Expectation of product of functions of random variables

Question

Note: I do not have a strong background in proofs, but I know that a version of these properties holds. So I likely will have missed out on some technical points below. Feedback on this is appreciated.

If $X$ and $Y$ are independent continuous random variables (each with finite expectation and densities $f_X(x)$ and $f_Y(y)$), the product of their expectation is the product of their individual expectation.

This can proved by starting from the definition: $$\mathbb{E}(XY)=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} xyf_X(x)f_Y(y) dy dx$$

Since $x$ is a constant with respect to $y$, we can move the $x$ terms outside of the integral: $$=\int_{-\infty}^{\infty} xf_X(x)\left(\int_{-\infty}^{\infty} yf_Y(y) dy\right) dx$$

Since $y$ is a constant with respect to $x$, we can move the $y$ terms outside of the integral: $$=\left(\int_{-\infty}^{\infty} xf_X(x)dx\right) \left(\int_{-\infty}^{\infty} yf_Y(y) dy\right)$$

Similarly (useful when showing that the mgf of a sum for independent random variables is the product of the mgfs), we can also show that: $$ \mathbb{E} \left[e^{tX} e^{tY}\right]=\mathbb{E} \left[e^{tX} \right]\mathbb{E} \left[e^{tY}\right]$$

But, in general when is is true that: $$ \mathbb{E} \left[g(X) h(Y)\right]=\mathbb{E} \left[g(X) \right]\mathbb{E} \left[h(Y)\right]$$

Is it always true, or are there any useful rules and properties on when it is (or is not) true? I think that replacing $x$ and $y$ in the proof above with $h(x)$ and $g(y)$ should work. But I am not whether any other nuance needs to be taken care of.

Note: This is an extension of a similar kind of property in the discrete case that I had proved some time back. (Similar as in the same idea would be used to prove the discrete version). I haven't taken the "proof" from any source.

Answer 1

As long as $g,h$ are measurable functions (which most are) then this is correct.

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Let $X_1,X_2$ be independent random variables taking values in spaces $E_1, E_2$, respectively, then for $A\subseteq E_1, B\subseteq E_2$ we have:

$$P_{X_1,X_2}(X_1 \in A,X_2 \in B) = P_{X_1}(X_1 \in A)P_{X_2}(X_2 \in B)$$

This is just your independence assumption. I'm writing it this way to make it clear how independence "transfers" to measurable functions on each.

Let $g,h$ be measurable functions $E_1 \to G,E_2 \to H$, respectively. And define $Y_1 = g(X_1),Y_2=h(X_2)$.

Then, for $Q\subseteq G,R\subseteq H$, we have by the measurability assumption:

$$P_{Y_1}(Y_1 \in Q) = P_{X_1}(X_1 \in Y_1^{-1}(Q))\;\text{and}\;P_{Y_2}(Y_2 \in R) = P_{X_2}(X_2 \in Y_2^{-1}(R))$$

Where $Y_1^{-1}(Q)$ is the preimage of $Y_1$ for set $Q$ (i.e., all the points in $E_1$ that are mapped to $Q$ by $Y_1$).

The reverse operation is just called the image of $X_1$ on $A$, etc. (i.e., just applying $X_1$ to all points in $A$ like any other set function).

Therefore,

$$P_{Y_1,Y_2}(Y_1 \in Q,Y_2 \in R) = P_{X_1,X_2}(X_1 \in Y_1^{-1}(Q),X_2 \in Y_2^{-1}(R))$$

But wait, we already know that the joint probability for $(X_1,X_2)$ factors due to independence assumption, therefore:

$$P_{X_1,X_2}(X_1 \in Y_1^{-1}(Q),X_2 \in Y_2^{-1}(R)) = P_{X_1}(X_1 \in Y_1^{-1}(Q))P_{X_2}(X_2 \in Y_2^{-1}(R)) = P_{Y_1}(Y_1 \in Q)P_{Y_2}(Y_2 \in R)$$

Since the random variables created by $g,h$ are independent, all the same rules hold for them as well.