Jensen's Inequality

For convex $f$ , $E [f (X)] \geq f (E [X])$ .

What is it?

Convex functions of random variables have expectations at least the function of the expectation; concave functions reverse the inequality.

It explains convexity value in options, why arithmetic returns overstate growth, and many inequality proofs.

Use it when comparing the average of a function to the function of the average.

Think of this method when…

A convex curve lies below its chords, so averaging first and applying $f$ later gives a smaller value.

f convex \Rightarrow E [f (X)] \geq f (E [X])

$X$ is 1 or 3 with equal probability. Compare $E [1/ X]$ with $1/ E [X]$ .

1
$E [1/ X] = \frac{1}{2} (1 + \frac{1}{3}) = \frac{2}{3}$ and $1/ E [X] = 1/2$ .
2
$1/ x$ is convex for $x > 0$ , consistent with Jensen.
$\frac{2}{3} > \frac{1}{2}$
Answer: $E [1/ X] = 2/3 > 1/2 = 1/ E [X]$

Is $E [max (S - K, 0)] \geq max (E [S] - K, 0)$ ?

6 QuantSolve questions are tagged with this method.