Mill's Inequality on normal distribution Given that $Z \sim N(0,1)$. Prove Mill's Inequality:
$$P(|Z|>t) \leq  \sqrt{\frac{2}{\pi}}\frac{e^ {\frac{-t^2}{2}}}{t} ~\forall t > 0$$
 A: 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
$\large P(|Z|>t)\leq \sqrt{\frac{2}{\pi}}\frac{e^{-\frac{t^2}{2}}}{t}$
