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Given A,B are independent identically distributed random variables

will $$E[A^2B]=E[A^2]E[B]$$

(uncorrelation) $$P_{A^2B}(a,b)=P_{A^2}(a)P_B(b)$$

(independence)

hold?

An initial thought is that one-to-one mapping of random variable will not destroy independence

However the mapping $$f(x)=x^2$$ apparently is not an 1-1 mapping.

Any proof toy example to show if the above two property still holds after mapping?

Hosein Rahnama
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Allen Kuo
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    Yes if A and B are independent then u(A) and v(B) are independent, for every functions u and v. To get more explanations, you might want to recall how you define the independence of some random variables A and B. – Did Dec 31 '15 at 09:16
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    $P(u(A)\in U,v(B)\in V)=P(A\in u^{-1}(U),B\in v^{-1}(V))=\dots$ – drhab Dec 31 '15 at 09:19

1 Answers1

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Remember that, if $f:\mathbb{R}\to\mathbb{R}$ is a Borel-measurable function, then $f(A)$ is a $\sigma(A)$-measurable function. So, in our case, $f(x)=x^2$ is a Borel-measurable function.

Hence, $A^2$ is a $\sigma(A)$-measurable function, and therefore, $\sigma(A^2)\subseteq\sigma(A)$. Now, using independence of $\sigma$-algebras $\sigma(A)$ and $\sigma(B)$ you can conclude.

sinbadh
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