u****h
发帖数: 2193 | 1 不是面试题，就是刚上随机过程的作业题。
假设公车和乘客在某个车站出现频率都服从poisson distribution, 公车来的rate是\
lambda, 乘客来的是\mu. 假设第一班车t=0来了，第二班t=0.5的时候来了，Wi是地i个
乘客的等待时间，W = (W1+W2...Wn)/n, 也就是n个乘客的等待时间的平均。求E[W|N>0
]
谢谢啦！ |
p*****k
发帖数: 318 | |
u****h
发帖数: 2193 | 3 不好意思大小写混淆了，其实就是总乘客个数
【在 p*****k 的大作中提到】
: what is N in E[W|N>0]?
|
p*****k
发帖数: 318 | 4 usfish, couple of things still seem confusing to me:
n is not fixed, right? are you asking for the average waiting time
for all the passengers arriving between t=0 and t=0.5? why do
we need the bus arrival distribution?
if possible, could you post the original version of the problem? |
u****h
发帖数: 2193 | 5 不好意思，其实这个是一个大题中的一个小题，所以我就把bus的rate也写上来了，其
实是没用的。
N其实是不fixed的。问题就是the average waiting time for all the passengers
arriving between t=0 and t=0.5
谢谢了
【在 p*****k 的大作中提到】
: usfish, couple of things still seem confusing to me:
: n is not fixed, right? are you asking for the average waiting time
: for all the passengers arriving between t=0 and t=0.5? why do
: we need the bus arrival distribution?
: if possible, could you post the original version of the problem?
|
p*****k
发帖数: 318 | 6 the answer is surprisingly (to me) simple: 0.5/2=0.25.
i have not found a simple argument, but by using the fact
that the time interval between any two passengers follows
independent exponential distribution, one can prove for
any particular n>0, E[W(n)|n]=0.25
e.g., when n=2, say the arrival time of these two passengers
are t1 and t1+t2. the probability of this happening is:
P2 = exp(-mu*T) * (mu*T)^2/2! following poisson distribution.
where T=0.5. the average waiting time W2=(T-t1)-t2/2 is t |
m******q
发帖数: 1 | 7 Maybe we can use this property. poisson arrival on a fixed time interval is
a uniform distribution. |
