- #1
collinback
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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.
[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.
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