Variable intensity with a Poisson Process? Customers arrive to a store according to the Poisson process $X={X(t): t>0}$ with the intensity $\lambda(t)=2t$.


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*Given that by the ending time $T$ the process has encountered $X(T)=10$ arrivals, find the conditional density for $W_1$, the first arrival time.

*Given that by $T$, the process has encountered $X(T)=10$ arrivals, find the joint conditional density for the first and last arrival times.



I recognize that this is a non-homogeneous Poisson process. 
There is a property that 
$$P[X(t+h)-X(t)=1]=\frac{(\lambda h)e^{-\lambda h}}{1!}$$
$$=\lambda h+o(h)$$
Can someone help me with setting up this problem in this context?
Thanks for any guidance. 
 A: For any $t>0$, let $$\Lambda(t) = \int_0^t \lambda(s)\ \mathsf ds. $$ Then $X(t)$ has $\mathsf{Pois}(\Lambda(t))$ distribution (where $\Lambda(t)=t^2$)). Now, the random variables $\{W_n\}$ are the arrival times of $X(t)$ if and only if $\{\Lambda(W_n)\}$ are the arrival times of a homogeneous Poisson process $\{N(t):t\geqslant 0\}$ with rate $1$. So conditioned on $X(T)=10$, the joint distribution of $\left(W_{1}^2, \ldots, W_{10}^2\right)$ is that of $\left(U_{(1)}, \ldots, U_{(10)}\right)$ where $U_n\stackrel{\mathrm{i.i.d.}}\sim\mathsf U(0,T^2)$. If $F$ is the cdf of $U_1$ then $U_{(1)}$ has density
$$f_{(1)}(t) = 10F'(t)(1-F(t))^9= \frac{10}{T^2}\left(1-\frac t{T^2}\right)^9\ \mathsf 1_{(0,\ T^2)}(t) $$
and $(U_{(1)}, U_{(10)})$ have joint density
$$f_{(1),\ (10)}(s,t) = \frac{90}{T^{20}}\left(t-s \right)^8\mathsf 1_{(0,t)\times(0,T^2)}(s,t). $$
Since the $W_n$ are nonnegative, $$\{W_n^2 \leqslant t^2\}=\{W_n\leqslant t\} $$ and we find the density of $W_1$ conditioned on $X(T)=10$ to be $$f_{W_1\mid N(T)=10}(t) = \frac{20}{T^2}t\left(1-\frac {t^2}{T^2}\right)^9\mathsf 1_{(0,T)}(t) $$ and the joint density of $(W_1, W_{10})$ conditioned on $X(T)=10$ to be
$$f_{(W_1, W_{10})\mid N(T)=10} = \frac{360}{T^{20}}st(t^2-s^2). $$
