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?

  • $\begingroup$ 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. $\endgroup$
    – Glen_b
    Commented Nov 14, 2021 at 0:21
  • $\begingroup$ @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 ? $\endgroup$
    – user226073
    Commented Nov 14, 2021 at 0:43

1 Answer 1



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

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 $$ \frac{x^{a/2-1}e^{-x/2}}{2^{a/2}\Gamma(a/2)} $$


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