Joint pdf of Poisson processes

Question

I recently stumbled upon this question:

Let T1 and T2 be the times for the first and second jump of a Poisson process with rate λ > 0. Find the joint probability density function of T1 and T2

After formulating that:

$ P(T_1 <= s, T_2 <= t) = P(X(s)>=1, X(t)>=2) = ... = (1-P(X(s) =0)) * (1 - P(X(t-s) = 0) = 1 - e^{- \lambda s }- e^{- \lambda (t-s) } + e^{- \lambda t }$

and then taking the derivative with respect to t and s to get the joint pdf I end up with

$ f(s,t) = \lambda^2 e^{-\lambda (t-s) }$

however the correct answer should be apparently be $ f(s,t) = \lambda^2 e^{-\lambda t }$

does anyone know where I do my mistake?

Thanks in advance

Answer 1

You've missed cases where $X(s) \ge 2$ while $X(t-s) = 0$.

\begin{align} P(T_1 \le s, T_2 \le t) &= P(X(s) \ge 1, X(t) \ge 2) \\ &= P(X(s) = 1, X(t-s) \ge 1) + P(X(s) \ge 2) \\ &= P(X(s) = 1) P(X(t-s) \ge 1) + P(X(s) \ge 2) \\ &= \lambda s e^{-\lambda s} (1 - e^{-\lambda(t-s)}) + (1 - e^{-\lambda s} - \lambda s e^{-\lambda s}) \\ &= -\lambda s e^{-\lambda t} + 1 - e^{-\lambda s}. \end{align}