We have been given this equality in a class, and I would like to prove it.
$$\int_a^\infty e^w \phi_{\mu,\sigma}(w)dw = e^{\mu+\frac{\sigma^2}{2}} \Phi(\frac{\mu+\sigma^2 – a}{\sigma})$$
Where $\phi$ is the normal pdf with mean $\mu$ and variance $\sigma$ and $\Phi$ is the standard normal (0,1) CDF.
The teacher's assistant refused to prove it but said, OK it is so easy, and can be done using first principles and a change in variables. Here is the proof I came up with.
$$e^z f(z) = \frac{1}{\sigma \sqrt{2 \pi}}e^{z-\frac{1}{2}\left(\frac{z – \mu}{\sigma}\right)^2}$$ And define the indefinite integral
(used Mathematica). $$\int \frac{1}{\sigma \sqrt{2 \pi}}e^{z-\frac{1}{2}\left( \frac{z – \mu}{\sigma}\right)^2} dz =
-\frac{1}{2} e ^{(\sigma^2 / 2 + \mu)} \text{Erf} \left(\frac{\sigma^2 + \mu – z }{\sqrt{2} \sigma} \right)$$The Erf function above is the Gauss error function, defined by
$\text{Erf}(z) = \frac{2}{\sqrt{\pi}}\int_0^z e^{-t^2} dt$. I then use
this to evaluate the definite integral, which uses the fact that the
Erf function has a value of -1 for $-z$ as $z$ tends to infinity.$$\int_{a}^\infty \frac{1}{\sigma \sqrt{2 \pi}}e^{z-\frac{1}{2}\left(
\frac{z – \mu}{\sigma}\right)^2} dz = -\frac{1}{2} e ^{(\sigma^2 / 2 +
\mu)} (-1)+ \frac{1}{2} e ^{(\sigma^2 / 2 + \mu)} \text{Erf}
\left(\frac{\sigma^2 +u – a }{\sqrt{2} \sigma} \right) $$ $$ =
\frac{1}{2} e ^{(\sigma^2 / 2 + \mu)}\left(1+ \text{Erf}
\left(\frac{\sigma^2 +u – a }{\sqrt{2} \sigma} \right) \right)$$The Erf is related to the standard normal cumulative distribution
function. Let $N(x)$ be the CDF for a standard normal distribution (
to match the notation used in Björk), then$$N(x)= \frac{1}{2}\left(1+ \text{Erf}\left( \frac{x}{\sqrt{2}}\right)
\right)$$So we surely have the following relation.
$$N \left( \frac{1}{\sigma} (\sigma^2 +u – a )
\right)= \frac{1}{2}\left(1+ \text{Erf}\left( \frac{\sigma^2 +u – a
}{\sqrt{2} \sigma}\right) \right)$$So all together for this step we have evaluated the following definite
integral. $$\int_{a}^\infty e^z f(z) dz = \int_{a}^\infty
\frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{1}{2}\left( \frac{z –
\mu}{\sigma}\right)^2} dz = e ^{(\sigma^2 / 2 + \mu)} N \left( \frac{1}{\sigma} (\sigma^2 +u – a ) \right)$$
The T.A. was suggesting that the way I used mathematica for the indefinite integral isn't needed and this conclusion can be made in a much "simpler" manner. Perhaps with just using basic principles of the PDF, CDF, etc. I am just not seeing how I can make this conclusion without evaluating the indefinite integral as I have done above, in terms of the Erf and then converting to the CDF.
Thanks!
Best Answer
\begin{align} \frac{1}{\sqrt{2\pi}\sigma}\int_a^\infty e^w e^{-\frac{(w-\mu)^2}{2\sigma^2}} dw &= \frac{1}{\sqrt{2\pi}\sigma}\int_a^\infty e^{-\frac{(w-\mu)^2 - 2\sigma^2w}{2\sigma^2}} dw \\ &= \frac{1}{\sqrt{2\pi}\sigma}\int_a^\infty e^{-\frac{(w^2+\mu^2-2(\mu+\sigma^2)w}{2\sigma^2}} dw \\ &= \frac{1}{\sqrt{2\pi}\sigma}\int_a^\infty e^{-\frac{(w-(\mu+\sigma^2))^2 + \mu^2 - (\mu + \sigma^2)^2}{2\sigma^2}} dw \\ &= e^{\mu + \frac{\sigma^2}{2}}\frac{1}{\sqrt{2\pi}\sigma}\int_a^\infty e^{-\frac{(w-(\mu+\sigma^2))^2}{2\sigma^2}} dw \end{align}
The result follows from the symmetry of Normal pdf i.e. $\Phi(-x) = 1 - \Phi(x).$