Question
A red light bulb has been flashing forever, according to a Poisson process with rate r . Similarly, a blue bulb has been flashing forever, , according to an independent Poisson process with rate b . Let us fix t to be 12 o'clock.
1. What is the expected length of the interval that t belongs to? That is, find the expected length of the interval from the last event before t until the first event after t . Here, an event refers to either bulb flashing.
2. What is the probability that t belongs to an RR interval? (That is, the first event before, as well as the first event after time t , are both red flashes.)
3. What is the probability that between t and t+1 , we have exactly two events: a red flash followed by a blue flash?
1. What is the expected length of the interval that t belongs to? That is, find the expected length of the interval from the last event before t until the first event after t . Here, an event refers to either bulb flashing.
2. What is the probability that t belongs to an RR interval? (That is, the first event before, as well as the first event after time t , are both red flashes.)
3. What is the probability that between t and t+1 , we have exactly two events: a red flash followed by a blue flash?
Answers
Anonymous
2/(r+b)
r^2/(r+b)^2
r*b*2*e^(r+b)/(r+b)^4
r^2/(r+b)^2
r*b*2*e^(r+b)/(r+b)^4
you
why 3) isnt r*b/(r+b)^2 ?
Anonymous
which is correct ? and reasoning
youtu
Why number 3 is not (rbe^-(r+b))/2 ? I get this other answer doing
(r/(r+b))*(b/(r+b))*(((r+b)^2)*e^-(r+b))/2
P(A/\B) = P(B)*P(A|B)
(r/(r+b))*(b/(r+b))*(((r+b)^2)*e^-(r+b))/2
P(A/\B) = P(B)*P(A|B)
mike
why isn't the answer to question one:
1/(r+b)
where does the 2 in 2/(r+b) come from
1/(r+b)
where does the 2 in 2/(r+b) come from
youtu
Because you have to think it as 2 Poissons, one from t to the next arrival and one from to to the previous arrival (reverse time). So you sum the expected length of the interval of both.
mike
thanks for your answer. that is weird. is the average time between flashes still (either color) equal to 1/(r+b)?
mike
thanks for your answer. that is weird. is the average time between flashes still equal to 1/(r+b)? but due to the fresh-start property t could occur at the beginning of time?
Anon1
I have the same answer as youtu for number 3. if you add red/blue as he posted to blue/red + red/red + blue/blue you get the total prob of 2 arrivals, which should be the case.