Hi I am having a trouble solving for this problem where I have to find
1) Joint distribution of $W_{1}$, $W_{r}$ for $r\geq2$.
2) $\operatorname{Cov}(W_{1},W_{r})$ for $r\geq2$.
[Notation explanation: $W_{r}=\min(t:N_{t}\geq r) $ is waiting time until the $r^\text{th}$ occurrence.
Here, $(N_{t})$, ${t\geq0}$ is a Poisson process with occurrence rate $\lambda>0$.]
How should I solve this? I did solve the case when $r=2$. In that case $\operatorname{pdf}_{W_{1},W_{2}}(t_{1},t_{2})=\lambda^2e^{-\lambda t_{2}}I(_{0<t_{1}<t_{2}})$
But as I tried to generalize it to case $r$, it became quite complicated and I got lost. I would really appreciate if someone could help.
Thanks.
Let $X_n = W_n - W_{n-1}$ for $n\geq 2$ and $X_1 = W_1.$ That is, $X_n$ are the interarrival times. We have that $X_n\overset{iid}{\sim} \text{Exp}(\lambda).$ Now denote the joint pdf of $(W_1, W_r)$ by $f_{1,r}.$ Then $$f_{1,r}(t_1, t_r) = f_{r|1}(t_r|t_1)\cdot f_{1}(t_1),$$ where $f_{r|1}$ is the pdf of $W_r$ conditional on $W_1$ and $f_1$ is the (marginal) pdf of $W_1.$ Then $f_1(t_1) = \lambda e^{-\lambda t_1}1\{t_1\geq 0\}.$ Now write $$W_r = X_1 + \sum_{i=2}^r X_i.$$ Then note that $X_1$ and $\sum_{i=2}^r X_i$ are indepdent and that the latter is $\Gamma(r-1, \lambda)$ distributed. Using that $X_1=W_1$, we have that conditional on $W_1,$ $W_r$ is the sum of $W_1$ and a $\Gamma(r-1, \lambda)$-distributed random variable. The conditional pdf is then a shifted Gamma pdf: $$f_{r|1}(t_r|t_1) = \frac{\lambda^{r-1}(t_r-t_1)^{r-2}}{\Gamma(r-1)}e^{-\lambda (t_r-t_1)} 1\{t_r\geq t_1\}.$$
So the product is $$f_{1r}(t_1, t_r) = \frac{\lambda^{r}(t_r-t_1)^{r-2}}{\Gamma(r-1)}e^{-\lambda t_r} 1\{t_r\geq t_1\geq 0\}.$$
For the covariance, note that $$Cov(W_1, W_r) = Cov(X_1, X_1) + \sum_{i=2}^n Cov(X_1, X_i) = Var(X_1) = \frac{1}{\lambda^2},$$ using independence of the $X_i.$