Let $Y_1,Y_2,\dots , i.i.d, \ \mathbb Z \ $valuable and integrable random variables. $Z_n:= \sum_{i=1}^n Y_i$. Derive $E(Z_n \mid \sigma(Z_k)), k,n \in \mathbb N_0$.
$$E(Z_n \mid \sigma(Z_k))=E(\sum_{i=1}^n Y_i \mid \sigma (Z_k))=n \cdot E(Y_1 \mid \sigma(Z_k))$$ by using the given properties. I am not sure how to continue now, should one consider two cases now ($n \le k$ and $n>k$)? Another point I am wondering is that if $Y_1$ is $\sigma (Z_k)$ measurable? But if so I could use the property that $E(X \mid G)=X$ if $X$ is $G$ measerable. I would appreciate any help!
Your last equation is wrong. Because ofc it matters if $i \le k$ or $i > k$.
For $k\le n$ we have:
$$\begin{align*}E(Z_n \mid \sigma(Z_k))&=E(\sum_{i=1}^n Y_i \mid \sigma (Z_k)) \\ &=E(\sum_{i=1}^k Y_i \mid \sigma (Z_k)) + E(\sum_{i=k+1}^n Y_i \mid \sigma (Z_k)) \\ &= Z_k + \sum_{i=k+1}^nE(Y_i \mid \sigma (Z_k)) \\ &= Z_k + (n-k)E[Y_1]\end{align*}$$ Where the last equation holds by the iid property.
For $n\le k$ by a symmetry argument we get:
$$E(Y_i \mid \sigma (Z_k)) = E(Y_j \mid \sigma (Z_k))$$ for $i,j \le k$ and so $$\begin{align*} kE(Y_i \mid \sigma (Z_k)) &= \sum_{i=1}^k E(Y_i \mid \sigma (Z_k)) \\ &= E(Z_k \mid \sigma (Z_k)) = Z_k\end{align*}$$ hence $$E(Y_i \mid \sigma (Z_k)) = \frac{Z_k}{k}$$ so $$E(Z_n \mid \sigma(Z_k)) = \frac{n}{k}Z_k$$ if $n \le k$