Let $t > 0$ and $X$ a random variable from Standard Gaussian Distribution. We then have the following tail bound
$$ P(X \geq t ) =\frac{1}{\sqrt{2\pi}} \int_{t} ^{\infty} \exp \left( \frac{-x^2} {2} \right) \ d x \leq \frac{1} {t\sqrt {2\pi}} \exp \left( \frac{-t^2} {2} \right). $$
I would like to derive the following, generalized tail bound where $X$ is drawn according to a Normal Distribution with mean $\mu$ and standard deviation $\sigma > 0$:
$$ P(X \geq \mu + \sigma t ) = \frac{1}{\sigma \sqrt{2\pi}} \int_{\mu +\sigma t} ^{\infty} \exp \left( \frac{-(x - \mu)^2} {2\sigma ^2} \right) \ d x \leq \frac{1} {t\sqrt {2\pi}} \exp \left( \frac{-t^2} {2} \right). $$
I came up with the following:
Proof. Let $z := \sigma x + \mu$. Then $ \frac{d}{dx} z = \sigma.$ So we get by substitution
$$ \int_{\mu +\sigma t} ^{\infty} \exp \left( \frac{-(z - \mu)^2} {2\sigma ^2} \right) \ d z = \int_{t} ^{\infty} \sigma\exp \left( \frac{-x^2} {2} \right) \ d x , $$ which, together with the original tail bound, yields the second bound. qed.
I am not very accustomed to probability theory and integrals. Can anyone point out mistakes that I made or can one actually generalize the bound that easily? Also, if you think one needs to operate a bit more carefully, let me know.
Thanks very much in advance!
