b*******d
发帖数: 32 | 1 You are on a subway station. There is an express train that arrives
every six minutes, on average. There is also a regular train which arrives
every three minutes, on average. The regular and express trains are moving
on separate tracks, independent from each other. Suppose you have just
arrived to the station. What is your expected waiting time?
My ans is 1 min, want to get confirmed | f******k
发帖数: 297 | 2 then it is quite complicated, since you have superposition of renewal
processes.
let me ask you. if the inter-arrival time is uniformly distributed with its
average t, what is the expected waiting time? don't tell me it is t/2.
【在 b*******d 的大作中提到】
: You are on a subway station. There is an express train that arrives
: every six minutes, on average. There is also a regular train which arrives
: every three minutes, on average. The regular and express trains are moving
: on separate tracks, independent from each other. Suppose you have just
: arrived to the station. What is your expected waiting time?
: My ans is 1 min, want to get confirmed
| D*******a
发帖数: 3688 | 3 what do you mean every 6 min on average?
or fixed 6 min strictly?
【在 b*******d 的大作中提到】
: You are on a subway station. There is an express train that arrives
: every six minutes, on average. There is also a regular train which arrives
: every three minutes, on average. The regular and express trains are moving
: on separate tracks, independent from each other. Suppose you have just
: arrived to the station. What is your expected waiting time?
: My ans is 1 min, want to get confirmed
| b*******d
发帖数: 32 | 4 It is t/2.
Because the distribution is uniform, let's consider the express train which
arrives every 6 minutes. The probability to wait time t is dt/6, therefore,
E(t)=\int_0^6 t dt/6= 3.
My computation for the problem is as follows:
For the express train (arrives every 6 min), the probability to wait time t is
dt/6, denoted as A
for the regular train (arrives every 3 min), the probability to wait time t is
dt/3 denoted as B
The probability to catch express OR regular train after time t is:
P(A
【在 f******k 的大作中提到】
: then it is quite complicated, since you have superposition of renewal
: processes.
: let me ask you. if the inter-arrival time is uniformly distributed with its
: average t, what is the expected waiting time? don't tell me it is t/2.
| b*******d
发帖数: 32 | 5 Let's say fixed 6 min strictly
【在 D*******a 的大作中提到】
: what do you mean every 6 min on average?
: or fixed 6 min strictly?
| f******k
发帖数: 297 | 6 hoho, seems i looked at a different problem.
【在 D*******a 的大作中提到】
: what do you mean every 6 min on average?
: or fixed 6 min strictly?
| s*****a
发帖数: 522 | 7 ～U(0,6] and ～U(0,3]
【在 D*******a 的大作中提到】
: what do you mean every 6 min on average?
: or fixed 6 min strictly?
| D*******a
发帖数: 3688 | 8 what do you mean every 6 min on average?
or fixed 6 min strictly?
【在 b*******d 的大作中提到】
: You are on a subway station. There is an express train that arrives
: every six minutes, on average. There is also a regular train which arrives
: every three minutes, on average. The regular and express trains are moving
: on separate tracks, independent from each other. Suppose you have just
: arrived to the station. What is your expected waiting time?
: My ans is 1 min, want to get confirmed
| s*****a
发帖数: 522 | 9 9/4
【在 b*******d 的大作中提到】
: You are on a subway station. There is an express train that arrives
: every six minutes, on average. There is also a regular train which arrives
: every three minutes, on average. The regular and express trains are moving
: on separate tracks, independent from each other. Suppose you have just
: arrived to the station. What is your expected waiting time?
: My ans is 1 min, want to get confirmed
| b*******d
发帖数: 32 | 10 I think the distribution of each train is uniform.
【在 f******k 的大作中提到】
: hoho, seems i looked at a different problem.
| D*******a
发帖数: 3688 | 11 他问的是一个renewal process, interarrival time~U[0,6]的
expected residual lifetime E[R]
这个不是3。根据renewal theory, E[R]=E[X^2]/2E[X] > E[X]^2/2E[X]=3.
你的原问题应该是两列车都是等间隔到达吧？这样的话，
time till the next train arrives is E[min(U[0,3],U[0,6])]
Let X=min(U1,U2)
F~(x)=P(X>x)=P(U1>x)P(U2>x)=(3-x)(6-x)/18
E[X]=\int_0^3 F~(x) dx=5/4
is
is
【在 b*******d 的大作中提到】
: It is t/2.
: Because the distribution is uniform, let's consider the express train which
: arrives every 6 minutes. The probability to wait time t is dt/6, therefore,
: E(t)=\int_0^6 t dt/6= 3.
: My computation for the problem is as follows:
: For the express train (arrives every 6 min), the probability to wait time t is
: dt/6, denoted as A
: for the regular train (arrives every 3 min), the probability to wait time t is
: dt/3 denoted as B
: The probability to catch express OR regular train after time t is:
| f******k
发帖数: 297 | 12 hoho, seems i looked at a different problem.
【在 D*******a 的大作中提到】
: 他问的是一个renewal process, interarrival time~U[0,6]的
: expected residual lifetime E[R]
: 这个不是3。根据renewal theory, E[R]=E[X^2]/2E[X] > E[X]^2/2E[X]=3.
: 你的原问题应该是两列车都是等间隔到达吧？这样的话，
: time till the next train arrives is E[min(U[0,3],U[0,6])]
: Let X=min(U1,U2)
: F~(x)=P(X>x)=P(U1>x)P(U2>x)=(3-x)(6-x)/18
: E[X]=\int_0^3 F~(x) dx=5/4
:
: is
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