If $X$ and $Y$ are independent. How about $X^2$ and $Y$? And how about $f(X)$ and $g(Y)$? I always have confusion about it. I feel ... yeah of course $f(X)$ and $g(Y)$ are independent, because $X$ and $Y$ are. But .. is it right?? Then how can I prove it?
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1If $f$ and $g$ is continuous, I think they are independent. – gaoxinge Jan 11 '14 at 11:34
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1$P(f(X)\in A,g(Y)\in B) = P(X\in f^{-1}(A), Y\in g^{-1}(B)))$ now use independence of X Y... – Lost1 Jan 11 '14 at 11:40
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1For any measurable function $f$: $\sigma(f(X))\subseteq \sigma(X)$. – Stefan Hansen Jan 11 '14 at 12:00
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@Lost1 : why does the equality hold? If f(X)=X^2 then f^-1 doesn't exist, I think... I am confused.. – syko Jan 11 '14 at 12:43
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@gaoxinge Continuity is irrelevant. – Did Jan 11 '14 at 13:36
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f^-1 is the pre image, not the inverse function. it should always mean this in analysis when you talk about f^-1 of a set. – Lost1 Jan 11 '14 at 14:52
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1Whoever said this question already has an answer at that linked question is using some really circular logic... – Nov 27 '17 at 17:26
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By definition we have that $X,Y$ are independent if $F(x,y) = F_X (x) F_Y(y)$. That is that $$ P\{X \le x , Y \le y \} = P \{X \le x \} P \{Y \le y \} $$ with this you can proof that, for any $A,B$ Borel sets $$ P\{X \in A , Y \in B \} = P \{X \in A \} P \{Y \in B \} $$ holds. Now let $g,h$ be measurable functions \begin{align*} F_{g(X), h(Y)} (x,y) =& P \{ g(X) \le x, h(Y) \le y \} = P \{ X \in g^{-1} (-\infty,x], Y \in h^{-1}(-\infty, y] \} \\ =& P \{X \in g^{-1} (-\infty,x] \} P \{Y \in h^{-1}(-\infty, y] \} = P \{g(X) \le x \} P \{h(Y) \le y \} \\ =& F_{g(X)}(x) F_{h(Y)}(y) \end{align*} This concludes the proof.
Regards,
D
Luigi Traino
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D G
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1Thank you! I think I need to study 'measurable function'. but here I have a question. If g(X)=X^2 then it doesn't have inverse function on R. right? then what's the difference between what you used for g^-1 , h^-1 from what I said about 'inverse function'? Actually I cannot understand the equality, "P{g(X)≤x,g(Y)≤y}=P{X∈g−1(−∞,x],Y∈h−1(−∞,y]}". why? – syko Jan 11 '14 at 12:33
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Another possible duplicate here http://math.stackexchange.com/questions/443659/if-x-and-y-are-independent-then-fx-and-gy-are-also-independent – Zbigniew Jan 11 '14 at 12:40
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D G what do you say re measurable? 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 12:29