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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 $\frac{1}{2}\frac{\exp{(-x/2)}}{\sqrt{2\pi x}}$?

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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.

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

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