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I am a statistics student who is simply trying to see if $X$ and $Y$ are independent then $\log X$ and $\log Y$ are independent as well.

I tried to look for information regarding this and I see people mentioning Borel functions all the time.

I am not familiar with its definition and with the courses I will be taking in the future it is not likely that I will ever study set theory or any topology classes that may lead me to fully learn this.

So, it boils down to this question.

Can any of the following elementary functions be an example where

"$X$ independent from $Y$ but $f(X)$ is dependent on $f(Y)$?"

$$f(x)=x^n$$ $$f(x) = a^x$$ $$f(x)=\log_a(x)$$ $$f(x)=x^{\frac{1}{n}}$$ $$f(x)...\text{Trig functions such as} \quad \sin(x), \cos(x).$$ $$f^{-1}(x)...\text{Inverse trig functions such as } \quad \arcsin(x), \arccos(x).$$

Intuitively I want to say that when $X$ and $Y$ are independent, invertible functions would produce independence in $f(X)$ and $f(Y)$ but I would like to know if that is the case. Or, at least, it is true for these elementary functions that are taught in highschool

Mars Plastic
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hyg17
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    Independence of $X$ and $Y$ are really the independence of information obtainable from $X$ and $Y$. Since composing a function to a RV can only result in 'loss of information', the composed RVs are still independent. – Sangchul Lee Jul 22 '19 at 23:32
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    Given independent random variables $X$ and $Y$, $f(X)$ and $g(Y)$ are independent whenever $f$ and $g$ are Borel measurable (note that $f$ does not necessarily have to equal to $g$). The main reason Borel measurability is required is to ensure that $f(X)$ and $g(Y)$ are themselves random variables and not some other weird mathematical object. The answer to your last question is yes since any function which is continuous or piecewise-continuous is Borel measurable. – parsiad Jul 22 '19 at 23:48
  • Thank you very much! This was exactly the type of answer that I was looking for. Just out of curiosity, though. In what type of course would one study Borel functions? – hyg17 Jul 23 '19 at 06:24
  • @hyg17 You'd study Borel functions in a course on real analysis, specifically one that covered a topic called measure theory. – grand_chat Jul 23 '19 at 17:41
  • For application, you need not worry about Borel measurability, since essentially all 'sensibly constructed' functions are Borel-measurable. Indeed, to create a function which is not Borel-measurable, you need to invoke a form of Axiom of Choice, a complete opposite of constructionistic approach. – Sangchul Lee Jul 24 '19 at 04:51
  • Thank you so much for your inputs. I am glad that I have a better understanding of the situation. – hyg17 Jul 25 '19 at 22:38

2 Answers2

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A more elementary definition of independence for real random variables:

$X$ and $Y$ are independent means $$P(X\le x, Y\le y)=P(X\le x)P(Y\le y)\tag1$$ for every $x$ and $y$.

So to show that $\log X$ and $\log Y$ are independent (assuming $X$ and $Y$ are positive random variables), it's enough to verify $$P(\log X\le x, \log Y\le y) = P(\log X\le x)P(\log Y\le y)\tag2$$ for every $x$ and $y$. To do so, write the LHS of (2) as $$P(X\le e^x, Y\le e^y)$$ which by (1) equals $$P(X\le e^x)P(Y\le e^y)$$ which in turn equals $$P(\log X\le x)P(\log Y\le y).$$ But this is the RHS of (2), and we're done! The same argument proves that $h(X)$ and $h(Y)$ are independent when $h$ is a monotonic function. If you're familiar with the concept of an "inverse image" $h^{-1}(A)$ for a given set $A$ you can establish this result for arbitrary "Borel measurable" functions $h$.

grand_chat
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  • Thank you very much. I was able to come to this conclusion myself after thinking a little, but it really helps to know that others confirm your logic. – hyg17 Jul 23 '19 at 06:26
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If $X$ and $Y$ are independent of each other, then knowing the value of one of them gives you no information about the distribution of the other.

As a result, it doesn't matter what function you apply - the distributions are still unrelated to each other and you can't gain information about one from the value of the other.

ConMan
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