Question

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.

a) 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]:

b)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:

Answer 1

a)

We are considering the time interval [0,2].

The red bulb flashes with the rate 1*2=2

If X=1, then the blue bulb flashes within the time interval [0,1].

Thus the rate will be 2*1=2.

If X=2, then the blue bulb flashes within the time interval [0,2].

Thus the rate will be 2*2=4.

Since it is given that it is a poisson process, the pdf of poisson distribution is given by,

When X=1

P(exactly 3 arrivals)= [P(3 red flash)*P(0 blue flash)]+[P(2 red flash}*P(1 blue flash)]+[P(1 red flash)*P(2 blue flash)]+[P(0 red flash)*P(3 blue flash)]

= [0.18*0.135]+[0.073*0.271]+[0.271*0.271]+[0.135*0.18]

= 0.195.

When X=2

P(exactly 3 arrivals)=0.089.

P(exactly 3 arrivals)= [P(X=1)*P(exactly 3 arrivals| X=1)]+[P(X=2)*P(exactly 3 arrivals| X=2)]

= [0.5*0.195]+[0.5*0.089]

= 0.142.

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