How to proof that $\Phi(-x)=-\Phi(x)$ for Normal Distribution

Question

I have been studying the normal distribution and frequently I see that $\Phi(-x)=1-\Phi(x)$ and I can proof it with the integral.

Otherwise, I reached out that $\Phi(-x)=-\Phi(x)$, is there any chance to prove it?

I have tried but there is nothing$\ldots$

Answer 1

If you mean by $\Phi$ the cumulative distribution function, then $\Phi(-x)=1-\Phi(x)$ holds. Recall that for any $x \in \mathbb{R}$, $$ \Phi(x) = \int_{-\infty}^{x}e^{-\frac{s^2}{2}}ds.$$

Then by using integration by substitution (link), \begin{align} 1- \Phi(x) &= 1- \int_{-\infty}^{x}e^{-\frac{s^2}{2}}ds \\ &= \int_{-\infty}^{\infty}e^{-\frac{s^2}{2}}ds - \int_{-\infty}^{x}e^{-\frac{s^2}{2}}ds \\ &= \int_{x}^{\infty}e^{-\frac{s^2}{2}}ds \\ &= \int_{-x}^{-\infty}e^{-\frac{(-t)^2}{2}}(-1)dt & \text{ substitute $s=-t$ }\\ &= -\int_{-x}^{-\infty}e^{-\frac{t^2}{2}}dt \\ &= - \left(- \int_{-\infty}^{-x}e^{-\frac{t^2}{2}}dt\right)& \text{ switch bounds of integral} \\ &= \int_{-\infty}^{-x}e^{-\frac{t^2}{2}}dt \\ &= \Phi(-x). \end{align}