Let $Z$ be a stochast with $EZ = 0$ and $VarZ = \sigma^{2}$. Show that for $u,v>0$ that the following inequality holds: Let $Z$ be a stochast with $EZ = 0$ and $VarZ = \sigma^{2}$. Show that for $u,v>0$ that the following inequality holds:
$P(Z\leq -u \space \text{or} \space Z\geq v) \leq \frac{4\sigma^{2} + (u-v)^{2}}{(u+v)^{2}}$.
The only thing I notice is that $EZ^{2} = \sigma^{2}$ and that the term on the left is the same as:
$P(Z\leq -u) + P(Z\geq v)$
I think it has something to do with the law of the large numbers but I'm still clueless....
Kees
 A: First of all, we note that
$$\mathbb{P}(Z \leq -u \, \text{or} \, Z \geq v) = \mathbb{P} \left( \left| Z - \frac{v-u}{2} \right| \geq \frac{1}{2}(u+v) \right).$$
By applying Markov's inequality (see below), we find
$$\begin{align*} \mathbb{P}(Z \leq -u \, \text{or} \, Z \geq v) &\leq \frac{4}{(u+v)^2} \mathbb{E} \left[ \left( Z- \frac{1}{2} (v-u) \right)^2 \right] \\ &= \frac{4}{(u+v)^2} \cdot \left( \text{var} \, Z + \frac{1}{4} (v-u)^2 \right) \\ &= \frac{4 \sigma^2+(v-u)^2}{(u+v)^2} \end{align*}$$
using that $\mathbb{E}Z=0$ in the penultimate line.


Theorem (Markov inequality) Let $X \in L^2$ a square-integrable random variable on a probability space $(\Omega,\mathcal{A},\mathbb{P})$. Then, $$\mathbb{P}(|X| \geq r) \leq \frac{1}{r^2} \cdot \mathbb{E}(X^2)$$ for any $r>0$.

Proof: $$\begin{align*} \mathbb{P}(|X| \geq r) &= \int 1_{|X| \geq r} \, d\mathbb{P} \leq \int \frac{X^2}{r^2} \cdot 1_{|X| \geq r} \, d\mathbb{P} \leq \frac{1}{r^2} \int X^2 \, d\, \mathbb{P} = \frac{\mathbb{E}(X^2)}{r^2} \end{align*}$$
