Consider $(X_n)_{n\geqslant1}$ i.i.d. Poisson with parameter $1$, $S_n = \sum\limits_{i=1} ^nX_i$ for every $n\geqslant1$, and $N_k = \min \{n : S_n > k\}$ for every $k\geqslant0$. Find $E[N_1| N_0]$.
I understand that if we reached $N_0$ with k = 1, then the expected value is distributed geometrically with parameter $ 1 - P(X_i = 0)$, plus $N_0$ because we need one more. But the problem is that we can reach both "milestones" when reaching the first one (if we get more than 1). So I don't know how to take that into account.
For every $n\geqslant1$, $$[N_0=n]=[X_n\ne0]\cap\bigcap_{k=1}^{n-1}[X_k=0]$$ hence, on the event $[N_0=n]$, $N_1=n$ on $[X_n\geqslant2]$ and $N_1=n+N'_0$ on $[X_n=1]$, where $$N'_0=\min\{k>n\mid X_k\ne0\}$$ Thus, on each $[N_0=n]$, $$N_1=n+N'_0\mathbf 1_{X_n=1}=N_0+N'_0\mathbf 1_{X_n=1}$$ where $N'_0$ is independent of $N_0$ and distributed as $N_0$. Furthermore, $$P(X_n=1\mid N_0=n)=P(X_n=1\mid X_n\ne0)=\frac{e^{-\lambda}\lambda}{1-e^{-\lambda}}$$ for every $n$, thus call the RHS $p$, and $$E(N_0)=\frac1{P(X_n\ne0)}=\frac1{1-e^{-\lambda}}$$ Finally, $$E(N_1\mid N_0)=N_0+E(N_0)p=N_0+\frac{e^{-\lambda}\lambda}{(1-e^{-\lambda})^2}$$