# Integration of a conditional probability

I encountered an integration problem when I read a statistics textbook, which involve the following relationship.

$\int_{-\infty}^{0}\phi(z;\beta,1) dz = 1-\phi(\beta)$

Here $\phi$ is a normal cdf. $\phi(z;\beta,1)$ represents a normal distribution of z with mean $\beta$ and standard deviation 1 and beta is also a random variable.

I wonder how to prove this relationship?

• This should be $\Phi(x\beta)$ in the right member of the first inequality. – Stéphane Laurent Jun 30 '17 at 13:06
• I had thought he means both sides are $\Phi$ which is quite hard. – Deep North Jun 30 '17 at 13:16

Let $X \sim {\cal N}(\mu, 1)$. Your integral is equal to $\Pr(X \leq 0)$ when $\mu = x\beta$. One has $X = Z + \mu$ where $Z \sim {\cal N}(0,1)$. Thus your integral is equal to $\Pr(Z \leq -\mu) = \Phi(-\mu) = 1 - \Phi(\mu)$.