Let $T_i$ be the $i$th arrival time of Poisson process of arrival rate $\lambda$, given $t>0$, how to calculate
$$
E\left(\sum_{i=1}^\infty T_i 1_{\{T_i<t\}}\right)
$$
I think since this is equal to
$$
E\left(\sum_{i=1}^{N_t} T_i\right)
$$
We can try
$$
E\left(E\left(\left. \sum_{i=1}^{N} T_i \right|N_t=N\right)\right)
=E\left(\sum_{i=1}^{N} E(T_i|N_t=N)\right)
$$
But I'm stuck at this step. Also making me more confused is that when I see a similar question on
There is a claim
Given $N_t$, the inter arrival times are uniformly distributed on $[0,t]$.
Hence, $T_k∼Beta(k,n−k+1)$
But support of beta distribution is just $[0,1]$, I don't understand why this claim is right.
Best Answer
That phrase is almost correct it should say
"Given $N_t$, the inter arrival times are uniformly distributed on $[0,t]$. Hence, $T_k/t∼Beta(k,n−k+1)$"
Please notice that is $T_k/t$ and not $T_k$ that is distributed as Beta. If you normalize the times by $t$ you'll get that the inter arrival times are uniformly distributed on $[0,1]$ and thus the distribution of $T_k/t$ is indeed Beta (do it as an excercise or check this)
Now with that in mind we have $$ E(T_k/t | N_t = N) = \frac{k}{N+1} $$ so $$ \sum_{k=1}^N E(T_k | N_t = N) = \sum_{k=1}^N t \, E(T_k/t | N_t = N) = t \frac{N}{2} $$
and I think you can take it from here.