# How to show independence and interpret the result

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 $p$ .

• What definitions of independence do you know? – whuber Jan 23 '13 at 15:01
• @whuber : I give the definitions in my question. Please see it. – Argha Jan 23 '13 at 15:20
• It's not going to lead you instantly to the answer, but did you notice that $X = UV$? – Glen_b -Reinstate Monica Jan 23 '13 at 15:24
• 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 $V$.) – whuber Jan 23 '13 at 15:25
• @Glen_b: Yes. The result is quite obvious . But I am unable to show the independence. By the way like your comment. – Argha Jan 23 '13 at 15:26

Assume $$X$$ is continuous with distribution function $$F$$.
Then for every $$t$$,
\begin{align} P(|X| 0)}{P(\text{sgn}(X)=1)} \\&=\small\frac{P(0 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|
Similarly, it holds that $$P(|X|
Thus the distribution of $$|X|$$ is independent of $$\text{sgn}(X)$$.