# Sum of series with normal CDF

1. Apr 18, 2012

### collinback

I am trying to calculate the following sum:

$vt\Phi(a)+\frac{(vt)^2}{2!}\Phi(\sqrt{2}a)+\frac{(vt)^3}{3!}\Phi(\sqrt{3}a)+\frac{(vt)^4}{4!}\Phi ( \sqrt{4}a)+\ldots$

where $\Phi$ is the standard normal CDF. $v,t,a$ are constants.

A relevant formula is $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots$. But this cannot be directly applied.

We can also transform $\Phi(\sqrt{n}a)$ from integrals with different upper bounds to integrals with identical upper bound in the following way:
$\Phi(\sqrt{n}a)=\int_{-\infty}^{\sqrt{n}a}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=\int_{-\infty}^{a}\frac{\sqrt{n}} { \sqrt{2\pi}}e^{-\frac{(\sqrt{n}x)^2}{2}}dx$

Therefore, the original question is now to calculate
$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{a}[vte^{-\frac{x^2}{2}}+\frac{(vt)^2}{2!}\sqrt{2}e^{-\frac{(\sqrt{2}x)^2}{2}}+\frac{(vt)^3}{3!}\sqrt{3}e^{-\frac{(\sqrt{3}x)^2}{2}}+\frac{(vt)^4}{4!}\sqrt{4}e^{-\frac{(\sqrt{4}x)^2}{2}}+\ldots]dx$

But still, I don't know how to calculate the sum in the brackets.

Please let me know if you have any progress or if you spot any error in the above transformation.

Last edited: Apr 18, 2012
2. Apr 18, 2012

### DonAntonio

Very hard to understand what you wrote as you didn't introduce the $...$ symbols correctly. Try again and check (with "preview post") before you send your post.

DonAntonio