Let $X$ be a real valued random variable.

Let $\operatorname{sgn}(x)$ be $1$ when $x>0$, $-1$ when $x<0$ and $0$ when $x=0$.

Why are $|X|$ and $\operatorname{sgn}(X)$ independent, when the density function of $X$ is symmetric with respect to $0$?

Are $|X|$ and $\operatorname{sgn}(X)$ independent, when the density function of $X$ is not necessarily symmetric with respect to $0$?



We write out part of the argument. Assume that $X$ has continuous symmetric distribution. Then $\Pr(X=0)=0$, and, for positive $x$, $$Pr(0\lt X\le x)=\frac{1}{2}\Pr(-x\le X \le x)=\frac{1}{2}\Pr(|X| \le x).$$ Thus $$\Pr((|X|\le x)\cap (\text{sgn}(X)=1))=\Pr((-x \le X \le x))\cap (\text{sgn}(X)=1))=\Pr(0\lt X \le x)=\frac{1}{2}\Pr(|X|\le x)=\Pr(|X|\le x)\Pr((\text{sgn}(X)=1)).$$ Almost exactly the same idea works when we replace $\text{sgn}(X)=1$ by $\text{sgn}(X)=-1$. The case $\text{sgn}(X)=0$ is trivial.

For the second part, we give an informal example and argument that we do not necessarily have independence when the distribution is not symmetric.

Suppose that $X$ has density function $\frac{1}{2}$ on $[-1,0]$, also $\frac{1}{2}$ on $[99,100]$, and $0$ elsewhere. If we know that $\text{sgn}(X)=1$, then we know that $|X|$ is big. So $|X|$ and $\text{sgn}(X)$ are not independent. We can write this out more formally in terms of conditional probabilities.

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    $\begingroup$ Conditional probabilities are not needed in the second part: if there is independence, then $P(X\in A)=\frac12P(X\in (A\cup-A))$ for every Borel set $A\subseteq\mathbb R_+$ hence (i) $P(X=0)=0$ and (ii) $P(X\in A)=P(X\in -A)$ for every Borel set $A\subseteq\mathbb R_+$. Thus, if there is a density, it is symmetric. $\endgroup$ – Did Jun 30 '12 at 6:08

If $X$ is a continuous random variables (absolutely continuous with respect to Lebesgue),

$$P(|X| \leq x|sgn(X)=1) = P(|X| \leq x|X > 0) = \frac{P(0 < X \leq x)}{P(X > 0)} = \frac{\int_{0}^{x}{f(x)dx}}{\int_{0}^{\infty}f(x)dx}$$

$$P(|X| \leq x|sgn(X)=-1) = P(|X| \leq x|X < 0) = \frac{P(-x \leq X < 0)}{P(X < 0)} = \frac{\int_{-x}^{0}{f(x)dx}}{\int_{-\infty}^{0}f(x)dx}$$

and $P(sgn(X)=0)$. If $f(x)$ is symmetric with respect to $0$, observe that $P(|X| \leq x|sgn(X)=1) = P(|X| \leq x|sgn(X)=-1)$ for any $x \geq 0$. Hence, $|X|$ is independent of sgn(x).

Now consider $X$ to be uniform in $(-1,2)$. Observe that $P(sgn(X)=1)=2/3$ and $P(sgn(X)=1||X|>1)=1$. Hence, $|X|$ and $sgn(X)$ are not independent.

Also observe that it was important for $X$ to be continuous. For example, consider $X $ uniform in $\{-1,0,1\}$. Its mass function is symmetric with respect to $0$ but $P(sgn(X)=0) = 1/3$ and $P(sgn(X)=0||X|=0)=1$ and, thus sgn(X) and |X| are not independent.

  • $\begingroup$ Isn't it only important for $X$ to be symmetric and have $P(X=0)=0$? (e.g. for $X$ uniform on $\{-1,1\}$, $\mathrm{sgn}(X)$ and $|X|$ are independent.) $\endgroup$ – huon Jun 30 '12 at 6:43
  • $\begingroup$ That's what did proved in his comment on André Nicola's answer, right? $\endgroup$ – madprob Jun 30 '12 at 6:46
  • $\begingroup$ yep seems so, I didn't notice that. :) $\endgroup$ – huon Jun 30 '12 at 6:48

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