I was confused by what it means when a Random variable is a deterministic function of another Random variable yet is independent of it? How is this possible?

Here's the question: Consider three random variables $X,Y$ and $Z$ such that $Z$ is independent of $X$ and of $Y$ and is a deterministic function of $(X,Y)$. Is $H(Z)=0$? If so prove it; if not, give a counter example.

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    $\begingroup$ Can you provide reference mentioning it and a broader context? Basically, if one random variable is a function of another one it is dependent of it (stats.stackexchange.com/questions/16321/… and stats.stackexchange.com/questions/231425/… ) $\endgroup$ – Tim Sep 18 '16 at 17:53
  • $\begingroup$ @Tim That's what i thought too . Here's the question . Consider three random variables X,Y and Z such that Z is independent of X and of Y and is a deterministic function of (X,Y). Is H(Z)=0 ? If so prove it ; if not , give a counter example $\endgroup$ – Christopher Sep 18 '16 at 17:59
  • $\begingroup$ please edit to add those details. Btw is it a homework? $\endgroup$ – Tim Sep 18 '16 at 18:38
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    $\begingroup$ Do you mean Z=H(X,Y)=0 or does H(Z) mean something else? $\endgroup$ – Patty Sep 18 '16 at 19:01
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    $\begingroup$ @Christopher still I think you should add [self-study] tag and check: stats.stackexchange.com/tags/self-study/info $\endgroup$ – Tim Sep 18 '16 at 19:10

A non trivial univariate real random variable that is a deterministic function of another random variable is not independent of it (see Mark L. Stone's answer for an example with a constant random variable). However, when more than two random variables are involved independence shows counterintuitive behaviours.

I'll give an example of $Z$ deterministic function of $X$ and $Y$ but independent from $X$ and $Y$.

Let $X$ and $Y$ be independent Bernouilli variables with $p=0.5$ (for example, $X$ and $Y$ are the results of tossing a coin each).

Let $f(X,Y)$ equal $1$ if $X=Y$ and $0$ if $X$ is different from $Y$.

Let $Z=f(X,Y)$.

You can easily see that $P(X=0)=0.5=P(X=0\mid Z=1)=P(X=0\mid Z=0)$ and that $P(X=1)=0.5=P(X=1\mid Z=1)=P(X=1\mid Z=0)$, proving that $X$ and $Z$ are independent, or using another definition of independence:

\begin{align*}P(x=0\text{ and }Z=0) &= P(X=0\text{ and }X \text{ different from }Y) \\ &= P(X=0\text{ and }Y=1) \\ &= 0.5 \cdot 0.5 = 0.25 \\\\ P(X=0) \cdot P(Z=0) &= 0.5 \cdot 0.5 = 0.25 \end{align*}

The same operation can be done for all values of $X$ and $Z$, thus proving that $P(X=a\text{ and }Z=b) = P(X=a)\cdot P(Z=b)$ for every value of $a$ and $b$.

Furthermore, the same proof holds for $Y$, therefore proving that $Y$ and $Z$ are independent and $X$ and $Z$ are independent. In fact, $X$, $Y$ and $Z$ are pairwise independent while $X$, $Y$ and $Z$ are not independent considered as a whole (not jointly independent). Interestingly, $Z$ is not independent of $(X, Y)$.

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    $\begingroup$ Your statement "An univariate real random variable that is a deterministic function of another random variable is not independent of it. " is incorrect, as shown in my answer. $\endgroup$ – Mark L. Stone Sep 18 '16 at 19:32
  • $\begingroup$ @Patty , the counter example i provided was an XOR . Seems like this is correct . Yes , Pere . +1 (Pairwise independence does not imply joint independence ) . $\endgroup$ – Christopher Sep 18 '16 at 19:34
  • $\begingroup$ Well, the correctness of your edited statement depends on your definition of non-trivial function. Consider any function $f(x)$ which is one to one. Apply it to a random variable $Y$ which is a constant with probability one. Then $f(Y)$ is independent of $Y$. $\endgroup$ – Mark L. Stone Sep 18 '16 at 19:39
  • $\begingroup$ Yes, I edited it again to apply "trivial" to the resulting random variable and not to the function. I'm not sure if there is a rigorous definition of trivial random variable, but I think it's clear enough - in fact, I'm not sure if there is a definition of random variable that includes something that always gives the same result (a deterministic random variable). However, since it points to your answer I think readers will easily understand what is the exception about. $\endgroup$ – Pere Sep 18 '16 at 19:44
  • $\begingroup$ @Pere I took the liberty of adding MathJax to your (great) answer to improve readability. I hope that is okay - otherwise reject the edit! $\endgroup$ – Therkel Sep 19 '16 at 7:12

A random variable which is a constant with probability 1 is independent of itself. I leave the trivial proof to you as an exercise.

Consider the deterministic function f(x) = x, applied to the random variable $Y$, which equals a constant, say $e^\pi$, with probability one. Therefore, the random variable $f(Y)$ is independent of $Y$. If you don't like that example because $f(Y)$ is the same as $Y$, then use the function $f(x) = 2x$. Same conclusion.

This provides a counterexample to the incorrect statement "An univariate real random variable that is a deterministic function of another random variable is not independent of it." in the first paragraph by the answer by @Pere .

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  • $\begingroup$ True , the link provided by @Tim serves as a good example of countering that $\endgroup$ – Christopher Sep 18 '16 at 19:34
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    $\begingroup$ @Mark L. Stone: I edited my answer to exclude the trivial case of a non random random variable. $\endgroup$ – Pere Sep 18 '16 at 19:39
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    $\begingroup$ This answer would be more interesting (rather than just pedantic) if you went on to describe the technical conditions (for "non-triviality") needed to make the statement true. $\endgroup$ – R.. GitHub STOP HELPING ICE Sep 19 '16 at 3:41

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