Conditional expectation of function from $\mathbb{R}^n$ to $\mathbb{R}$

Question

Let $f$ be a 1-Lipschitz function from $\mathbb{R}^n$ to $\mathbb{R}$, and let $\{X_i\}$ be an independent sequence of random variables, each with variance $\leq 1$.

Suppose we are given a function $g(x_1, \dots, x_{k}) := E\left(f\left(x_1, \dots, x_{k},X_{k+1}, \dots, X_{n}\right)\right)$ from $\mathbb{R}^k \rightarrow \mathbb{R}$ with $k < n$. Finally, we define another function $v(x_1, \dots, x_{k-1}) := Var(g(x_1, \dots, x_{k-1},X_{k}))$.

I realize that this is a pretty convoluted construction, but I believe everything so far is well-defined.

Now, I have a few questions about what happens if I replace $x$s with random variables. In particular, what is $g(X_1, \dots, X_k)$? I am interpreting it as $$g(X_1, \dots, X_k) = E(f(X_1, \dots, X_n)|X_1, \dots, X_k),$$ is this correct?

What about $Ev(X_1, \dots, X_{k-1})$? I am interpreting this as $$Ev(X_1, \dots, X_{k-1}) = E\left(Var(g(X_1, \dots,X_{k})|X_1, \dots, X_{k-1})\right),$$ is this correct?

If the above interpretations are correct, how do I evaluate $E\left(Var(g(X_1, \dots,X_{k})|X_1, \dots, X_{k-1})\right)$? I would like to show that it is less than 1.

Intuitively, this should follow from the fact that $f$ is $1$-Lipschitz and that $Var(X_{k}) \leq 1$.

I want to formalize the idea that $X_1, \dots, X_{k-1}$ are "fixed" (since they are being conditioned on), and that $X_{k+1}, \dots, X_n$ are also "fixed" (since they don't vary with $g$), leaving only $X_k$ to "vary", but I am having an extremely difficult pinning this down rigorously.

Answer 1

Here is a well-known result about conditioning:

Let $X$ and $Y$ take values in $(S,\mathcal{S})$ and let $\mathcal{F}$ be a $\sigma$-field s.t. $X$ is $\mathcal{F}$-measurable and $Y$ is independent of $\mathcal{F}$. If $\varphi$ is a function with $\mathsf{E}|\varphi(X,Y)|<\infty$, then $$ \mathsf{E}[\varphi(X,Y)\mid \mathcal{F}]=\psi(X)\quad\text{a.s.}, $$ where $\psi(x):=\mathsf{E}\varphi(x,Y)$.

In your case, letting $W_k:=[X_1,\ldots, X_k]$, $$ g(W_k)=\mathsf{E}[f(W_n)\mid W_k] $$ and $$ v(W_{k-1})=\mathsf{E}[(g(W_k))^2\mid W_{k-1}]-\left(\mathsf{E}[g(W_k)\mid W_{k-1}]\right)^2. $$ Therefore, \begin{align} \mathsf{E}v(W_{k-1})&=\mathsf{E}\left(\mathsf{E}[f(W_n)\mid W_k]\right)^2-\mathsf{E}\left(\mathsf{E}[f(W_n)\mid W_{k-1}]\right)^2 \\ &=\mathsf{E}[\operatorname{Var}(\mathsf{E}[f(W_n)\mid W_k]\mid W_{k-1})]. \end{align}

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The function $g$ is $1$-Lipschitz on $\mathbb{R}^k$ because \begin{align} |g(x)-g(y)|&\le\mathsf{E}|f(x_1,\ldots,x_k,W_{k+1},\ldots,W_n) \\ &\qquad-f(y_1,\ldots,y_k,W_{k+1},\ldots,W_n)|\le \|x-y\|. \end{align}

Consequently,

\begin{align} \mathsf{E}v(W_{k-1})&=\mathsf{E}\operatorname{Var}\left(g(W_k)-g(\mathsf{E}[W_k\mid W_{k-1}])\mid W_{k-1}\right) \\ &\le \mathsf{E}[g(W_k)-g(\mathsf{E}[W_k\mid W_{k-1}])]^2 \\ &\le \mathsf{E}\|W_k-\mathsf{E}[W_k\mid W_{k-1}]\|^2=\operatorname{Var}(X_k). \end{align}