Question

Let \(\{N(t);t\geq0\}\) be a homogeneous Poisson process with rate \(3\), and let \(T_1\) denote the first arrival time. Then which of the following statements is/are correct?

• A. \(E(T_1)=3\)
• B. \(E(T_1)=\dfrac{1}{3}\)
• C. \(E\big(T_1\mid N(2)=4\big)=\dfrac{6}{5}\)
• D. \(E\big(T_1\mid N(3)=4\big)=\dfrac{3}{5}\)

Hint:

T1 is exponential with mean 1 over the rate unconditionally; given N(t)=m, T1 is the minimum of m iid Uniform(0,t) variables with mean t/(m+1).

Answer 1

The Correct Option is B, D

Step 1: Find the unconditional mean of T1.
For a Poisson process with rate 3, the first arrival time \(T_1\) has an Exponential distribution with rate 3, so its mean is
\[ E(T_1)=\frac{1}{3}. \]
So (A) is FALSE and (B) is TRUE.

Step 2: Recall the order statistics property of a Poisson process.
Given \(N(t)=m\), the m arrival times up to time t have the same distribution as the order statistics of m independent \(\text{Uniform}(0,t)\) random variables. In particular, \(T_1\) given \(N(t)=m\) is the minimum of m iid \(\text{Uniform}(0,t)\) variables.

Step 3: Use the minimum-of-uniforms formula.
For m iid \(\text{Uniform}(0,t)\) random variables, the expected minimum is
\[ E(\text{minimum})=\frac{t}{m+1}. \]

Step 4: Check (C).
Here \(t=2,\ m=4\), so
\[ E\big(T_1\mid N(2)=4\big)=\frac{2}{4+1}=\frac{2}{5}. \]
This does not equal \(6/5\), so (C) is FALSE.

Step 5: Check (D).
Here \(t=3,\ m=4\), so
\[ E\big(T_1\mid N(3)=4\big)=\frac{3}{4+1}=\frac{3}{5}. \]
This matches exactly, so (D) is TRUE.

Final Answer:
T1 has mean 1/3 unconditionally, and given N(3)=4 its conditional mean works out to 3/5. \[ \boxed{\text{(B) and (D)}} \]