# Why does $\mathbb{P}\left(X < -z\right) = \alpha \Rightarrow -z = \chi^2_{1 - \alpha}(2n)$ hold?

Assume $$X_i$$ are generated by $$\Gamma(\theta_0,n)$$ distribution, and $$S_n = \sum X_i$$.

Further, it is known that $$2 \theta_0 S_n$$ follows a $$\chi^2(2n)$$ distribution, $$\theta_0$$ is known, $$\theta_1 > \theta_0$$.

My question: why does the following statement hold:

$$\mathbb{P}_{\theta_0}\left(2 \theta_0 S_n < -\frac{d2 \theta_0}{\theta_1 - \theta_0}\right) = \alpha \Rightarrow -\frac{d 2 \theta}{\theta_1 - \theta_0} = \chi^2_{1 - \alpha}(2n)$$ for some constant $$d$$ and a fixed value $$\alpha>0$$.

My reasoning would be, that since $$2\theta_0 S_n$$ ~ $$\chi_2(2n)$$, the statement would be equivalent to

$$F_{\chi^2}\left(-\frac{d2\theta_0}{\theta_1 - \theta_0}\right) = \alpha \Rightarrow F^{-1}F_{\chi^2}\left(-\frac{d2\theta_0}{\theta_1 - \theta_0}\right) = F^{-1}(\alpha)=\chi^2_{\alpha}(2n)$$

Why does the last quantile turn to $$\chi^2_{1 - \alpha}(2n)$$ instead of $$\chi^2_{\alpha}(2n)$$?

What am I missing?

• Remember that by definition, $P(\chi^2 (2n)>\chi^2_{1-\alpha}(2n))=1-\alpha$. – StubbornAtom Apr 17 at 21:58
• @StubbornAtom oh, so that is the definition of $\chi_{1-\alpha}^2$. I assumed it was $F^{-1}(1-\alpha)$. Yes, then $P(X < \chi^{2}_{1-\alpha}) = \alpha$ and everything makes sense. Thank you! Do you want to post an answer, or should I just delete this? – Nutle Apr 17 at 22:06
• Do not delete the post. – StubbornAtom Apr 17 at 22:32

$$\chi^2_{\alpha,p}$$ always means the upper $$100(1-\alpha)\%$$ point or equivalently the $$(1-\alpha)$$th quantile/fractile of a chi-square distribution with $$p$$ degrees of freedom. Same goes with fractiles of other distributions.
As such, for $$X\sim \chi^2_p$$ and any $$\alpha\in (0,1)$$ this means $$P(X>\chi^2_{\alpha,p})=\alpha$$