Claim. Let $Z_1, Z_2$ be two independent and identically distributed random variables. Then we have:
$$
\mathbb E[Z_1|Z_1+Z_2] =\frac{Z_1+Z_2}{2}.
$$
Proof.
To see this, I have proceeded as follows. From the general properties of conditional expectation, we have:
$$
\mathbb E[Z_1+Z_2|Z_1+Z_2] =Z_1+Z_2.
$$
Now, again from the general properties of c.e. (linearity), I can write:
$$
\mathbb E[Z_1+Z_2|Z_1+Z_2] =\mathbb E[Z_1|Z_1+Z_2] +\mathbb E[Z_2|Z_1+Z_2] =2\mathbb E[Z_1|Z_1+Z_2]
$$
The last equality because $Z_1$ has the same distribution as $Z_2$. Putting the two equalities together we find immediately the identity stated in the Claim.
Question What I don't find clear is where exactly the condition of independence of the two random variables is used. I know that the result is not true if they aren't independent, but I don't see where this conditoin is needed. The only point I can think of is in the last step of the chain of equalities, where we use the fact that $\mathbb E[Z_1|Z_1+Z_2]=\mathbb E[Z_2|Z_1+Z_2]$ but it seems to me that this holds just beacuse of the same distribution of $Z_1, Z_2$ and does not require independence.
And yet, the result is not true if $Z_1, Z_2$ are not independent. So, where is the independece condition used?
Best Answer
We need only show that, for any Borel set $A \in \mathbb{R}$, \begin{align*} \int_{Z_1+Z_2 \in A} Z_1 dP = \int_{Z_1+Z_2 \in A} Z_2 dP. \end{align*} We denote by $F$ the common cumulative distribution function of $Z_1$ and $Z_2$. Then, from the independence assumption, \begin{align*} \int_{Z_1+Z_2 \in A} Z_1 dP &= E\left(Z_1 \pmb{1}_{Z_1+Z_2 \in A} \right)\\ &=\iint_{\mathbb{R}^2} x \pmb{1}_{x+y \in A} dF(x) dF(y). \end{align*} Analogously, \begin{align*} \int_{Z_1+Z_2 \in A} Z_2 dP &= E\left(Z_2 \pmb{1}_{Z_1+Z_2 \in A} \right)\\ &=\iint_{\mathbb{R}^2} u \pmb{1}_{v+u \in A} dF(u) dF(v)\\ &=\iint_{\mathbb{R}^2} u \pmb{1}_{u+v \in A} dF(u) dF(v). \end{align*} That is, \begin{align*} \int_{Z_1+Z_2 \in A} Z_1 dP = \int_{Z_1+Z_2 \in A} Z_2 dP. \end{align*} In other words, \begin{align*} E\left(Z_1 \mid Z_1+Z_2 \right) = E\left(Z_2 \mid Z_1+Z_2 \right). \end{align*}