Starting at time 0, a red bulb flashes according to a Poisson process with rate λ=1 . Similarly, starting at time 0, a blue bulb flashes according to a Poisson process with rate λ=2 , but only until a nonnegative random time X , at which point the blue bulb “dies." We assume that the two Poisson processes and the random variable X are (mutually) independent.
1) Suppose that X is deterministically equal to 1. What is the expected total number of flashes (of either color) during the interval [0,2] ?
Expected total number of flashes:
2) Suppose that X=∞ (i.e., the blue bulb never dies). What is the expected value of the time of the first flash (of either color)?
Expected value of the time of the first flash:
3) In the time interval [0,X] , there are exactly 5 flashes. What is the probability that exactly 2 of them were red?
Probability that exactly 2 of the 5 flashes were red:
4) Suppose that X is equal to either 1 or 2, with equal probability. Write down an expression for the probability that there were exactly 3 arrivals during the time interval [0,2] .
(Enter e for the constant e . You may use standard notation for this numerical entry even though there will be no parser below the answer box. Enter an exact answer or a numerical answer accurate to at least 3 decimal places.)
Probability that there were exactly 3 arrivals during the time interval [0,2] :
5) Suppose that X is an exponential random variable with parameter (and mean) equal to 1. Find the MAP estimate of X , given that there were exactly 5 blue flashes.
MAP estimate of X :
Hint for 3: see 9.23.7 and use combination and conditional probabilities.
1) 4
2) 1/3
3) 80/243
1) The expected total number of flashes during the interval [0,2] is equal to the sum of the expected number of red flashes and the expected number of blue flashes.
The expected number of red flashes during the interval [0,2] can be calculated using the Poisson process with rate λ=1. The rate parameter λ is equal to the expected number of events in a unit time interval. Therefore, the expected number of red flashes during the interval [0,2] is λ * t = 1 * 2 = 2.
The expected number of blue flashes during the interval [0,1] is λ * t = 2 * 1 = 2. After time X=1, the blue bulb dies and no more blue flashes occur.
Therefore, the total expected number of flashes during the interval [0,2] is 2 + 2 = 4.
Expected total number of flashes: 4
2) If X=∞, the blue bulb never dies, and hence blue flashes keep occurring indefinitely. The expected value of the time of the first flash is the same as the expected value of the minimum of two exponential random variables with rates λ=1 for the red bulb and λ=2 for the blue bulb.
The expected value of an exponential random variable with rate λ is given by 1/λ. Therefore, the expected value of the time of the first flash is 1/λ + 1/λ = 1/1 + 1/2 = 3/2.
Expected value of the time of the first flash: 3/2
3) The number of red flashes in the interval [0,X] follows a Poisson distribution with rate λ=1. The number of blue flashes in the interval [0,X] also follows a Poisson distribution with rate λ=2.
Given that exactly 5 flashes occurred in the interval [0,X], the probability that exactly 2 of them were red can be calculated using the PMF of the Poisson distribution.
The probability mass function (PMF) of a Poisson distribution with rate λ is given by P(X=k) = (λ^k * e^(-λ))/k!, where k is the number of events.
In this case, we want to calculate the probability that exactly 2 flashes were red. Therefore, k=2 and λ=1.
P(X=2) = (1^2 * e^(-1))/2! = (e^(-1))/2 ≈ 0.18394.
Probability that exactly 2 of the 5 flashes were red: 0.18394
4) If X can be either 1 or 2 with equal probability, we can use the law of total probability to calculate the probability that there were exactly 3 arrivals during the time interval [0,2].
Let A1 be the event that X=1 and A2 be the event that X=2. The probability of each event is P(A1) = P(A2) = 0.5.
The total probability of 3 arrivals during the time interval [0,2] can be calculated as:
P(3 arrivals) = P(3 arrivals | A1) * P(A1) + P(3 arrivals | A2) * P(A2)
From question 1, we know that the expected number of arrivals during [0,1] is 2. Therefore, the probability of 3 arrivals during [0,2] if X=1 is given by the PMF of a Poisson distribution with rate λ=2:
P(3 arrivals | A1) = (2^3 * e^(-2))/3! = (8 * e^(-2))/6
Similarly, if X=2, the probability of 3 arrivals during [0,2] can be calculated as:
P(3 arrivals | A2) = (3^3 * e^(-3))/3! = (27 * e^(-3))/6
Therefore, the probability that there were exactly 3 arrivals during the time interval [0,2] is:
P(3 arrivals) = (8 * e^(-2))/6 * 0.5 + (27 * e^(-3))/6 * 0.5
Probability that there were exactly 3 arrivals during the time interval [0,2]: (8 * e^(-2) + 27 * e^(-3))/12
5) If X is an exponential random variable with parameter (and mean) equal to 1, we want to find the maximum a posteriori (MAP) estimate of X given that there were exactly 5 blue flashes.
The MAP estimate of X can be found by maximizing the posterior probability given the observed data. In this case, the observed data is exactly 5 blue flashes.
The posterior probability can be calculated using Bayes' theorem:
P(X=x | 5 blue flashes) = P(5 blue flashes | X=x) * P(X=x) / P(5 blue flashes)
The term P(5 blue flashes | X=x) can be calculated using the PMF of the Poisson distribution with rate λ=2:
P(5 blue flashes | X=x) = (2^5 * e^(-2))/5!
The term P(X=x) is the probability density function (PDF) of an exponential distribution with parameter λ=1:
P(X=x) = λ * e^(-λx) = e^(-x)
The term P(5 blue flashes) can be calculated using the law of total probability:
P(5 blue flashes) = P(5 blue flashes | X=1) * P(X=1) + P(5 blue flashes | X=2) * P(X=2)
From question 1, we know that the expected number of blue flashes during [0,1] is 2. Therefore, the probability of 5 blue flashes during [0,1] if X=1 is given by the PMF of a Poisson distribution with rate λ=2:
P(5 blue flashes | X=1) = (2^5 * e^(-2))/5!
Similarly, if X=2, the probability of 5 blue flashes during [0,2] can be calculated as:
P(5 blue flashes | X=2) = (3^5 * e^(-3))/5!
Therefore, the probability that there were exactly 5 blue flashes is:
P(5 blue flashes) = (2^5 * e^(-2))/5! * 0.5 + (3^5 * e^(-3))/5! * 0.5
Substituting all the calculated values into Bayes' theorem, we can find the MAP estimate of X.
0.13925
1) 4
2) 1/3
3) 1/3
4) ?????
5) ?????