Assume that $X_j$ and $Y_j$ are both independent poisson distributed RV with the same rate $\lambda>0$ for all $j=0,1,2,....$.
Now define $U_j$ such that $U_j(\omega)=Y_{X_{j}(\omega)}(\omega)$ for all $\omega \in \Omega$.
I now want to find $\mathbb{E}[U_j]$.
My approach has been the following argument:
Since $X_j\sim Pois(\lambda)$ and IID, we know that for some $j\in \{0,1,2...\}$ that $X_j\in \{0,1,2...\}$, hence will we have that:
$\mathbb{E}[U_j]=\mathbb{E}[Y_{X_j}]=\mathbb{E}[Y_k]=\lambda$ for some $k\in \{0,1,2...\}$ since all $Y_j$ and $X_j$ are IID.
I just feel that this is a bit to easy....
Careful, $X_j$ is a random variable that is nonconstant as $\omega$ varies.
I think there is no way around it: You have to compute. This is done as follows:
$$ \mathbb E[Y_{X_j}] = \sum_{k=0}^n \mathbb P(X_j = k) \mathbb E[Y_k|X_j = k] \overset{\text{independence}}{=} \lambda \sum_{k=0}^\infty \frac{\lambda^k e^{-\lambda}}{k!} = \lambda. $$
(Apparently, your solution was correct anyway.)