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Suppose that we have three functions $f$, $g$, and $h$, such that for all $x$ and $y$, $h(x, y) = g(f(x, y, \cdot))$, where $\cdot$ represents an input where a random variable goes (ignored in the informal version below), and $g$ is the expectation of an expression involving that random variables (this makes $g$ and $h$ deterministic functions). We are given $\frac{\partial h(x, y)}{\partial x}$, and want to find $\frac{\partial h(x, y)}{\partial y}$.

Informal Version (more formal version given below):

We are given the partial derivative of $h$ w.r.t. $x$,

$$\frac{\partial h(x, y)}{\partial x} = \mathbb E \left[\sum_{t=0}^\infty \text{[some stuff involving random variables and } f(x, y)] \frac{\partial f(x, y)}{\partial x}\right],$$

Where the "some stuff" is all independent of $x$ and $y$ given the output of $f(x, y)$. We want to prove that we can substitute $\partial y$ for $\partial x$:

$$\frac{\partial h(x, y)}{\partial y} = \mathbb E \left[\sum_{t=0}^\infty \text{[some stuff involving random variables and } f(x, y)] \frac{\partial f(x, y)}{\partial y}\right].$$

Intuition/Main Question:

Intuitively, we can argue that since $x$ and $y$ only affect $h$ via $f$ (that is, $h$ is conditionally independent of $x$ and $y$ given $f(x, y)$), and we should be able to simply replace $\frac{\partial f(x, y)}{\partial x}$ with $\frac{\partial f(x, y)}{\partial y}$ in the definition of $\frac{\partial h(x, y)}{\partial x}$ to get $\frac{\partial h(x, y)}{\partial y}$ (see the informal chain rule argument in the appendix below for more intuition).

How can we formalize the argument above to prove this property about $\frac{\partial h(x, y)}{\partial y}$? I suspect there some theorem or well-known result that immediately gives us what we need, but I do not know what this theorem is. Thank you.

Appendix (not necessary for answering the question, but may be helpful):

More Formal Version:

We are given that partial derivative of $h$ w.r.t. $x$,

$$\frac{\partial h(x, y)}{\partial x} = \mathbb E \left[\sum_{t=0}^\infty d(f(x, y, Z_t), Z_t) \frac{\partial f(x, y, Z_t)}{\partial x}\right],$$

where, $d$ is some function, the $Z_t$'s are random variables influenced by $f(x, y, \cdot)$ and, for all $t$, $Z_t$ is conditionally independent of $x$ and $y$ given $Z_{t-1}$ and $f(x, y, Z_{t-1})$. We need to find $\frac{dh(x, y)}{dy}.$ We'd like to show that

$$\frac{\partial h(x, y)}{\partial y} = \mathbb E \left[\sum_{t=0}^\infty d(f(x, y, Z_t), Z_t) \frac{\partial f(x, y, Z_t)}{\partial y}\right].$$

More informal intuition involving the chain rule:

Consider the following informal argument, ignoring the sum, expectation, and random variables for a moment:

$$\frac{\partial h}{\partial x} = \frac{\partial h}{\partial f} \frac{\partial f}{\partial x} = d(f(x, y)) \frac{\partial f(x, y)}{\partial x},$$

so $\frac{\partial h}{\partial f} = d(f(x, y)).$ Substituting this definition of $\frac{\partial h}{\partial f}$ to find $\frac{\partial h}{\partial y}$:

$$\frac{\partial h}{\partial y} = \frac{\partial h}{\partial f} \frac{\partial f}{\partial y} = d(f(x, y)) \frac{\partial f(x, y)}{\partial y}.$$

Q.E.D. (except for the troublesome sum, expectation, and random variables).

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