How Do You Calculate the Sum Involving Series and Normal CDF?

[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.

 

[DonAntonio]

I am trying to calculate the following sum:\Phi

vt\Phi (a) + \frac{2}{2!} \Phi(\sqrt{2}a) + \frac{(vt)³}{3!} ]\Phi(\sqrt{3}a)+\frac{(vt)⁴}{4!}]\Phi[/itex](\sqrt{4}a)+\ldots

Of course an relevant formula is that e^x=1+x+\frac{(x)²}{2!}+\frac{(x)³}{3!}+\frac{(x)⁴}{4!}+\ldots. But this cannot be directly applied.

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