Joint density function Poisson Process We did an example in class that I'm not sure how we came up with the answer.
The problem is:
If I let X(t) be a Poisson process of rate $\lambda$.  I'm supposed to validate the identity 
{$W_1>w_1, W_2>w_2$} if and only if {$X(w_1)=0,X(w_2)-X(w_1)=0  \hspace{3pt}\textrm{or}\hspace{3pt}  1$}
We were supposed to use this to determine the joint upper tail probability
Pr{$W_1>w_1, W_2>w_2$} = Pr{$X(w_1)=0,X(w_2)-X(w_1)=0\hspace{3pt}\textrm{or}\hspace{3pt}   1$}
=$e^{-\lambda w_1}[1+\lambda(w_2 - w_1)]$ $e^{-\lambda (w_2-w_1)}$
Then we were supposed to differentiate twice to obtain the joint density function:
$f(w_1,w_2)=\lambda^2$ $exp(-\lambda w_2)$ for $0<w_1<w_2$.
In class, all he wrote was:
$$\int_0^1\int_0^1 f_{w_1,w_2} (w_1',w_2') \,\mathrm dw_1'\,\mathrm dw_2'$$
=$[1 + \lambda(w_2-w_1)]$ $e^{-\lambda w_2}$
I'm not sure how he went from the integral to the answer, any suggestions?
Thanks in advance!
 A: Obviously, 
$$P(W_1 > w_1, W_2 > w_2 ) = \int_{0}^{1}\int_{0}^{1}f(x, y)\ \text dx\text dy$$ 
cannot be correct because the lhs is a function of $w_1$ and $w_2$ while the rhs, being a real number, isn't.
Furthermore, recall that by the definition of probability density functions, if $f(w_1, w_2) = \lambda^2e^{-\lambda w_2}{\bf 1}_{\{0<w_1<w_2\}}$ is the joint density function for the random variables $W_1, W_2$, then

$$
P(W_1 > w_1, W_2 > w_2 ) = \int_{w_2}^{\infty}\int_{w_1}^{\infty}f(x, y)\ \text dx\text dy 
$$

Indeed, for $0<w_1<w_2$, 
$$
\begin{eqnarray*}
\int_{w_2}^{\infty}\int_{w_1}^{\infty}f(x, y)\ \text dx\text dy &=&  \int_{w_2}^{\infty}\int_{w_1}^{\infty}\lambda^2e^{-\lambda y}{\bf 1}_{\{0<x<y\}}\text dx\text dy \\&=& \lambda^2\int_{w_2}^{\infty}e^{-\lambda y}\Big(\int_{w_1}^{\infty}{\bf 1}_{\{0<x<y\}}\text dx\Big)\text dy \\
&=& \lambda^2\int_{w_2}^{\infty}e^{-\lambda y}(y-w_1)\text dy \\ 
&& \\
&=& e^{-\lambda w_2}(1+\lambda w_2) - \lambda w_1 e^{-\lambda w_2} \\
&& \\
&=& e^{-\lambda w_2} (1 + \lambda(w_2 - w_1))\,,
\end{eqnarray*}
$$
which is the form of $P\Big(W_1 > w_1, W_2 > w_2 \Big)$ deduced from the Poisson process $\{X_t\}_{t\geqslant0}$. This makes obvious that $\int_{0}^{1}\int_{0}^{1}f(x, y)\ \text dx\text dy$ is not relevant for our immediate computations, and certainly cannot evaluate as suggested in your question.
