I have a normal Distribution $X \sim N(\mu, \sigma)$. Is there an easy way to give an asymptotic estimate with small error (I would prefer with relative error $\rightarrow 0$) for $P[X \geq k]$?
We have that $k=O(1)$, $\mu=o(1)$, $ \sigma \sim \mu$.
We would write then:
$$P[X \geq k] = \int_k^{\infty} \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(x-\mu)^2}{\sigma^2}}dx$$
Can we say something like
$P[X \geq k ] \sim f(\mu, \sigma, k)$?
Or at least $P[X \geq k] \leq f(\mu,\sigma,k)$ and $P[X \geq k] \leq g(\mu, \sigma, k)$ for some closed-form functions $f$, $g$?
Thank you very much..
For the $N(0,1)$ distribution we have that: $$\mathbb{P}[X>k]=\frac{1}{\sqrt{2\pi}}\int_{k}^{+\infty}e^{-x^2/2}dx =\frac{e^{-k^2/2}}{\sqrt{2\pi}}\int_{0}^{+\infty}e^{-kx}\cdot e^{-x^2}\,dx<\frac{e^{-k^2/2}}{k\sqrt{2\pi}}.\tag{1}$$ You can easily adapt this argument to the $N(\mu,\sigma)$ distribution.
For an efficient continued fraction approximation of the LHS in $(1)$, have a look at this page.