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If I know that two continuous random variables $X$ and $Y$ are independent, are $X^2$ and $Y$ necessarily independent? Are $X^2$ and $Y^2$ also independent?

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Are these discrete or continuous? – user21436 Jan 21 '12 at 18:58
Continuous. (I edited the question to specify.) – jamaicanworm Jan 21 '12 at 18:59
If $f,g\colon\mathbb R\to\mathbb R$ are Borel-measurable functions and $X$ and $Y$ are independent then so are $f(X)$ and $g(Y)$. – Davide Giraudo Jan 21 '12 at 18:59
Thanks! What is Borel-measurable? – jamaicanworm Jan 21 '12 at 20:48
It mean measurable with respect to the Borel $\sigma$-algebra: If you don't know what it is, I guess independence between two random variables $X$ and $Y$ has been defined by $P(X\leq s,Y\leq t)=P(X\leq s)P(Y\leq t)$ for all real numbers $s$ and $t$. – Davide Giraudo Jan 21 '12 at 21:45
up vote 2 down vote accepted

As we will see, the result has nothing to do with the fact that $X$ and $Y$ are continuous or not.

We will use the following definition of independence: $X$ and $Y$ are independent if $$\forall s,t\in\mathbb R\quad P(X\leq s,Y\leq t)=P(X\leq s)P(Y\leq t).$$ (in fact it's equivalent to $P(X\in A,Y\in B)=P(X\in A)P(Y\in B)$ for all $A,B$ Borel measurable subsets of $\mathbb R$). With the first definition, we have for $s\geq 0, t\in\mathbb R$: \begin{align*} P(X^2\leq s,Y\leq t)&=P(|X|\leq \sqrt s,Y\leq t)\\ &=P(X\leq\sqrt s,Y\leq t)-P(X< -\sqrt s,Y\leq t)\\ &=P(X\leq \sqrt s)P(Y\leq t)-P(X<-\sqrt s)P(Y\leq t)\\ &=P(|X|\leq \sqrt s)P(Y\leq t)\\ &=P(X^2\leq s)P(Y\leq t), \end{align*} and the last inequality is true if $s<0$, since all these probabilities are $0$. It shows that $X^2$ and $Y$ are independent.

For the independence of $X^2$ and $Y^2$, we have shown that if $X_1$ and $X_2$ are independent then so are $X_1^2$ and $X_2$. So apply it to $X_1=Y$ and $X_2=X^2$.

If we work with the second definition, and take $f,g$ two Borel measurable functions and $A,B$ two Borel-measurable sets, since $f^{-1}(A)$ and $g^{-1}(B)$ are still Borel measurable, we have \begin{align*}P(f(X)\in A,g(Y)\in B)&=P(X\in f^{-1}(A),Y\in g^{-1}(B))\\ &=P(X\in f^{-1}(A))P(Y\in g^{-1}(A))\\ &=P(f(X)\in A) P(g(Y)\in B). \end{align*} So $f(X)$ and $g(Y)$ are still independent.

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Nice, except that $[X^2\leqslant s]\ne[X\leqslant\sqrt{s}]$ in general. // Unrelated: regarding the comments above, you might wish to stress that this has nothing to do with $X$ and $Y$ being continuous or discrete or whatever. – Did Jan 22 '12 at 9:39
@DidierPiau you are right, I missed an $|\cdot|$ that I will add readily. – Davide Giraudo Jan 22 '12 at 9:43
Right, but now one wonders how you deal with the probability of the event $[|X|\leqslant\sqrt{s},Y\leqslant t]$ since the notion of independence you recalled at the beginning of the post is not adapted. – Did Jan 22 '12 at 11:43
Yes, I have to give more details (hope it will convince jamaicanworm to study measure theory!). – Davide Giraudo Jan 22 '12 at 12:16
Regarding your hope: yes. – Did Jan 22 '12 at 13:22

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