Question

Let {X(t); t >= 0) be a Poisson process having rate parameter lambda = 1. For the random variable, X(t), the number of events occurring in an interval of length t. Determine the following.

(a) Pr(X(3.7) = 3|X(2.2) >= 2)

(b) Pr(X(3.7) = 1|X(2.2) <2)

(c) E(X(5)|X(10) = 7)

Answer 1

We are given the distribution here as:

$X \sim Poisson(\lambda = 1)$

a) Note that poisson distribution follows memoryless property,

The probability that P(X(3.7) = 3 | X(2.2) >= 2) is computed using Bayes theorem here as:

$P(X (3.7) = 3 | X(2.2) \geq 2) = \frac{P(X(3.7) = 3, X(2.2)\geq 2)}{P(X(2.2) \geq 2)}$

$P(X (3.7) = 3 | X(2.2) \geq 2) = \frac{P(X(2.2) = 2)P(X(3.7 - 2.2) = 1) +P(X(2.2) = 3)P(X(3.7 - 2.2) = 0) }{1 - P(X(2.2) = 0) - P(X(2.2) = 1)}$$P(X (3.7) = 3 | X(2.2) \geq 2) = \frac{P(X(2.2) = 2)P(X(1.5) = 1) +P(X(2.2) = 3)P(X(1.5) = 0) }{1 - P(X(2.2) = 0) - P(X(2.2) = 1)}$

Now using Poisson probability function, we get here:

$P(X (3.7) = 3 | X(2.2) \geq 2) = \frac{\frac{2.2^2}{2} e^{-2.2}*1.5e^{-1.5}}{1 - e^{-2.2} - 2.2e^{-2.2}}$

$P(X (3.7) = 3 | X(2.2) \geq 2) = \frac{0.0897}{0.6454} = 0.1390$

Therefore 0.1390 is the required probability here.

b) The probability here is computed in similar way as:

$P(X(3.7) = 1 | X (2.2) < 2) = \frac{P(X(2.2) = 1)P(X(1.5) = 0) + P(X(2.2) = 0)P(X(1.5) = 1)}{P(X(2.2) = 0) + P(X(2.2) = 1)}$

$P(X(3.7) = 1 | X (2.2) < 2) = \frac{2.2e^{-2.2}e^{-1.5} + 1.5e^{-1.5}*e^{-2.2}}{e^{-2.2} + 2.2e^{-2.2}}$

$P(X(3.7) = 1 | X (2.2) < 2) = \frac{2.2e^{-1.5} + 1.5e^{-1.5}}{3.2} = 0.2580$

Therefore 0.2580 is the required probability here.

c) The expected value of X(5) is computed by using the fact that the arrivals should be uniformly distributed across the time range of 10.

Therefore E(X(5) | X(10) = 7) = 7/2 = 3.5

Therefore 3.5 is the expected number of process here.