Show $z_\alpha = -z_\alpha$.

Question

Given $Z\sim N(0,1)$ and the standard normal upper quantile $z_\alpha = \Phi^{-1}(1-\alpha)$ I need to show that $$ z_\alpha=-z_\alpha$$ The hint given was to use the fact that the density function ($\phi$) is an even function and show $\Phi(x)=1-\Phi(-x)$.

Showing that was not too difficult: $$\begin{align*} \Phi(x)=&\int_{-\infty}^{\infty}\phi(x)dx -\int_{x}^{\infty}\phi(x)dx\\ =&\int_{-\infty}^{\infty}\phi(x)dx -\int_{-\infty}^{-x}\phi(x)dx\quad\text{because $\phi$ is even}\\ =& 1-\Phi(-x) \end{align*} $$

The next hint was to now substitute $x$ for $z$ but I do not see how to arrive at $ z_\alpha=-z_\alpha$

The only other conclusions I came to is that $\Phi(z_\alpha) = 1-\alpha $ an then $\Phi(-z_\alpha )=\alpha$?

Any push in the right direction would be appreciated!

Edit:

I realized substituting it into what was proven does show something that is:

$\Phi(z_\alpha) = 1-\Phi(-z_\alpha)$ which gives us $1-\alpha = 1-\alpha$

But that did not really achieve any result.

Answer 1

Assuming the problem is asking you to show $z_\alpha = -z_{1-\alpha}$:

• By the definition of $z_\alpha$ and $z_{1-\alpha}$, you have $\Phi(z_{1-\alpha}) = \alpha$ and $\Phi(z_\alpha) = 1-\alpha$.
• By the work that you have done, you have $\Phi(z_{1-\alpha}) = 1 - \Phi(-z_{1-\alpha})$.
• Thus $\Phi(-z_{1-\alpha}) = 1-\alpha$, and finally $-z_{1-\alpha} = z_\alpha$.