Steps to show that $\Phi(−z) = 1−\Phi(z)$

Question

The solution is to let $w = -v$. I determined that the steps are:

$$ \begin{split} \Phi(-z) &= \int_{-\infty}^{-z} \frac{1}{\sqrt{2\pi}}e^{\frac{-w^2}{2}}dw\\ &= \int_{\infty}^{z} \frac{1}{\sqrt{2\pi}}e^{\frac{-v^2}{2}}(-dv)\\ &= \int_{z}^{\infty} \frac{1}{\sqrt{2\pi}}e^{\frac{-v^2}{2}}dv\\ &= 1 - \int_{-\infty}^{z} \frac{1}{\sqrt{2\pi}}e^{\frac{-v^2}{2}}dv \end{split} $$

the first 3 of these make sense to me. You first make the substitution, then you flip the limits of integration to cancel out the negative. But, the last step in the book I don't understand. I see that the limits are flipped back, and in this way they match the function $\Phi(z)$. However, I don't understand how to add a $1$ such that the integral is subtracted from it. Thanks in advance!

Answer 1

This uses the fact \begin{align*} 1 &= \lim_{z \rightarrow \infty} \Phi(z) \\ &= \int_{-\infty}^\infty \; \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-v^2/2} \,\mathrm{d}v \\ &= \int_{-\infty}^z \; \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-v^2/2} \,\mathrm{d}v + \int_z^\infty \; \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-v^2/2} \,\mathrm{d}v \text{,} \end{align*} where $z$ is any real number. This is the relation used on the third and fourth lines of your display.

Equivalently, $\Phi$ is a cumulative density function and the total probability must be $1$.

Answer 2

An alternative mode to prove the equation consist to use the proprety of the gaussian variables; in fact if $Z\sim N(\mu,\sigma)$ then $aZ\sim N(a\mu,\sqrt{\lvert a\rvert}\sigma)$.

If $Z\sim N(0,1)$ then $-Z\sim N(0,1)$, $$\phi(-z)=\mathbb{P}(Z\leq-z)=1-\mathbb{P}(Z\geq-z)=1-\mathbb{P}(-Z\leq z)=1-\mathbb{P}(Z\leq z)=1-\phi(z)$$