The Notebook

**E[f(x)] \leq f(E[f(x)]) **

i.e.

**\sum_{i=1}^k p_{i} f(x_{i}) \leq f(\sum_{i=1}^k p_{i} f(x_{i})) **

**Preliminary Definitions on Concave and Convex**

**\texttt{concave: }$f((1-\lambda)a + \lambda b) \geq (1-\lambda)f(a) + \lambda f(b)$**

**\texttt{convex: }$f((1-\lambda)a + \lambda b) \leq (1-\lambda)f(a) + \lambda f(b)$**

**Proof**

Let

We can write:

**f(X) \leq f(E[X]) + (X - E[X])f^{\prime}(E[X])**

This inequality is true for all

**E[f(x)] \leq f(E[X]) + f^{\prime}(E[X])E[(X - E[X])] = f(E[X])**

since we can say that

**E[X] = \mu**, **E[(X - E[x])] = E[X - \mu] = E[X] - E[\mu] = \mu - \mu = 0**

- https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/cv1/t
- http://www.sef.hku.hk/~wsuen/teaching/micro/jensen.pdf