Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose $Y_1,Y_2,\ldots, Y_n$ are independent, where $Y_i$ is a continuous valued random variable with a density $p_{Y_i}(y_i)$ on its domain $\mathcal{D}_i\subseteq \mathbb{R}.$ Can one show that for any continuous function $f:\mathbb{R}^n\to\mathbb{R},$ that is non-constant on $\mathcal{D}_1\times \mathcal{D}_2\times\ldots \times \mathcal{D}_n,$ we must have that $f(Y_1,Y_2,\ldots, Y_n)$ is not independent of $Y_i$ for at least one $i$?

Note: It is possible that $f(Y_1,Y_2,\ldots, Y_n)$ is independent of $Y_i$ for some $i.$ For instance, if $Y_1$ is distributed uniformly on $[0,1]$ and $Y_2,Y_3$ are normally distributed with mean 0 and variance 1, then $Z:=f(Y_1,Y_2,Y_3) = Y_1\cdot Y_2+\sqrt{1-Y_1^2}\cdot Y_3$ is independent of $Y_1.$

Note 2: One can construct counterexamples if $Y_1,Y_2,\ldots, Y_n$ are allowed to be discrete valued. For instance, if $Y_i$ is a Bernoulli random variables with parameter $\frac{1}{2},$ then $f(Y_1,Y_2,\ldots, Y_n)$ defined as the XOR of all the $Y_i$'s is independent of each $Y_i.$

share|improve this question
More generally, in the discrete case such functions are called "correlation immune" and are important in cryptology and elsewhere. –  Brendan McKay Aug 6 '13 at 2:34
add comment

1 Answer 1

up vote 3 down vote accepted

First define $f:[0,1]\rightarrow [0,1]$ as follows:

$$f(x,y)=2(y-x)-1 \hbox{ if } 1\ge y-x \ge 1/2$$ $$f(x,y)=-2(y-x)+1 \hbox{ if } 1/2\ge y-x \ge 0$$ $$f(x,y)=2(y-x)+1 \hbox{ if } 0\ge y-x \ge -1/2$$ $$f(x,y)=-2(y-x)-1 \hbox{ if } -1/2\ge y-x \ge -1$$

Then let $X$ and $Y$ be uniformly (and independently) distributed on $[0,1]$. It's easy to check that $f(X,Y)$ is independent of both $X$ and $Y$.

share|improve this answer
Thanks Steven. That's a nice construction. –  Hedonist Aug 6 '13 at 6:39
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.