Expectation of sum of arrival times of Poisson process in $[0, t]$

Question

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

Compound Poisson process: calculate $E\left( \sum_{k=1}^{N_t}X_k e^{t-T_k} \right)$, $X_k$ i.i.d., $T_k$ arrival time

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.

Answer 1

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.