If $X$ is symmetrically distributed about zero , then show that $U=|X|$ and $$V = \left\{ \begin{array}{} +1, & X \geq0 \\ -1, & X<0 \end{array} \right .$$ are independently distributed and interpret the result.

Here distribution is not given only it is known that it symmetrically distributed . I think this result holds for continuous and discrete distribution . How can I show independence and interpret the result . Please help

Edit: Here independence means statistical independence.

Defn1: Two random variables $X$ and $Y$ are independent if and only if $$ F_{XY}(x,y)=F_X(x).F_Y(y) \forall x,y \in R$$ where $ F_{XY}(x,y)$ is their joint distribution function and $F_X(x)$ and $F_Y(y)$ are their marginal distribution functions.

I also know the independence with PDF/PMF.

Defn2: Two random variables $X$ and $Y$ , forming an absolutely continuous random vector,are independent if and only if $$f_{XY}(x,y)=f_X(x).f_Y(y) \forall x,y \in R$$ where $ f_{XY}(x,y)$ is their joint probability density function and $f_X(x)$ and $f_Y(y)$ are their marginal probability density functions.

For discrete case $f$ is replaced by $p$ .

  • $\begingroup$ What definitions of independence do you know? $\endgroup$
    – whuber
    Jan 23, 2013 at 15:01
  • $\begingroup$ @whuber : I give the definitions in my question. Please see it. $\endgroup$
    – Argha
    Jan 23, 2013 at 15:20
  • 1
    $\begingroup$ It's not going to lead you instantly to the answer, but did you notice that $X = UV$? $\endgroup$
    – Glen_b
    Jan 23, 2013 at 15:24
  • $\begingroup$ OK, have you tried to apply each definition to your problem? To do so, you will need to obtain the CDF or PDF of both $U$ and $V$. Obviously they must be related to the CDF and PDF of $X$. Precisely how? (If you work this through carefully you will be able to find a class of random variables $X$ for which the conclusion of the problem is false! Take a close look at the asymmetry in the definition of $V$.) $\endgroup$
    – whuber
    Jan 23, 2013 at 15:25
  • $\begingroup$ @Glen_b: Yes. The result is quite obvious . But I am unable to show the independence. By the way like your comment. $\endgroup$
    – Argha
    Jan 23, 2013 at 15:26

1 Answer 1


The result "sign and magnitude are independently distributed" is certainly true for continuous distributions symmetric about zero. Notably, it can be applied in the derivation of the test statistic in Wilcoxon's signed-rank test for symmetry.

Assume $X$ is continuous with distribution function $F$.

Then for every $t$,

\begin{align} P(|X|<t\mid \text{sgn}(X)=1)&=\small\frac{P(-t<X<t\,,\,X> 0)}{P(\text{sgn}(X)=1)} \\&=\small\frac{P(0<X<t)}{P(X> 0)} \\&=\small\frac{F(t)-F(0)}{1-F(0)} \\&=2F(t)-1 & \small[\,\because F(0)=1/2\,] \\&=F(t)-(1-F(t)) \\&=F(t)-F(-t) & \small[\,\because F(t)+F(-t)=1\,] \\&=P(|X|<t) \end{align}

Similarly, it holds that $P(|X|<t\mid \text{sgn}(X)=-1)=P(|X|<t)$

Thus the distribution of $|X|$ is independent of $\text{sgn}(X)$.


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