Tricky Poisson Formula Rearrangement

#1
Hello,

I am currently doing some bus stop designs. I have not done statistics for a while but remember once doing poisson distributions at Uni.

The problem I have, I think, requires a specific arrangement of the poisson formula for cumulative distribution.

If I have x number of buses arrive per hour and they each stay on average y seconds how many bus stops would I require so that the number of buses at the stops exceeds the number of stops z%?
e.g. buses per hour = 30
buses wait time = 60 seconds
poisson random distribution for arrival times
probability of exceeding number of bus spaces = 5%
bus spaces = ?

I would appreciate any guidance to solving this problem. I have researched formulas but can't seem to find one that allows me to perform this calculation.

Many Thanks!
 

BGM

TS Contributor
#2
Let \( X_i \) be the arrival time of the \( i \)-th bus

and \( Y_i \) be the amount of time that the \( i \)-th bus stay in the bus stop.

Then the number of buses stay in the bus stop at time \( t \)

\( M(t) \triangleq \sum_{i=1}^{N(t)} \mathbf{1}\{t < X_i + Y_i\} \)

where \( N(t) \) is the number of buses arrived before time \( t \)

The key trick here is to observe that \( M(t) \) is a symmetric functional of the arrival time, as we assume \( Y_i \) are i.i.d.

Therefore, since the arrival time conditional on the number of arrivals is distributed like the ordered-uniform on \( (0, t] \) , by the symmetric property they are distributed like the independent uniform on \( (0, t] \)

Therefore, \( M(t) \stackrel {d} {=} \sum_{i=1}^{N(t)} \mathbf{1}\{t < U_i + Y_i\}
\)
where \( U_i \) are i.i.d. uniform on \( (0, t] \) and independent of \( N(t) \)

The exact distribution of \( M \) maybe is difficult to obtain, but at least you may evaluate the desired probabilities numerically. If you want the expectations then it is already in a closed form solution.
 
#3
Ok. Well this is interesting but pretty much over my head. I have my old stats book out to try and find the solution I am after but it doesn't seem to work. I am attempting trial and error solutions at the moment.
If the mean rate of buses using the bus stop is 30/60 per minute, and the mean stopping time is say 30 seconds and there are 2 bus spaces available how would I calculate how often the number of bus spaces is exceeded?
I think I need to use this formula: P(X < x) = ((vt)^x)/x!*exp(-vt) where x = 0, 1 2,...
 

BGM

TS Contributor
#4
Yes of course you need to use the pmf of Poisson, because \( N(t) \) is a Poisson process.

So using law of total probability, conditioning on \( N(t) = n \), we obtain

\( \Pr\{M(t) = m\} = \sum_{n=0}^{+\infty} \Pr\{M(t) = m|N(t) = n\}\Pr\{N(t) = n \}\)

The latter part is the Poisson pmf, while the first part is the Binomial pmf, because

\( M(t)|N(t) = n \sim \text{Binomial}(n, p) \)
where \( p = \Pr\{U + Y > t\} \)

All things are ready, I think you need to specify the distribution of the staying time of
each bus as well. Otherwise at most you can give a bound for the approximation.
 

BGM

TS Contributor
#6
I have check my text again. If you are just want to ask for the long-run distribution,

then \( M(t) \) has a Poisson distribution with mean \( \lambda \mu \),

where \( \lambda \) is the mean arrival rate and \( \mu \) is the mean staying time.

I have to correct 1 point for my previous post:

Actually you can show the unconditional distribution of \( M(t) \) is Poisson by evaluating the above sum.

So if you can determine the \( p \), then you know the whole distribution of \( M(t) \)

\( M(t) \sim \text{Poisson}(\lambda p t) \)

and thus you can evaluate the corresponding stops needed.