Question

Let \( \{ N(t): t \geq 0 \} \) be a homogeneous Poisson process with the intensity/rate \( \lambda = 2 \). Let} \[ X = N(6) - N(1), \quad Y = N(5) - N(3), \quad W = N(6) - N(5), \quad Z = N(3) - N(1). \] Then which one of the following options is correct?

• A. \( {Cov}(W, Z) = 2 \)
• B. \( Y + Z \sim {Poisson}(10) \)
• C. \( \Pr(Y = Z) = 1 \)
• D. \( {Cov}(X, Y) = 4 \)

Hint:

For Poisson processes, increments in disjoint intervals are independent. Use this property to calculate covariances and verify the distributions of sums of Poisson random variables.

Answer 1

The Correct Option is D

The Poisson process has independent increments, meaning that the random variables corresponding to the number of events in disjoint time intervals are independent. 
Step 1: Analyzing the increments.
The increments \( X = N(6) - N(1) \), \( Y = N(5) - N(3) \), \( W = N(6) - N(5) \), and \( Z = N(3) - N(1) \) all correspond to the number of events in specific time intervals. \( X \) counts the number of events in the interval \( [1, 6] \), with length 5.
\( Y \) counts the number of events in the interval \( [3, 5] \), with length 2.
\( W \) counts the number of events in the interval \( [5, 6] \), with length 1.
\( Z \) counts the number of events in the interval \( [1, 3] \), with length 2. Each of these variables follows a Poisson distribution with parameter \( \lambda \times {length of the interval} \), so:
\( X \sim {Poisson}(10) \)
\( Y \sim {Poisson}(4) \)
\( W \sim {Poisson}(2) \)
\( Z \sim {Poisson}(4) \)
Step 2: Covariance of \( X \) and \( Y \). The covariance of \( X \) and \( Y \) is calculated by considering the overlap between the time intervals of \( X \) and \( Y \). Both \( X \) and \( Y \) share the interval \( [3, 5] \). The covariance of two Poisson random variables with overlapping intervals is equal to the length of the overlapping interval times the rate \( \lambda \). The length of the overlapping interval is 2, and the rate \( \lambda = 2 \), so: \[ {Cov}(X, Y) = \lambda \times {overlap length} = 2 \times 2 = 4 \] Thus, the correct answer is \( \boxed{(D)} \). 
Step 3: Verifying other options.
Option (A) is incorrect because \( {Cov}(W, Z) = 0 \), as \( W \) and \( Z \) correspond to independent intervals.
Option (B) is incorrect because \( Y + Z \sim {Poisson}(8) \), not \( {Poisson}(10) \).
Option (C) is incorrect because \( Y \) and \( Z \) are independent, so \( \Pr(Y = Z) \neq 1 \).