$\frac{1}{\sqrt{2\pi}}\int_a^\infty e^{-\frac{1}{2}(x-\mu)^2}dx$ - Normal Distribuition

Question

I have read in one of my finance books (Asset Pricing - John H. Cochrane) that there is this identity:

\begin{equation} \begin{split} \frac{1}{\sqrt{2\pi}}\int_a^\infty e^{-\frac{1}{2}(x-\mu)^2}dx &= \Phi(\mu-a) \end{split} \end{equation}

I was unable to find any explanation in the book or on the internet, could someone please explain why this works or give me a resource I could validate it from?

Answer 1

It seems that you have a random variable $X$ which is normally distributed with mean $μ$ and unit variance $σ^2=1$. So, $$f_X(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac12(x-μ)^2}$$ for $x\in \mathbb R$. As usual you can standardize this random variable by

$$Z=\frac{X-μ}{σ}=\frac{X-μ}{1}=X-μ \sim N(0,1)$$ i.e. $Z$ has the standard normal distribution with cdf $\Phi(\cdot)$. So,

\begin{align}\int_{a}^{+\infty}f_X(x)dx=P(X\ge a)&=P(X-μ \ge a-μ)=1-P(X-μ\le a-μ)\\[0.2cm]&\overset{(*)}=P(X-μ \le -(a-μ))=P(Z \le μ-a)\\[0.2cm]&=\Phi(μ-a)\end{align} where the (*) equality is true due to symmetry of the normal distribution.

Answer 2

Using the definition of the cumulative normal distribution function $\Phi$:

$$ \Phi(\mu-a) = {1\over \sqrt{2\pi}}\int_{-\infty}^{\mu - a} e^{-{1\over 2}t^2 }dt = - {1\over \sqrt{2\pi}}\int_{\mu - a}^{-\infty} e^{-{1\over 2}t^2 }dt = {1\over \sqrt{2\pi}}\int_{-\mu + a}^{\infty} e^{-{1\over 2}t^2 }dt $$ $$= {1\over \sqrt{2\pi}}\int_{ a}^{\infty} e^{-{1\over 2}({t-\mu})^2 }dt $$