A) 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: ________
B) 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] :
______________
C) 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 : _________
1. The expected total number of flashes during the interval [0,2] can be calculated by adding the expected number of red flashes and the expected number of blue flashes.
The expected number of red flashes can be calculated using the formula for the expected value of a Poisson distribution, which is λ * t, where λ is the rate and t is the time interval. In this case, λ = 1 and t = 2, so the expected number of red flashes is 1 * 2 = 2.
The expected number of blue flashes is given by the product of the probability that the blue bulb is still flashing at time 1 (since X is deterministically equal to 1) and the expected number of blue flashes per unit time. The probability that the blue bulb is still flashing at time 1 is equal to the survival function of the exponential distribution with rate 2, evaluated at 1. Since the exponential distribution has a mean of 1/2, the rate is equal to 1/(1/2) = 2. The survival function of the exponential distribution is e^(-λx), so the probability that the blue bulb is still flashing at time 1 is e^(-2*1) = e^(-2).
The expected number of blue flashes per unit time is given by the rate, which is 2.
Therefore, the expected number of blue flashes is e^(-2) * 2.
Adding the expected number of red flashes and the expected number of blue flashes gives the expected total number of flashes:
Expected total number of flashes = 2 + e^(-2) * 2.
2. Since X is equal to infinity, the blue bulb never dies and keeps flashing indefinitely. In this scenario, the expected value of the time of the first flash is simply the expected value of the minimum of two exponentially distributed random variables.
The expected value of an exponential random variable with rate λ is 1/λ. For the blue bulb, λ = 2, so the expected value of the time of the first flash for the blue bulb is 1/2.
For the red bulb, λ = 1, so the expected value of the time of the first flash for the red bulb is 1/1 = 1.
Taking the minimum of these two values (since we are interested in the time of the first flash regardless of the color), the expected value of the time of the first flash is min(1, 1/2) = 1/2.
3. Given that there are exactly 5 flashes in the time interval [0, X], we will consider the two possibilities for X: 1 and 2.
If X = 1, there can be at most 1 red flash in the interval [0, X], since the blue bulb dies at time 1. The probability of exactly 2 red flashes in this case is 0.
If X = 2, there can be at most 2 red flashes in the interval [0, X]. The probability of exactly 2 red flashes is given by the product of the complementary probabilities of no red flashes and exactly 2 red flashes.
The probability of no red flashes is e^(-λt), where λ = 1 and t = 2, so it is e^(-2).
The probability of exactly 2 red flashes is given by the Poisson probability mass function with rate λ = 1 and k = 2, which is (e^(-λt) * (λt)^k) / k!.
Therefore, the probability that exactly 2 of the 5 flashes were red can be calculated for each possibility of X and then combined using the law of total probability:
Probability that exactly 2 of the 5 flashes were red = P(X = 1) * 0 + P(X = 2) * (e^(-2) * (1*2)^2) / 2!