Given a Poisson process with rate $\lambda$, the following is to be determined: $E[N_3|N_1=3]$ and $E[\sum_{i=1}^6S_i|N_1=3]$ where $S_i$ is the interarrival time.
For the first question, I know that the process has independent increments, so $N_1$ of events in [0,1) is independent of $N_3-N_1$. Then I think I basically have to calculate the increments in the interval between $t=1$ and $t=3$ using $E[N_t-N_s]=\lambda(t-s)$ and adding that to $N_1$. So $E[N_3|N_1=3]=3+2\lambda$.
Would this be correct? I have no idea how to approach the second expectation.
Ok we can write: $$E[\sum_{i=1}^6 S_i|N_1=3]=E[\sum_{i=1}^4 S_i|N_1=3]+E[S_5+S_6|N_1=3]$$ now note that $S_5,S_6$ do not depend on the event that at time $1$ we already have $3$ arrival (the fact that $N_1=3 $ doesn't tell nothing about how much time we will have to wait from the fourth to the fifth arrival and the same is valid for the sixth).
Consider $\sum_{i=1}^4 S_i$ the time of the fourth arrival (from the start) and $S_4$ the time between the third and the fourth arrival. Now we know that at time $t=1$ there have been already $3$ arrival but not $4$, and we also know that we have the memoryless property of intertime arrival (they are distributed as iid exponential) so the fact that we have waited some amount of time doesn't tell us nothing about how much more time we will have to wait which is $E[S_4|S_4>t]=t+ E[S_4]$.
In formula:
$E[\sum_{i=1}^4 S_i|N_1=3]=E[\sum_{i=1}^4 S_i|\sum_{i=1}^3S_i \leq 1 ,S_4 > 1-\sum_{i=1}^3S_i]=1+E[S]$ where $S$ has the same distribution of the interarrival times.
So we have: $$E[\sum_{i=1}^6 S_i|N_1=3]=E[\sum_{i=1}^4 S_i|N_1=3]+E[S_5+S_6|N_1=3]=1+\frac{1}{\lambda}+E[S_5+S_6]=1+\frac{3}{\lambda}.$$
The only difficult part is to notice that we have already waited $1$ and that we expect to wait $E[S]$ more for the fourth arrival, in fact even if we could have already waited $x$ for the fourth arrival that does not tell us that then "the item will arrive sooner".