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?

  • 1
    $\begingroup$ Remember that by definition, $P(\chi^2 (2n)>\chi^2_{1-\alpha}(2n))=1-\alpha$. $\endgroup$ – StubbornAtom Apr 17 at 21:58
  • $\begingroup$ @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? $\endgroup$ – Nutle Apr 17 at 22:06
  • $\begingroup$ Do not delete the post. $\endgroup$ – 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 $$P(X>\chi^2_{\alpha,p})=\alpha$$


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