# Different Versions of Jensen's Inequality?

According to wikipedia:

Jensen's inequality is

$${\displaystyle f(tx_{1}+(1-t)x_{2})\leq tf(x_{1})+(1-t)f(x_{2}).}$$

With this, I can easily prove Inequality of arithmetic and geometric means, in other words to prove that:

$${\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\geq {\sqrt[ {n}]{x_{1}\cdot x_{2}\cdots x_{n}}}\,,$$

My problem is that my teacher didn't define Jensen's inequality in the same way above but rather:

Let $$X$$ be a discrete random variable with finite expected value and let $$h:\mathbb{R} \to \mathbb{R}$$ be a convex function. then: $$h(E[X])\leq E[h(X)]$$

How can I reach (prove) the first version using my professor's version of the inequality, so I can prove the Inequality of arithmetic and geometric means?

• Take $X$ with distribution concentrated on the set $\{x_1, x_2\}$ with masses $t$ and $1-t.$ Commented Jan 13, 2022 at 21:15
• Proving that the definition of convexity implies the expectation variety is not entirely straightforward. Commented Jan 13, 2022 at 21:17
• @copper.hat I suspect the reciprocal of what OP is asking, i.e. what you are comenting, is not true for general discrete variables but rather finite discrete. Commented Jan 13, 2022 at 21:18
• @WilliamM. I suspect you are correct, but I wanted the OP to realise that, while equivalent, it is not immediate (unless you are familiar with some facts about convex functions). Commented Jan 13, 2022 at 21:21

The way your prof stated Jensen's inequality is more general. The inequality $$f(tx_1+(1-t)x_2)\leq tf(x_1)+(1-t)f(x_2),\quad t\in[0,1]\quad [1]$$ is a special case of
$$f(E[X])\leq E[f(X)]\quad \text{Jensen's inequality}$$
for $$f$$ convex, where the random variable $$X$$ has two-point support $$\{x_1,x_2\}$$ with respective probability masses $$t,1-t.$$ However, it is important to note that $$[1]$$ is actually the definition of $$f$$ being a convex function, so one doesn't require Jensen's inequality to justify it.