# Properties of $\chi^2(1)$ multiplied by a real value "$a$"

Is it true that the $$\chi^2$$ distribution with $$k=1$$ (noted $$\chi^2(1)$$) multiplied by a real value $$a$$ is equal to $$\chi^2(a)$$ ?

If not, is there a particular distribution for $$a\cdot\chi^2(1)$$?

If yes, it doesn't make sense since the summing index is no longer integer, does it?

Positive value for $$a$$ is also mandatory, no?

• It will have a gamma distribution, as your tags suggest. You can figure out which by starting with thinking about $a=1$ (to figure out the shape parameter) while recognizing that $a$ is a scale parameter. Commented Nov 14, 2021 at 0:21
• @Glen_b . The answer to my question is no : that will be a $\Gamma(\dfrac{1}{2},2a)$ distribution with convention (shape/scale) and so, this won't be a $\chi^2$ distribution anymore. Do you agree ?
– user226073
Commented Nov 14, 2021 at 0:43

# TL;DR

$$a>0, X\sim\chi^{2}(k)\implies aX\sim\Gamma\left( \frac{k}{2}, 2a\right)$$

## The full story

The assertion is false. You can see it as follows. Take $$k\geq 1$$, and consider $$X\sim \chi^{2}(k)$$ (read as: X is distributed as chi-square with k degrees of freedom). Take any constant $$a>0$$: what is the distribution of the random variable $$aX$$? You can show that $$aX\sim \Gamma\left( \frac{k}{2}, 2a\right)$$. Indeed,

$$P(aX\leq t) = P\left(X \leq \frac{t}{a} \right)$$ (note that here we use the fact that $$a>0$$ otherwise the equality is not valid). \begin{align*} P\left(X \leq \frac{t}{a}\right) &= \int_{0}^{ \frac{t}{a} } \frac{x^{k/2-1}e^{-x/2}}{2^{k/2}\Gamma(k/2)}dx \\ &\text{use the substitution } x= \frac{l}{a} \\ &=\int_{0}^{t} \frac{(l/a)^{k/2-1}e^{-l/(2a)}}{2^{k/2}\Gamma(k/2)} \frac{1}{a} dl \\ &=\int_{0}^{t} \frac{l^{k/2-1}e^{-l/(2a)}}{(2^{k/2}a^{k/2-1})\Gamma(k/2)} \frac{1}{a} dl \\ &=\int_{0}^{t} \frac{l^{k/2-1}e^{-l/(2a)}}{(2^{k/2}a^{k/2})\Gamma(k/2)} dl \\ &=\int_{0}^{t} \frac{l^{k/2-1}e^{-l/(2a)}}{(2a)^{k/2}\Gamma(k/2)} dl \\ &=P\left(Y \leq t\right) \end{align*} where $$Y$$ is distributed as $$\Gamma\left( \frac{k}{2}, 2a\right)$$.

Thus, take $$k=1$$ and you have that $$a>0, X\sim\chi^{2}(1)\implies aX\sim\Gamma\left( \frac{1}{2}, 2a\right).$$

The latter has pdf equal to $$\text{pdf } \Gamma\left( \frac{1}{2}, 2a\right)=\frac{x^{1/2-1}e^{-x/(2a)}}{(2a)^{1/2}\Gamma(1/2)}$$ which is different from the pdf of $$\chi^2(a)$$, which is equal to $$\frac{x^{a/2-1}e^{-x/2}}{2^{a/2}\Gamma(a/2)}$$