# Quantile of Chi-squared = inverse Quantile of inverse Chi-squared distribution?

I used to think that the following held true:

If $Y\sim \chi^2$, then $\frac{1}{Y}\sim \operatorname{inv}\chi^2$, the inverse $\chi^2$-distribution. Let $\chi^2_\alpha$ denote the $\alpha$-quantile of $Y$, then $\frac{1}{\chi^2_\alpha}$ is the $\alpha$-quantile of $\frac{1}{Y}$, meaning $$P\left(Y\leq \chi^2_\alpha\right)=P\left(\frac{1}{Y}\leq\frac{1}{\chi^2_\alpha}\right)=\alpha.$$ Now I figured out that this cannot be true, right? But what is the relationship between the two quantiles then? (Or is it true after all?)

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You need to change "$\le$" to "$\ge$": $$\Pr\left(Y\le\chi^2_\alpha\right) = \Pr\left(\frac 1 Y \ge \frac{1}{\chi^2_\alpha}\right).$$ (The fact that $Y$ is always positive is essential here; this inversion of inequalities doesn't happen if the number on one side of the inequality is positive and the one on the other side is negative.)