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Sum of series with normal CDF

  1. Apr 18, 2012 #1
    I am trying to calculate the following sum:

    [itex]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[/itex]

    where [itex]\Phi[/itex] is the standard normal CDF. [itex]v,t,a[/itex] are constants.

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

    We can also transform [itex]\Phi(\sqrt{n}a)[/itex] from integrals with different upper bounds to integrals with identical upper bound in the following way:
    [itex]\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[/itex]

    Therefore, the original question is now to calculate
    [itex]\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[/itex]

    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. jcsd
  3. Apr 18, 2012 #2


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

    DonAntonio
     
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