0

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?

Jose Avilez
  • 12,710
David
  • 9
  • Take $X$ with distribution concentrated on the set ${x_1, x_2}$ with masses $t$ and $1-t.$ – William M. Jan 13 '22 at 21:15
  • Proving that the definition of convexity implies the expectation variety is not entirely straightforward. – copper.hat Jan 13 '22 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. – William M. Jan 13 '22 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). – copper.hat Jan 13 '22 at 21:21

1 Answers1

1

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.

Golden_Ratio
  • 12,591