# The probability density function of half-chi-square distribution

Let $$X$$ be a random variable from a chi-square distribution with 1 degree of freedom. The probability density function (pdf) of $$X$$ is $$f(x) = \frac{\exp{(-x/2)}}{\sqrt{2\pi x}}$$, $$x>0$$.

In the case of half-chi-square distribution with 1 degree of freedom ($$\frac{1}{2} \chi^{2}_{1}$$), is the pdf provided by $$\frac{1}{2}\frac{\exp{(-x/2)}}{\sqrt{2\pi x}}$$?

Here's a useful general result.

Theorem. Suppose $$X$$ is a random variable with cumulative distribution function $$F_X$$, let $$a, b \in \mathbb{R}$$ with $$b > 0$$, and let $$Y = a + b X$$.

1. The cumulative distribution function of $$Y$$ (call it $$F_Y$$) is given by $$F_Y(y) = F_X\left(\frac{y - a}{b}\right)$$ for all $$y \in \mathbb{R}$$.
2. If $$X$$ is also continuous with density $$f_X$$, then the density of $$Y$$ (call it $$f_Y$$) is given by $$f_Y(y) = \frac{1}{b} f_X\left(\frac{y - a}{b}\right)$$ for all $$y \in \mathbb{R}$$.

Your question is the special case where $$X \sim \chi^2_1$$ (whose density you know), and you're interested in the density of $$Y = \frac{1}{2} X$$ (here $$a = 0$$ and $$b = 1/2$$). The theorem above will allow you to compute the density of $$Y$$.

Proof of 1. Just compute: \begin{aligned} F_Y(y) &= P(Y \leq y) \\ &= P(a + b X \leq y) \\ &= P\left(X \leq \frac{y - a}{b}\right) && \text{(*)}\\ &= F_X\left(\frac{y - a}{b}\right). \end{aligned} (*) Here we used the assumption that $$b > 0$$. If $$b < 0$$, the inequality would change direction, and if $$b = 0$$, we wouldn't be able to divide by $$b$$ at all.

Proof of 2. Differentiating the result of the first part, we get \begin{aligned} f_Y(y) &= \frac{d}{dy} F_Y(y) &&\text{(**)}\\ &= \frac{d}{dy} F_X\left(\frac{y - a}{b}\right) \\ &= \frac{1}{b} F^\prime_X\left(\frac{y - a}{b}\right) &&\text{(chain rule)}\\ &= \frac{1}{b} f_X\left(\frac{y - a}{b}\right). &&\text{(**)} \end{aligned} (**) Here we used the fact that a density of a continuous random variable is the derivative of that random variable's cumulative distribution function, which is a consequence of the definition of the density and the fundamental theorem of calculus.

• You might want to re-emphasize the $b>0$ assumption since it is critical to both parts 1. and 2. – Dilip Sarwate Feb 8 '19 at 11:29
• @DilipSarwate good suggestion; I've updated my answer – Artem Mavrin Feb 8 '19 at 17:40