Question

Let \(\{N(t),t\geq 0\}\) be a Poisson process with rate \(\lambda=1\). Then \(E[N(8)N(5)]\) equals (in integer).

Hint:

For a Poisson process, increments over disjoint intervals are independent. Always split \(N(t)\) as \(N(s)+[N(t)-N(s)]\) when \(t>s\).

Answer 1

Correct Answer: 45

Step 1: Use independent increments of Poisson process.
For a Poisson process,
\[ N(8)=N(5)+\bigl(N(8)-N(5)\bigr) \] Let
\[ M=N(8)-N(5) \] Then \(M\) is independent of \(N(5)\).
Also, since the rate is \(\lambda=1\),
\[ N(5)\sim Poisson(5) \] and
\[ M\sim Poisson(3) \]

Step 2: Expand the required expectation.
\[ E[N(8)N(5)] = E\left[\left(N(5)+M\right)N(5)\right] \] \[ = E[N(5)^2]+E[MN(5)] \] Since \(M\) and \(N(5)\) are independent,
\[ E[MN(5)] = E[M]\cdot E[N(5)] \]

Step 3: Compute the terms separately.
For \(N(5)\sim Poisson(5)\),
\[ E[N(5)]=5 \] and
\[ Var(N(5))=5 \] Therefore,
\[ E[N(5)^2] = Var(N(5))+\{E[N(5)]\}^2 \] \[ = 5+25 \] \[ = 30 \] Also,
\[ E[M]=3 \] Hence,
\[ E[MN(5)] = 3\times 5 = 15 \]

Step 4: Final calculation.
\[ E[N(8)N(5)] = 30+15 = 45 \]

Step 5: Final conclusion.
Hence,
\[ \boxed{45} \]