Is the density function of $|X|$ given by $f_{|X|}(x) = 2f_X(x)$?

Question

Let $X$ be a continuous random variable with density $f_X$ such that $X$ has the same distribution as $-X$. Evaluate if the following statement is true or false and bubble in your solution:

Statement: The density of the random variable $|X|$ is given by $f_{|X|}(x) = 2f_X(x)$ for all $x\in\mathbb R$.

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I just wanted to ensure that my reasoning was sound on this question. For the density of $X$ to have the same distribution as $-X$ we must have something symmetric. Ie. we can essentially "fold" the part of the distribution left of the $y$-axis over the positive $x$ values which is why I think this would be correct. I just wanted to make sure that I did not miss any hidden caveat of the question.

Answer 1

Your intuition is right.

To formalize just notice that if $x > 0$: $$ P(|X| \leq x) = 1 - P(|X| > x) = 1 - (P(X > x)+P(X<-x)) = 1 - (P(X > x)+P(-X > x)) $$

So, $$ P(|X| \leq x) = 1 - 2P(X > x) = 1 - 2(1- P(X \leq x)) $$

In terms of the cdfs: $$ F_{|X|} (x) = 2F_X(x) - 1\text{ when } x> 0 $$

Clearly, $F_{|X|} (x) = 0$ when $x\leq 0$.

We can take the derivative when $x>0$ and so: $$ f_{|X|}(x) = \begin{cases} 2f_X(x)\text{, if }x > 0\\ 0\text{, if }x\leq 0 \end{cases}$$

Answer 2

The density function of $\lvert X\rvert$ is given by $f_{\small \lvert X\rvert}(x)=2f_{\small X}(x)$ for all $x\in\Bbb R$.

I just wanted to ensure that my reasoning was sound on this question. For the density of X to have the same distribution as -X we must have something symmetric. Ie. we can essentially "fold" the part of the distribution left of the y-axis over the positive x values which is why I think this would be correct.

What about the negative values for $x$?

Consider that if the statement was correct, then $f_{\small\lvert X\rvert}$ would not be a valid density function, because taking the integral for all values in $\Bbb R$ would give:

$$\int_\Bbb R f_{\small\lvert X\rvert}(x)\mathrm d x=2\int_\Bbb R f_{\small X}(x)\mathrm d x\\=2$$

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You have the correct approach, but note that the folding maps a positive value and a negative value for $X$ onto a single positive value for $\lvert X\rvert$, and no values for $X$ will map to a negative for $\lvert X\rvert$ (which, of course, cannot be negative).

Perhaps the use of the token $x$ as the value for $\lvert X\rvert$ is confusing.

Let $Y=\lvert X\rvert$. Then when $Y=y$, for any $y\in[0..\infty)$, there are two values for $X$, which are $\{y,-y\}$.

$$f_{\small Y}(y)~{= ( f_{\small X}(-y)+f_{\small X}(y) )\,\mathbf 1_{y\geqslant 0}\\=2 f_{\small X}(y)\,\mathbf 1_{y\geqslant 0} }$$

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Thus we have a valid probability density function when we include the support. $$\begin{align}f_{\small \lvert X\rvert}(x) &= 2 f_{\small X}(x)\,\mathbf 1_{x\geqslant 0} \\[2ex]\int_\Bbb R f_{\small\lvert X\rvert}(x)\,\mathrm d x&=\int_0^\infty 2 f_{\small X}(x)\,\mathrm d x\\[1ex]&= 1\end{align}$$