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Let $Phi$ be standard Gaussian CDF and $u > 0$. What is good u-bound for $int_0^1Phi(u/r - ur)dr$ as a function of $u$?

Mathematics Asked on December 20, 2021

Let the function $Phi(x) := (1/sqrt{2pi})int_{-infty}^x e^{-t^2/2}dt$ be the standard Gaussian CDF. For $u > 0$, define $I(u) := int_0^1 Phi(u/r-ur)dr$.

Question. What are good upper-bounds for $I(u)$ in terms of $u$ ?

One Answer

A good simple upper bound for $u > 1$:

Using integration by parts and then the substitution $w = frac{u}{r} - ur$, we have begin{align} I(u) &= frac{1}{2} + frac{1}{2usqrt{2pi}} int_0^infty mathrm{e}^{-frac{1}{2}w^2} (sqrt{w^2+4u^2} - w)mathrm{d} w\ &le frac{1}{2} + frac{1}{2usqrt{2pi}} int_0^infty mathrm{e}^{-frac{1}{2}w^2} left(2u + frac{w^2}{4u} - wright)mathrm{d} w\ &= 1 - frac{1}{2usqrt{2pi}} + frac{1}{16u^2}. end{align} The upper bound $I_1(u) = 1 - frac{1}{2usqrt{2pi}} + frac{1}{16u^2}$ is good if $u > 1$. For example, $frac{I_1(1)-I(1)}{I(1)} approx 0.009176113058$, $frac{I_1(2)-I(2)}{I(2)} approx 0.0007002868670$, and $frac{I_1(10)-I(10)}{I(10)} approx 0.000001187547307$.

Answered by River Li on December 20, 2021

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