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?
Show 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).
Step 1: Identify the marginal law of T1.
In a Poisson process of rate $\mu=3$, the gaps between arrivals are iid $\text{Exponential}(\mu)$, so $T_1\sim\text{Exponential}(3)$ and $E(T_1)=1/\mu=1/3$. This rules out (A) and confirms (B).
Step 2: Build the conditional density of T1 given N(t)=m.
Given $N(t)=m$, the joint law of the m arrival times matches m iid $\text{Uniform}(0,t)$ points sorted in order. The smallest of m iid $\text{Uniform}(0,t)$ points has density
\[ f_{(1)}(x)=\frac{m}{t}\left(1-\frac{x}{t}\right)^{m-1},\qquad 0<x<t. \]
Step 3: Integrate to get the mean.
\[ E(T_1\mid N(t)=m)=\int_0^t x\cdot\frac{m}{t}\left(1-\frac{x}{t}\right)^{m-1}dx. \]
Substituting $u=1-x/t$ turns this into a standard Beta-type integral, and it works out to the known closed form $\dfrac{t}{m+1}$.
Step 4: Plug in the numbers.
For (C), with $t=2,m=4$: $E(T_1\mid N(2)=4)=\dfrac{2}{5}$, not $\dfrac{6}{5}$, so (C) is FALSE.
For (D), with $t=3,m=4$: $E(T_1\mid N(3)=4)=\dfrac{3}{5}$, which matches exactly, so (D) is TRUE.
Final Answer:
The minimum of the uniform order statistics gives mean t/(m+1), confirming (B) and (D).
\[ \boxed{\text{(B) and (D)}} \]
