Joint density function Poisson Process

Question

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!

Answer 1

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.