# Integrating a half-normal pdf

The PDF of the half-normal distribution is:

$$f(x;\sigma)=\frac{\sqrt{2}}{\sigma\sqrt{\pi}}\exp\left(-\frac{x^2}{2\sigma^2}\right)\,\quad x>0\,.$$

The integral $$\int_{0}^{\infty}f(x;\sigma)\mathrm{d}x=\int_{0}^{\infty}\frac{\sqrt{2}}{\sigma\sqrt{\pi}}\exp\left(-\frac{x^2}{2\sigma^2}\right)\mathrm{d}x$$

should be equal to one. The code used to check this:

Integrate[(Sqrt[2/Pi]/s)*Exp[-(x^2)/(2*s^2)], {x, 0, Infinity},
Assumptions -> Re[s^2] > 0]


Mathematica gives $\frac{1}{\sqrt{\sigma}}$ which doesn't make sense. What am I doing wrong?

## migrated from math.stackexchange.comApr 14 '14 at 5:35

This question came from our site for people studying math at any level and professionals in related fields.

• And it is equal to $1$. – André Nicolas Apr 14 '14 at 3:33
• See the Mathematica code I added to the question. – user142972 Apr 14 '14 at 3:48
• Luckily not my problem, I can't afford Mathematica. Conceivably a bug. More likely an input typo. – André Nicolas Apr 14 '14 at 4:22

Using Assuming[Element[s, Reals] && s > 0, ...] tells Mathematica that Re[s^2]>1 and also s itself is real to allow it to simplify final result to 1.

Mathematica did not simplify $\sqrt{\frac{1}{s^2}}$ to $\frac{1}{s}$ since it did not know if $s$ was real or not and if it was real, if it is positive or not. That is why you did not get the result 1 expected.

Assuming[Element[s, Reals] && s > 0, Integrate[(Sqrt[2/Pi]/s)*Exp[-(x^2)/(2*s^2)],
{x, 0, Infinity}]]

(* 1 *)

• thanks this is the right answer, but for this OP, it might be helpful to provide a bit more textual explanation. – Verbeia Apr 14 '14 at 6:19
• Take s in the denominator outside the integral and you need no assumptions. It just gives 1. – Bill Watts Mar 5 at 8:33

Here are some variants:

td = TruncatedDistribution[{0, Infinity}, NormalDistribution[0, s]];
Assuming[Element[s, Reals] && s > 0,
Integrate[Sqrt[2/Pi] Exp[-x^2/(2 s^2)]/s, {x, 0, Infinity}]]
Integrate[Sqrt[2/Pi] Exp[-x^2/(2 s^2)]/s, {x, 0, Infinity},
Assumptions -> {Element[s, Reals], s > 0}]
Integrate[PDF[td, x], {x, 0, Infinity},
Assumptions -> {Element[s, Reals], s > 0}]
Integrate[
PDF[HalfNormalDistribution[Sqrt[2/Pi]/s], x], {x, 0, Infinity},
Assumptions -> {Element[s, Reals], s > 0}]


Note all the integrals yield 1. As @Nasser has observed the assumptions need to be declared.