# Upper bound for distribution function of the standard normal distribution

Could use a little help here,

Let: \begin{align*} \Phi (x) := \int_{-\infty}^x \dfrac{1}{\sqrt{2\pi}}e^{-\dfrac{t²}{2}}\, dt \end{align*}

and I would like to have a result for $x\leq 0$, that it holds: \begin{align*} \Phi (x) \leq e^{-\dfrac{x²}{2}} \end{align*}

Does this hold? If yes, how could one show that?

Wikipedia reference for these formulas.

The cumulative distribution is related to the error function:

$$\Phi(x) = \frac{1}{2} \operatorname{erfc}\left( -\frac{x}{\sqrt{2}} \right)$$

and the complementary error function satisfies

$$\operatorname{erfc}\left(x\right) \leq \exp(-x^2) \qquad \qquad x > 0$$

Combining these gives

$$\Phi(x) \leq \frac{1}{2} \exp\left( -\frac{x^2}{2} \right) \qquad \qquad x < 0$$