I have been studying normal distribution and frequently I see that $$\Phi(1)=1-\Phi(-1)$$ where $\Phi(x)=\int_{-\infty}^{x}\phi(u)du$ is the cdf of a standard normal variable. How one can prove equality above? and is this true in general? i.e $\Phi(x)=1-\Phi(-x)$?
Note: I'm pretty beginner with probability so excuse me if this turns out to be very trivial.
Thanks
It's actually true for any $x$ that $\Phi(x) = 1-\Phi(-x).$ The easiest way to see this is to sketch the region each of these represents on the bell curve (LHS is everything but the right tail, RHS is everything but the left tail.)
To see this with the integrals, first write $$ \Phi(x) = \int_{-\infty}^x\phi(u)du = \int_{-\infty}^\infty\phi(u)du-\int_{x}^{\infty}\phi(u)du$$ then note two important facts. For one $\int_{-\infty}^\infty\phi(u)du = 1,$ and since $\phi(x) = \phi(-x),$ $\int_x^\infty\phi(u)du = \int_{-\infty}^{-x}\phi(u)du = \Phi(-x).$ Note the only special property of the normal distribution we used was that it is symmetric, so this holds for the CDF of any symmetric RV.
For a Standard Normal Distribution we already know that the total probability, \begin{align*} P&=1\\ \Rightarrow\int_{-\infty}^{\infty}\!f(x)\ dx&=1\\ \Rightarrow\int_{-\infty}^{-1}\!e^{-x^2/2}\ dx+\int_{-1}^{\infty}\!e^{-x^2/2}\ dx&=1\qquad\left[\text{as }e^{-x^2/2}\text{ is a even function, }\right.\\ \Rightarrow\int_{1}^{\infty}\!e^{-x^2/2}\ dx+\int_{-1}^{\infty}\!e^{-x^2/2}\ dx&=1\qquad\left.\int_{-\infty}^{-1}\!e^{-x^2/2}\ dx=\int_{1}^{\infty}\!e^{-x^2/2}\ dx.\right]\\ \Rightarrow\Phi(1)+\Phi(-1)&=1\\ \Rightarrow\Phi(1)&=1-\Phi(-1) \end{align*}