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In advanced probability we can just say:

\begin{align} & P(h(X) \in A, g(Y) \in B) \\[6pt] = {} & P(X \in h^{-1}(A), Y \in g^{-1}(B)) \\[6pt] = {} & P(X \in h^{-1}(A)) P(Y \in g^{-1}(B)) \\[6pt] = {}& P(h(X) \in A) P(g(Y) \in B) \end{align}

where $X,Y \in (\Omega, \mathscr F, \mathbb P)$ and are $(\mathbb R, \mathscr B(\mathbb R))$-valued, $A,B$ are Borel and $h,g$ are some kinds of functions (Borel? Bounded and Borel? $\mathscr F$-measurable?)

But what about in basic probability? Is there a double integral way to approach this?

The hint given is to show that

$$P(h(X) \in (-\infty, a], g(Y) \in (-\infty, b])$$

$$ = P(h(X) \in (-\infty, a]) P(g(Y) \in (-\infty, b])$$

I can sort of use the same argument as in advanced probability but then I would have to use

$$h^{-1}(-\infty, a]$$

Is that acceptable to use in basic probability?

I think

$$h^{-1}(-\infty, a] = \{x \mid h(x) \in (-\infty, a]\}$$

So $$X \in h^{-1}(-\infty, a] = \operatorname{Range}(X) \cap \{x \mid h(x) \in (-\infty, a]\} = \{\omega \in \Omega \mid X(\omega) \in \{x \mid h(x) \in (-\infty, a]\}\}$$

Regardless, I think $X$ and $Y$ are absolutely continuous continuous with pdfs.

I think there should be a double integral way to approach this.

Edit 4.5 years later: by the way, what's the condition for $h$ and $g$ to do this please? See here: Prove that for independent random variables $X_i$, we have $f_i(X_i)$ are independent.

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  • I wonder if your "advanced" argument only looks advanced because of al the notation? $\qquad$ – Michael Hardy Jul 12 '16 at 00:46
  • @MichaelHardy function inverse of an interval is acceptable? Prerequisite for the class is only calc I and II – BCLC Jul 12 '16 at 04:13
  • Actually the "advanced" part is just the stuff about Borel sets. With students who haven't had that, if you instead just say "sets", they probably won't suspect a thing. You can say $f^{-1}(A) = { x : f(x)\in A}$. If they haven't seen that before, I suppose it should be introduced. $\qquad$ – Michael Hardy Jul 12 '16 at 04:27
  • @MichaelHardy is there perhaps a double integral way to go about this? – BCLC Jul 12 '16 at 04:34
  • @MichaelHardy by the way, what's the condition for $h$ and $g$ to do this please? See here: https://math.stackexchange.com/questions/3944284/prove-that-for-independent-random-variables-x-i-we-have-f-ix-i-are-indepe – BCLC Dec 11 '20 at 12:31

1 Answers1

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Idk about pdf a thought that works for both elementary and advanced is I think to use $\pi$-systems instead of $\sigma$-algebras.

The sets $$\{f_i(X_i) \in (-\infty,x_i]\}_{i \in I}$$ are independent because the sets

$$\{X_i \in f_i^{-1}(-\infty,x_i]\}_{i \in I}$$

are independent. I guess the only conditions on the $f_i:A_i \to \mathbb R$'s are

  1. $A_i$ contains the image of $X_i$
  2. $E[f_i(X_i)]$ is defined, i.e. $\int_{\Omega} |f_i(X_i)| d\mathbb P$ (aka $\int_{\mathbb R} |f_i(x_i)| h_{X_i} d_{x_i}$) $<\infty$
  3. $f_i^{-1}(-\infty,x_i]$ is ... still an interval $(-\infty,g_i]$ for some $g_i$, maybe $g_i$ = $f(x_i)$
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