# 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 – Henry Apr 23 '15 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 $1$. – Brian Tung Apr 24 '15 at 17:38
• It's called Folded Normal Distribution. en.wikipedia.org/wiki/Folded_normal_distribution – Mr.M Nov 22 '16 at 9:49

You can to take the cdf as a starting point.

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

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

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

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

$$=2 \cdot P(X \leq x)-1=2\cdot F(x)-1=2 \cdot \lim_{a \to -\infty} \int_{a}^x \frac{1}{\sqrt{2\cdot \Pi}}\cdot e^{-\frac{{(t-\mu)}^2}{2 \sigma ^2}} \, dt-1$$

$$F(x)-F(a)-1$$

Differentiating :

$$\frac{dF(x)}{dx}-\frac{dF(a)}{dx}=2\cdot f(x)-0-0=2\cdot f(x)=2\cdot \frac{1}{\sqrt{2\cdot \Pi}}\cdot e^{-\frac{{(x-\mu)}^2}{2 \sigma ^2}}$$

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

• \Pi $\to$ \pi. – Did Dec 5 '18 at 8:53