Prove fact about $\chi^2$ distribution.

Question

Prove that if $X \sim \chi^2_{n-1}$, and if $F$ is the distribution function of the variable. Then,

$$\mathbb{P}\{F^{-1}(\alpha/2) \leq X \leq F^{-1}(1-\alpha/2)\} = 1 - \alpha$$

I.e. this is needed to find the confidence interval for the variance of a population proportion.

Attempt: $$\mathbb{P}\{F^{-1}(\alpha/2) \leq X \leq F^{-1}(1-\alpha/2)\}$$$$ = \mathbb{P}\{X \leq F^{-1}(1-\alpha/2)\} - \mathbb{P}\{F^{-1}(\alpha/2) \leq X\} $$$$= \mathbb{P}\{X \leq F^{-1}(1-\alpha/2)\} - (1-\mathbb{P}\{X \leq F^{-1}(\alpha/2)\})$$$$ = F(F^{-1}(1-\alpha/2)) - (1- F(F^{-1}(\alpha/2))) $$$$=1 - \alpha/2 - (1- \alpha/2) = 0$$

and this is wrong.

Where is my mistake?

Answer 1

I tried working this out myself: $$\begin{align}\mathbb{P}(F^{-1}(\alpha/2) \leq X \leq F^{-1}(1-\alpha/2)) &= \mathbb{P}(X \leq F^{-1}(1-\alpha/2))-\mathbb{P}(X \color{red}{\leq} F^{-1}(\alpha/2)) \\ &= F(F^{-1}(1-\alpha/2))-F(F^{-1}(\alpha/2)) \\ &= 1-\alpha/2 - \alpha/2 \\ &= 1-\alpha\text{.} \end{align}$$ It looks like you messed up on the second line of equations. See the red for the difference above.