Probability Distribution of a standard normal distribution with absolute value

Z is a standard normal and we have to prove that

$$P(|z| \geq s) \leq {\sqrt{\frac{2}{\pi}}} {\frac{e^{\frac{-s^{2}}{2}}}{s}}$$

As far as I understand I have to use the formula for distribution function of a standard normal random variable, but I do not know how to deal with the ≥ inequality in the probability function as I have only seen examples with the ≤ inequality.

• "I do not know how to deal with the ≥ inequality": Well, you can always use $P(X \geq z) = 1-P(X < z)$... – Florian Apr 25 '17 at 8:20
• Consider the complement event: $P(|z| \ge s) = 1- P(|z|< s)$. – Crostul Apr 25 '17 at 8:21
• @Florian Sorry but, what for? – Did Apr 25 '17 at 8:59
• @Crostul Sorry but, what for? – Did Apr 25 '17 at 8:59
• @Did: well you had asked how to convert $\geq$ into $\leq$, this is how. But as you see in the answer by Mau314, this was not even necessary to begin with. – Florian Apr 26 '17 at 8:39

I suggest to do it without considering the complement; you can prove this by studying explicitly the corresponding integral: For $s\geq 0$, we have $$P(|Z|\geq s)=2\int_s^\infty \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right)\mathrm{d}x.$$ Now via transformation you can artificially introduce a factor that corresponds to $\frac{1}{x}$. This factor can be bounded by $\frac{1}{x}\leq\frac{1}{s}$ and you can integrate the remaining $\exp(\cdot)-$term as usual.
In more detail, choose the transformation $z:=\frac{x^2}{2}$. Then the right hand side above is equal to $$\frac{\sqrt{2}}{\sqrt{\pi}}\int_{s^2/2}^\infty \frac{1}{\sqrt{2 z}}\exp(-z)\mathrm{d}z.$$ Now, using $\frac{1}{\sqrt{2z}}\leq\frac{1}{s}$ we obtain $$P(|Z|\geq s)\leq \frac{\sqrt{2}}{\sqrt{\pi}}\frac{1}{s}\int_{s^2/2}^\infty \exp(-z)\mathrm{d}z=\frac{\sqrt{2}}{\sqrt{\pi}}\frac{1}{s}\exp\left(-\frac{s^2}{2}\right).$$