Pedestrian at a road crossing

[Mehmood_Yasir]

Homework Statement

Pedestrians approach to a signal for road crossing in a Poisson manner with arrival rate $\lambda$ per sec. The first pedestrian arriving the signal pushes the button to start time $T$, and thus we assume his arrival time is $t=0$, and he always see $T$ wait time. A GREEN light is flashed after $T$, and all arrived pedestrians within $T$ must cross. Process repeats. What is the expected wait of the LAST pedestrian crossing the road?

Homework Equations

$P_k=\frac { {(\lambda T)}^k e^{-\lambda T} } {k!}$

The Attempt at a Solution

If $t_l$ is the arrival time of last pedestrian, the stay time is T minus $t_l$. then this stay time is also exponentially distributed with same parameter $\lambda$. For expectation value, the exponential pdf can be integrated over the interval from 0 to T. I am not sure if this statement is correct that T minus $t_L$ is also exponentially distributed, because $t_L$ is arrival time of last, the next exponential in original may have larger value than $T$.

 

[tnich]

Homework Statement

Pedestrians approach to a signal for road crossing in a Poisson manner with arrival rate $\lambda$ per sec. The first pedestrian arriving the signal pushes the button to start time $T$, and thus we assume his arrival time is $t=0$, and he always see $T$ wait time. A GREEN light is flashed after $T$, and all arrived pedestrians within $T$ must cross. Process repeats. What is the expected wait of the LAST pedestrian crossing the road?

Homework Equations

$P_k=\frac { {(\lambda T)}^k e^{-\lambda T} } {k!}$

The Attempt at a Solution

If $t_l$ is the arrival time of last pedestrian, the stay time is T minus $t_l$. then this stay time is also exponentially distributed with same parameter $\lambda$. For expectation value, the exponential pdf can be integrated over the interval from 0 to T. I am not sure if this statement is correct that T minus $t_L$ is also exponentially distributed, because $t_L$ is arrival time of last, the next exponential in original may have larger value than $T$.

You already know the wait time pdf of the kth pedestrian $f(t|k)$ from a previous thread. If you know the probability p(k) that the kth pedestrian is the last pedestrian, then you can get the pdf of the last pedestrian's waiting time.

 

[StoneTemplePython]

Homework Statement

Pedestrians approach to a signal for road crossing in a Poisson manner with arrival rate $\lambda$ per sec. The first pedestrian arriving the signal pushes the button to start time $T$, and thus we assume his arrival time is $t=0$, and he always see $T$ wait time. A GREEN light is flashed after $T$, and all arrived pedestrians within $T$ must cross. Process repeats. What is the expected wait of the LAST pedestrian crossing the road?

- - - - -
If this is "merely" for expected values, an easier approach may be to use conditioning (specifically Law of Total Expectation).

e.g. Suppose $n = 0$ arrivals happen in $(0, t]$ -- i.e. no arrivals once the button is pushed. The expected wait time of the last person, i.e. the button pusher, is $t$.

Now for all other cases, $n \gt 0$:
you can ignore the button pusher and look at this in terms of 'regular' Poisson mechanics. Conditioning on $N(t) = n$, all arrivals are uniformly distributed in $(0, t]$. There are some technical nits on $n!$ orderings but pick one without loss of generality -- because we're actually looking at order statistics here. The easy problem is then to find the CDF of time until first arrival given $N(t) = n$

You should get (with $U$ referring to the i.i.d. uniform random variables that exist in this conditional world -- there are n of them)

$Pr\{\text{Arrival 1's time} \gt \tau \big \vert N(t) =n \} = 1 - F_{min}U = \big[1 - F_U(u)\big]^n$

(note this was the underpinning of a recent thread: https://www.physicsforums.com/threa...f-ind-r-v-that-follows-distribution-f.946651/ and working through the CDFs for minimum and maximum amongst N i.i.d. random variables -- order statistics-- is an exercise worth doing.)

you can integrate this complementary CDF to get the expected time between Light going off and first 'real' arrival. Thus you have $E\big[ X_1 \big \vert N(t) =n\big]$, then using a symmetry argument, or the fact that this process is time reversible (why?), you have thus found $E\big[ X_n \big \vert N(t) =n\big]$.

Putting it all together with Law of Total Expectation (and hopefully note spoiling the result), should give:

$E\big[\text{Last Ped's wait time}\big] = p_\lambda(n=0, t) \Big(t\Big) + \sum_{n=1}^{\infty}p_\lambda(n, t) \Big(E\big[ X_n \big \vert N(t) =n\big]\Big)$

When you're all done, you should see there's some nice structure in the factorials in the denominator that should allow a closed form for the series.

- - - -
What I've said above is all discussed in a lot more detail here:

https://ocw.mit.edu/courses/electri...ring-2011/course-notes/MIT6_262S11_chap02.pdf

which is something I've mentioned to you a while back. I am not supposed to give this whole thing away, though, so I leave the reading and a lot more lifting, to you.