If we have a Poisson process of rate $x$, and we want to find the joint density of $T_1$ (first arrival time) and $T_2$ (second arrival time), then
$$
\begin{align}
P(T_1 < a,T_2 < b) &= P(T_2 < b \vert T_1 < a)P(T_1 < a)\\
&= P(T_2-T_1 < b-a)P(T_1 < a)\\
&= (1-e^{-x(b-a)})(1-e^{-xa})\quad \textrm{since interarrival times are i.i.d exp(x)}\\
&= 1 – e^{-xa} + e^{-xb} – e^{-x(b-a)}
\end{align}
$$
Then I differentiate this with respect to $a$, then differentiate with respect to $b$, and this gives the density to be
$$
f(a,b) = x^2e^{-x(b-a)}
$$
however the answer should be
$$
x^2e^{-xb}
$$
What is my mistake?
Best Answer
Line 1 is incorrect. $\{T_2-T_1 \le b-a, T_1 \le a\}$ is a subset of $\{T_1 \le a, T_2 \le b \}$. For example $T_1=0$ and $T_2=b$ is not in $\{T_2-T_1 \le b-a, T_1 \le a\}.$ Instead, condition on a given value of $T_1.$ $$P(T_1\le a,T_2\le b)=\int_0^a P(T_2-T_1\le b-t \space | \space T_1=t)f_{T_1}(t)dt$$ This becomes $$\int_0^a (1-e^{-x(b-t)} )xe^{-xt}dt$$ and evaluates as $$1-e^{-xa}-xae^{-xb} $$ on the range $0<a<b.$ If you only need the joint density, you can immediately write it down: $$f_{T_1,T_2}(a,b)= f_{T_1,T_2-T_1}(a,b-a)=f_{T_1}(a)f_{T_2-T_1}(a,b-a)=x^2e^{-xb} , \space 0<a<b. $$ (To be precise, you would need to go through the change of variable formalism with the Jacobean, but you may not have studied that yet.)