribution of $\lambda S$ should depend on $\lambda\bar t$ only. The random variable $S$ should converge in distribution to $0$ when $\lambda\to0$, for any given $\bar t$, and also when $\bar t\to0$, for any given $\lambda$.

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Item a asks for the probability of her waiting $0$ time units, which is oddly trivial to calculate if $S$ is phrased as I did. I'll change that in the question. –  Luke Jun 24 '11 at 16:32
    
Luke: Regarding the modifications you made to my post, I have nothing against police character niceties such as using mathbb for the symbol E of expectations or adding thin spaces here and there, but I must object to replacing $E(S)$ by $E(S|T_1\le \bar t)$ at places where the former is meant and correct and the latter is false. As a consequence I rolled back your edits. (I also think that the least one can do in such circonstances is to inform the author of the post of the heavy mathematical modifications one has in mind.) –  Did Jun 24 '11 at 19:13
    
@Didier Piau Sorry, I thought there was no change in the meaning. So the analogous change I made in the question is wrong as well? –  Luke Jun 24 '11 at 20:44
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@Theo: What is happening here? Twice now, Luke modified my answer. The modified version is wrong, again. Obviously Luke does not know the difference between E(X;A) (the expectation of X on the set A only) and E(X|A) (the expectation of X conditioned by the event A). When X is a constant x the former is xP(A) and the latter is x. These two quantities are different. Obviously, Luke does not even care to ask before vandalizing the answer. Does anybody believes in the displayed equation with an integral sign as it appears now? .../... –  Did Jun 29 '11 at 6:12
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.../... It is false. I am tired with spending time on this matter and fighting to keep my answer mathematically correct. This has to stop. Flagging for attention. –  Did Jun 29 '11 at 6:12

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