Consider the following exercise
If $X_1,\dots,X_n$ are independent with same distribution and $S_n=X_1+\cdots+X_n$, prove that
$$\mathbb{E}[X_i \mid S_n]=\frac{S_n}{n}$$
Proof: Since the variables have the same distribution we have that:
$$S_n=\mathbb{E}[S_n \mid S_n]=\mathbb{E}\left[\sum_{i=0}^n X_i \mid S_n \right]=\sum_{i=0}^n\mathbb{E}[X_i \mid S_n]=n \cdot \mathbb{E}[X_i \mid S_n]$$
I'm trying to figure out why the fact that the variables have the same distribution implies that they have the same conditional expectation with respect to $S_n$. I tried to see if the expected values of the variables on the sets of the form $S_n^{-1}(A)$ (with $A\subseteq \mathbb{R}$ borel set) are equal (I mean that $\mathbb{E}[X_1\mathbf{1}_{S_n^{-1}(A)}]=\mathbb{E}[X_i\mathbf{1}_{S_n^{-1}(A)}]$), because this fact would give me the thesis, but I don't know how to do it. Any suggestion on how to demonstrate the equality between the conditional expectations?
Best Answer
The exercise is a direct application of the following proposition, with $X=(X_1,X_2,\cdots,X_n)$ and $Y=(X_{i_1},X_{i_2},\cdots,X_{i_n})$ :
If $X=(X_1,X_2,\cdots,X_n)$ and $Y=(Y_1,Y_2,\cdots,Y_n)$ are identically distributed and $f$ and $g$ are measurable functions, then: $$ E[f(X)\mid g(X)=z]=E[f(Y)\mid g(Y)=z] \tag{1}$$ Proof: $$\int E[f(X)\mid g(X)]I_A(g(X))dP_{\Omega}=\int f(X)I_A(g(X))dP_{\Omega} $$ But: $$\int E[f(X)\mid g(X)]I_A(g(X))dP_{\Omega}=\int E[f(X)\mid g(\mathbf x) ]I_A(g(\mathbf x))dP_X$$ And: $$\int f(X)I_A(g(X))dP_{\Omega}=\int f(\mathbf x)I_A(g(\mathbf x))dP_X$$ Therefore, for $(1)$ to be valid, it is enough that: $$\int E[f(X)\mid g(\mathbf x)]I_A(g(\mathbf x))dP_X=\int f(\mathbf x)I_A(g(\mathbf x))dP_X$$ But this last equation, and therefore $E[f(X)\mid g(\mathbf x)]$, only depends on the distribution of $X$.