Discussion
Let \(\{X(t), t \ge 0\}\) be a time-homogeneous Poisson process with rate \(\lambda\). Let T be a positive continuous r.v. distributed independently of \(\{X(t), t \ge 0\}\). Then :
Answer: C
\(\mathrm{Cov}(X(T), T) = \) \(\lambda\,\mathrm{Var}(T)\).
Condition on T: given \(T = t\), \(X(t)\) is Poisson with mean \(\lambda t\), so \(E[X(T) \mid T] = \lambda T\). Then
\[ \begin{aligned} \mathrm{Cov}(X(T), T) &= E[T\,E(X(T) \mid T)] - E[X(T)]E[T] \\ &= \lambda E[T^{2}] - \lambda (E[T])^{2} = \lambda\,\mathrm{Var}(T) \end{aligned} \]
The covariance is non-negative, so D is wrong; it depends on the distribution of T, so a constant 1/2 is wrong; and \(\lambda/(\lambda + E(T))\) has no basis in this calculation.
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