I need to solve this exercise:
Let $\xi$ be a random variable defined on a probability space $(\Omega, F,\mathbb{P}).$ Prove that:
- $\inf_{a\in\mathbb{R}}\mathbb{E}(\xi - a)^2 = \mathbb{D}\xi,$ where $\mathbb{D}\xi = \mathbb{E}(\xi - \mathbb{E}\xi)^2$,
- $\inf_{a\in\mathbb{R}}\mathbb{E}|\xi - a| =\mathbb{E}|\xi - m_\xi|,$ where $m_\xi$ is a median of random variable $\xi$, i.e. $m_\xi$ is a point under condition $\mathbb{P}(\xi<m_\xi)\leq\frac{1}{2}\leq\mathbb{P}(\xi\leq m_\xi)$.
I know how to prove the first one since $(\xi -a )^2 = (\xi - \mathbb{E}\xi + \mathbb{E}\xi - a)^2 = (\xi - \mathbb{E}\xi)^2 + 2(\xi - \mathbb{E}\xi)(\mathbb{E}\xi -a) + (\mathbb{E}\xi -a)^2.$ Applying this to the first problem we will get that this expression is smallest when $a = \mathbb{E}\xi$ and we will receive that $\inf_{a\in\mathbb{R}}\mathbb{E}(\xi - a)^2 = \mathbb{E}(\xi - \mathbb{E}\xi)^2 = \mathbb{D}\xi$. Can somebody explain me how I need to solve 2nd problem?
