Let $X$ a Poisson distribution with parameter $3$, Y have an exponential distribution with parameter $1$, $Z$ have a uniform distribution on $[1,3]$, and all these random variables are independent. Find the mathematical expectation of the random variable $\mathbb E(\mathbb E(X+Y+Z\mid X-Y+Z)\mid X+Y-Z )$
I guess I can use the tower property $\mathbb E[X] = \mathbb{E}[\mathbb E [X | Y]]$. But I can't see next steps.
It would be an extremely challenging problem to calculate: $$\mathbb{E}(\mathbb{E}(X+Y+Z\mid X-Y+Z)\mid X+Y-Z)$$
Fortunately, the problem as stated is asking for the unconditional expectation of that doubly iterated conditional expectation, so we are looking for: $$\mathbb{E}(\mathbb{E}(\mathbb{E}(X+Y+Z\mid X-Y+Z)\mid X+Y-Z))$$
Applying the tower rule to the outer pair of iterated expectations removes one condition: $$\mathbb{E}(\mathbb{E}(X+Y+Z\mid X-Y+Z))$$ and reapplying the tower rule to the remaning pair of iterated expectations gives: $$\mathbb{E}(X+Y+Z)$$ The answer then follows from the means of the distributions. Note that independence is not actually needed.
If the above is unclear, note that we've really just applied the tower rule $E(E(T\mid W))=E(T)$ to the random variable $T=E(U\mid V)$ to get: $$E(E(E(U\mid V)\mid W))=E(E(U\mid V))$$ and a second application of the tower rule to the right-hand side gives the following identity, true for any random variables $U$, $V$, and $W$: $$E(E(E(U\mid V)\mid W)) = E(U)$$ In particular, taking $U=X+Y+Z$, etc., gives the result.