Assuming a Poisson process $N_t$ and denoting wait times $S_k$ (i.e. times until the $k$-th jump), I want to find the expected value of their sum: $$ \mathrm{E}\left[\sum_{k=1}^{N_t} S_k\right]\;. $$ It is an exercise in O. Calin's An Informal Introduction to Stochastic Calculus with Applications and I actually know how to do it. For instance using the integrated Poisson process $$ U_t = \int_0^t N_s \,\mathrm{d}s $$ and the result $$ U_t = tN_t - \sum_{k=1}^{N_t} S_k\;. $$ Taking the expected value of both sides and using its linearity gives $$ \frac12 \lambda t^2 = \lambda t^2 - \mathrm{E}\left[\sum_{k=1}^{N_t} S_k\right], $$ so the answer is $\frac12\lambda t^2$.
The problem is that I can calculate the expected value using a different straighforward method, getting a different (apparently wrong) answer. And the question is why.
In the second method we note that $N_t=n$ for different $n$ are disjoint events. So we can calculate the conditional expectations $$ \mathrm{E}\left[\sum_{k=1}^{N_t} S_k \middle| N_t = n\right] $$ and sum over $n$ with the corresponding proabilities: $$ \mathrm{E}\left[\sum_{k=1}^{N_t} S_k\right] = \sum_{n=0}^\infty \mathrm{Pr}(N_t=n)\times\mathrm{E}\left[\sum_{k=1}^{N_t} S_k \middle| N_t = n\right]\;. $$ Using $$ S_1 + S_2 + S_3 + \dots + S_n = nT_1 + (n-1)T_2 + \dots + 2T_{n-1} + T_n $$ and independence of inter-arrival times $T_k$ we get for the conditional expectation $$ \mathrm{E}\left[\sum_{k=1}^{N_t} S_k \middle| N_t = n\right] = \frac{n(n+1)}{2\lambda}\;. $$ This is in fact the result of another exercise (3.11.4). The probability is just Poisson $$ \mathrm{Pr}(N_t=n) = \mathrm{e}^{-\lambda t} \frac{(\lambda t)^n}{n!}\;. $$ Putting it together: $$ \mathrm{E}\left[\sum_{k=1}^{N_t} S_k\right] = \sum_{n=0}^\infty \mathrm{e}^{-\lambda t} \frac{(\lambda t)^n}{n!} \times \frac{n(n+1)}{2\lambda} = \mathrm{e}^{-\lambda t} \frac t2 \sum_{n=0}^\infty \frac{(\lambda t)^n}{n!} (n+2) = \frac12\lambda t^2 + t\;. $$ So there is an extra $t$.
Now I am stuck and can't find neither an error in the calculation, nor a reason why the summation over $n$ could be wrong.
I am not very proficient in probability so I cannot help you further than this but I think it is helpful:
I believe the incorrect step is when you take the conditional expectation following $$S_{1}+\dots+S_{n} = nT_1+\dots+T_{n}.$$ In these notes from MIT on page 92 we find $$E[T_{i}|N_{t}=n] = \frac{t}{n+1}$$ so that $$E\left[\sum_{k=1}^{N_{t}}S_{k}\bigg| N_{t}=n\right] =\frac{nt}{2}.$$ This then yields the correct result: $$E\left[\sum_{k=1}^{n}S_{k}\bigg|N_{t}=n\right]=e^{-\lambda t}\frac{t^{2}\lambda}{2}\sum_{n=0}^{\infty}\frac{(\lambda t)^{n}}{n!} = \frac{t^{2}\lambda}{2}.$$