# P.d.f of the absolute value of a normally distributed variable

I came across this question as an exercise, had a brief idea, but didn't know how to proceed.

Let $$X \sim N(0, 1)$$. What is the p.d.f of $$|X|$$ ?

I know the final p.d.f looks just like the right half of the original pdf, but extended vertically for a factor of 2.

Could someone please show mathematically what the p.d.f of variable |X| is going to be, and explain briefly? Thank you.

• Since the original distribution is symmetric about $0$, the density for positive values of the absolute value is double the density of the original random variable for the same value, while the density for negative values of the absolute value is $0$. This particular case is called a half-normal distribution Apr 23, 2015 at 14:18
• To add on to Henry's comment (which provides the intuition behind @calculus's answer): Remember that the PDF must have an area under the curve of $1$. Since the PDF of the normal distribution is symmetric around the $y$-axis, the area under the curve of the right half is only $1/2$, and therefore must be scaled up by a factor of $2$ to bring the area up to $1$. Apr 24, 2015 at 17:38
• It's called Folded Normal Distribution. en.wikipedia.org/wiki/Folded_normal_distribution
– Mr.M
Nov 22, 2016 at 9:49

You can take the cdf as a starting point.

$$P(|X| \leq x)=P(-x \leq X \leq x)$$

$$X\sim \mathcal N(0,1)$$

$$=P(X \leq x)-P(X \leq -x)$$

$$=P(X \leq x)-\left[ 1-P(X \leq x) \right]$$

$$=2 \cdot P(X \leq x)-1=2\cdot F(x)-1=2 \cdot \lim\limits_{a \to -\infty} \int_{a}^x \frac{1}{\sqrt{2\cdot \pi}}\cdot e^{\frac{t^2}2} \, dt-1$$

$$=2\cdot F(x)-2\cdot F(a)-1$$

Differentiating :

$$2\cdot \frac{dF(x)}{dx}-2\cdot \frac{dF(a)}{dx}-0=2\cdot f(x)-0-0=2\cdot f(x)$$

$$=2\cdot \frac{1}{\sqrt{2\cdot \pi}}\cdot e^{-\frac{{x}^2}{2 }}, \quad \forall \ x\geq 0$$

$$-1$$ and $$F(a)$$ are constants.

• Thanks for the answer, but for what range of x is this formula valid? Is this range greater or equal to 0? Why pdf of x for x smaller than 0 is 0? Nov 9, 2020 at 17:22
• @curiouscppnoob You´re welcome. Yes, the domain of $f_{|X|}(x)$ is $x>0$. If you have a negative x-value at the normal distribution then $-x$ is positive. Nov 9, 2020 at 19:05
• Interesting that $P(|X|)= P(\frac{| X- \mu)|}{\sigma})$ when X ~N(0,1) so that the answer makes sense by symmetry.
– MSIS
Jun 11, 2021 at 20:12
• The pdf is incorrect, instead of twice the same term, it should be equal to the sum of two terms, one with $(x - \mu)^2$ in the exponent and one with $(x + \mu)^2$. Oct 18, 2023 at 3:35
• @callculus42 why have the $\mu$ symbol at all if it can only take the value 0? Oct 19, 2023 at 20:50