conditional expected value - Poisson process plus random variable I've struggled with this actuary excercise for a while and I don't know how to do it:
Each claim can be characterized by two random variables $(T,D)$, where $T$ is the moment of reporting the claim and $D$ describes how long it takes to settle the claim (so  $T+D$ is the moment of settlement of the claim). We numerate claims according to their reporting times, so that $0<T_1<T_2< ...$, of course similar inequalities don't necessary hold for $(T_1+D_1),(T_2+D_2), ..$
Assuming, that:


*

*$T_1, (T_2-T_1), (T_3-T_4), ... $ and $D_1, D_2 ...$ are all independent,

*$T_1, (T_2-T_1), (T_3-T_4), ... $ are exponentially distributed with $\lambda=1$,

*Each $D_i$ can be equal to $1,2,3,4$ with the same probability $\frac{1}{4}$.


Calculate the expected value of time of the settlement, provided that the claim is reported but not yet settled at $t=4$:
$E(D|T<4<T+D)$
I tried to calculate $P(T<4<T+d)$ for fixed d, but I can't find distribution of $T$.
 A: The r.v. $T$ in $T<4<T+D$ refers to any $T_i$ such that $T_i<4<T_i+D_i.$ First we note this condition is the same as $4-D_i<T_i<4.$ If $D_i=d,$ then $4-d<T_i<4$ is Poisson with rate $=\lambda d=d.$
We define a “success” claim of type $d, d=1,2,3,4$ as those values of  $T_i, D_i$ with:
$$(4-d<T_i<4)\cap(D_i=d).$$
By the disaggregation property of Poisson Processes, the success claims are made up of 4 independent Poisson Processes. The rate for type $d$ is $\lambda d \frac 14 = \frac d4,$ given that $\lambda=1.$ By the aggregation property, all the successful claims $\{S_i\}$ form a Poisson process with rate $1/4+1/2+3/4+1=5/2.$
Now we want to select a successful claim at random. What is the probability that this is a type $d$ claim?  Since the 4 types of successful claims are independent Poisson r.v., it is well-known that the probability that an arbitrary element of $\{S_i\}$ is a type $d$ is the ratio of the rates: $\frac{d/4}{5/2}=\frac{d}{10}.$
Therefore the expected value of $D$ given we have selected a successful claim is:
$\sum_{d=1}^4 {d}\frac d{10}=3.$
A: Since $T$ is exponentially distributed with rate parameter $1$. [Exponential distribution is memoryless.]
$\begin{align}
\mathsf P(4-n < T < 4) & = (1-e^{-4})-(1-e^{n-4})
\\[1ex]
& = (e^{n}-1)e^{-4} 
\end{align}$
