# Upper bound for the gaussian measure of an epsilon strip.

I have a question concerning the normal probability distribution: Suppose that $X\sim N(\mu,\sigma)$ is a normal distributed random variable with mean $\mu$ and variance $\sigma$. Let $\varepsilon>0$. Can one make a good guess on what might be an upper bound for

$$\frac{\text{Probability}\{0<X<\varepsilon\}}{\text{Probability}\{0<X\}}\quad (=\text{Probability}\{X<\varepsilon|X>0\}).$$

in terms of $\varepsilon$? My initial guess would be something like $\mathcal{O}(\varepsilon)$. It seems like an elementary question to me, but I haven't found an answer so far.

• Rewriting as integrals gives: $\frac{\int\limits_t^{t+\varepsilon}\exp\left(\frac{-x^2}{2}\right)\; dx}{\int\limits_{t}^{\infty}\exp\left(\frac{-x^2}{2}\right)\; dx}$, where $t=-\frac{\mu}{\sigma}$. I easily can bound the numerator by $\exp(-\mu^2/2) \varepsilon/2$. But I can't find a good bound for the fraction itself. – user131370 Feb 25 '14 at 15:04
If you write $F(u)=\int_0^u e^{-\frac{x^2}{2}}dx$, then $F$ is $\mathcal C^1$ with derivative $e^{-\frac{u^2}2}$, and $$\int_t^{t+\varepsilon}e^{-\frac{x^2}{2}}dx =F(t+\varepsilon)-F(t) .$$ So $\int_t^{t+\varepsilon}e^{-\frac{x^2}{2}}dx$ is equivalent to $\varepsilon \, F'(t)=e^{-\frac{t^2}2}\,\varepsilon$ as $\varepsilon\to 0$. Your denominator seems to be a constant with respect to $\varepsilon$. So your probability is indeed $O(\varepsilon)$.