Mill's Inequality on normal distribution

Question

Given that $Z \sim N(0,1)$. Prove Mill's Inequality:

$$P(|Z|>t) \leq \sqrt{\frac{2}{\pi}}\frac{e^ {\frac{-t^2}{2}}}{t} ~\forall t > 0$$

Answer 1

For $Z\sim \mathcal{N}(0,1)$ (i.e. a standard Normal distirbution) and $t>0$ we have

$P(|Z|>t)=P(Z^{-}<-t)+P(Z^{+}>t)$

where $Z^{-}$ are negative values of $Z$ and $Z^{+}$ are positive values of $Z$.

Now, the distribution of $Z$ is symmetric around $0$, so $P(Z^{-}<-t)=P(Z^{+}>t)$, leading to

$P(|Z|>t)=2P(Z^{+}>t)$

As part of the inequalities involved in proving Markov's inequality (see third slide in here http://www.math.leidenuniv.nl/~gugushvilis/STAN4.pdf), we have (for probability density function $f(x)$)

$\int_t^\infty xf(x)dx \geq t\int_t^\infty f(x)dx=tP(x>t)$

substituting the values with our standard Normal distribution, we have

$\large\frac{1}{\sqrt{2\pi}}\int_{t}^{\infty}xe^{-\frac{x^2}{2}}dx\geq \frac{t}{\sqrt{2\pi}}\int_{t}^{\infty}e^{-\frac{x^2}{2}}dx=tP(Z^{+}>t)=\frac{t}{2}P(|Z|>t)$

Integrating the left hand side results in

$\large\frac{1}{\sqrt{2\pi}}\int_{t}^{\infty}xe^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2\pi}}[-e^{-\frac{x^2}{2}}]_t^{\infty}=\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}$

Therefore

$\large \frac{t}{2}P(|Z|>t)\leq \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}$

leading to

$\large P(|Z|>t)\leq \sqrt{\frac{2}{\pi}}\frac{e^{-\frac{t^2}{2}}}{t}$