Conditional expectation of independent variables

Question

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?

Answer 1

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*}

Answer 2

Let $(Z_1,Z_2)$ be uniformly distributed on $\{(-1,0),(0,1),(1,-1)\}$. Then $Z_1$ and $Z_2$ are identically distributed but dependent. The value of $Z_1+Z_2$ fixes the (different) values of $Z_1$ and $Z_2$, so $\mathbb E[Z_1\mid Z_1+Z_2]\ne\mathbb E[Z_2\mid Z_1+Z_2]$.

Answer 3

In addition to the identical distribution independence is used to obtain that $(Z_1,Z_1+Z_2)$ and $(Z_2,Z_1+Z_2)$ are identically distributed which is needed to conclude $\mathbb E(Z_1 \mid Z_1+Z_2) = \mathbb E(Z_2 \mid Z_1+Z_2)$.

Edit. The distribution of $(Z_1,Z_1+Z_2)$ is the image (push forward) of $\mathbb P^{(Z_1,Z_2)}= \mathbb P^{Z_1} \otimes \mathbb P^{Z_2}$ (here independence is used) under the transformation $(u,v)\mapsto (u+v,v)$.

Answer 4

Just as another remark: a neat corollary which follows from this statement and the strong law of large numbers is that whenever $X_i$ is an iid sequence of r.v. such that the SLLN holds (e.g. $X_i \in L^1$ by Kolmogorov's SLLN), then $\mathbb{E}(X_1 \mid S_n) \to \mathbb{E}(X_1)$ $\mathbb{P}$-almost surely.

Answer 5

If $Z_1$ and $Z_2$ are i.i.d. then they are exchangeable, i.e. $(Z_1,Z_2)$ is distributed as $(Z_2,Z_1)$. Exchangeability is what you need to prove the claim. Indeed, it ensures that $\mathbb{E}(Z_1\mid Z_1+Z_2)=\mathbb{E}(Z_2\mid Z_1+Z_2)$.