I have these pair of numbers $ (a, b) = (\frac{4}{9}, \frac{1}{9}) $ and $(c, d) = (\frac{1}{2}, \frac{1}{6}) $. (Number mean nothing, just for illustration and simplification)

Note that - (a, b) are pair of numbers which represent $((E(e_1))^2, (E(e_2))^2) $ and (c, d) are pair for $((E(e_1^2)), (E(e_2^2))) $

Where, E is the expectation.

Clearly, c>a and d>b (by Jensen inequality) But when I divide $\frac{a}{b}$ and $\frac{c}{d}$, I get $\frac{a}{b}$ > $\frac{c}{d}$.

So, does Jensen's inequality flips when two Jensen inequality are divided?

Is there a property which ensure that this will be true always.

Please help

  • $\begingroup$ What is the context? What are $e_1$ and $e_2$? How do you define "superior"? Why are you surprised that $\frac{a}{b}>\frac{c}{d}$? $\endgroup$ Sep 20, 2023 at 17:56
  • $\begingroup$ Individually a and b are smaller, and it's not like b is too small (as difference is same) yet ratio comes out to be bigger with a/b. $e_1$ and $e_2$ are random variable with uniform distribution over the interval (0,1). Is it because I have taken uniform distribution that I am getting these result. Had it been other distribution, results would have been so starking? $\endgroup$ Sep 20, 2023 at 23:25
  • $\begingroup$ If both $e_1$ and $e_2$ are Uniform[0,1], then their squared expected values are the same, $a=b=0.5^2$, and the expected values of their squares are the same, $c=d=\frac{1}{3}$. Do you mean sample means instead of expected values? $\endgroup$ Sep 21, 2023 at 6:37
  • $\begingroup$ I don't understand the question. Jensen's inequality is a statement about expectations over concave/convex functions. I would ask the question as if X is a RV and we know $f(x)>g(x)$ (where all these functions are concave/convex) then what can we say about $\frac{\mathbb{E}[f(x)]}{\mathbb{E}[g(x)]}$ and $\mathbb{E}\left[\frac{f(x)}{g(x)}\right]$. In this case you'd just note that Jensen's inequality for $f(x)$ and $g(x)$ and find that ratios provide indeterminant equality. $\endgroup$
    – EconJohn
    Sep 22, 2023 at 5:35

2 Answers 2


If $X$ and $Y$ are independent positive random variables then the following are true by Jensen's Inequality: \begin{eqnarray*} \mathbb{E}\left(\dfrac{X}{Y}\right) \underbrace{=}_{\text{By Independence}} \mathbb{E}\left(X\right)\mathbb{E}\left(\dfrac{1}{Y}\right)\underbrace{\geq}_{\text{By Jensen's}} \dfrac{\mathbb{E}\left(X\right)}{\mathbb{E}\left(Y\right)} & \ldots (1)\end{eqnarray*} \begin{eqnarray*} \mathbb{E}\left(\dfrac{X^2}{Y^2}\right) = \mathbb{E}\left(\left(\dfrac{X}{Y}\right)^2\right) \underbrace{\geq}_{\text{By Jensen's}} \left(\mathbb{E}\left(\dfrac{X}{Y}\right)\right)^2 \underbrace{\geq}_{\text{By (1)}} \left(\dfrac{\mathbb{E}\left(X\right)}{\mathbb{E}\left(Y\right)}\right)^2 = \dfrac{\left(\mathbb{E}\left(X\right)\right)^2}{\left(\mathbb{E}\left(Y\right)\right)^2}\end{eqnarray*}

However, if you are interested in comparing the ratios $\dfrac{\mathbb{E}\left(X^2\right)}{\mathbb{E}\left(Y^2\right)}$ and $\dfrac{\left(\mathbb{E}\left(X\right)\right)^2}{\left(\mathbb{E}\left(Y\right)\right)^2}$, then the inequality depends on the choice of the distribution of random variables. When $X$ and $Y$ are identically distributed then $\dfrac{\mathbb{E}\left(X^2\right)}{\mathbb{E}\left(Y^2\right)}=\dfrac{\left(\mathbb{E}\left(X\right)\right)^2}{\left(\mathbb{E}\left(Y\right)\right)^2}$. When $X \sim \text{Unif}(0,1)$ and $Y=X+1\sim\text{Unif}(1,2)$, then $\dfrac{\mathbb{E}\left(X^2\right)}{\mathbb{E}\left(Y^2\right)}=\dfrac{\mathbb{E}\left(X^2\right)}{\mathbb{E}\left(X^2+1+2X\right)}=\dfrac{1}{7}>\dfrac{1}{9}=\dfrac{\left(\mathbb{E}\left(X\right)\right)^2}{\left(\mathbb{E}\left(Y\right)\right)^2}$. Reversing the roles of $X$ and $Y$ i.e. when $X \sim \text{Unif}(1,2)$ and $Y=X-1\sim\text{Unif}(0,1)$, then $\dfrac{\mathbb{E}\left(X^2\right)}{\mathbb{E}\left(Y^2\right)}=\dfrac{\mathbb{E}\left(X^2\right)}{\mathbb{E}\left(X^2+1-2X\right)}=7 < 9=\dfrac{\left(\mathbb{E}\left(X\right)\right)^2}{\left(\mathbb{E}\left(Y\right)\right)^2}$.

  • 1
    $\begingroup$ Thank you! This was what I was looking for! $\endgroup$ Sep 22, 2023 at 21:42

A bit of a "non answer" but it appears that you cant say anything about how the a ratio of expectations is related to an expectation of ratios.

To illustrate, let $f(X)$ and $g(X)$ be concave functions and $X$ a random variable. By Jensens inequality we can say

$$\mathbb{E}[f(X)]\leq f\left(\mathbb{E}[X]\right)$$ and $$\mathbb{E}[g(X)]\leq g\left(\mathbb{E}[X]\right)$$

Note that the reciprocal of the second inequality gives us:

$$\frac{1}{\mathbb{E}[g(X)]}\geq \frac{1}{g\left(\mathbb{E}[X]\right)}$$

If we multiply this equation by the first equation we can say nothing about this relationship.

$$\frac{\mathbb{E}[f(X)]}{\mathbb{E}[g(X)]} \lesseqqgtr \frac{f\left(\mathbb{E}[X]\right)}{g\left(\mathbb{E}[X]\right)} $$.

Alternatively if $f(X)$ is concave and $g(X)$ is convex we can state:

$$\frac{\mathbb{E}[f(X)]}{\mathbb{E}[g(X)]} \leq\frac{f\left(\mathbb{E}[X]\right)}{g\left(\mathbb{E}[X]\right)} $$.

and if $f(X)$ is convex and $g(X)$ is concave we can say:

$$\frac{\mathbb{E}[f(X)]}{\mathbb{E}[g(X)]} \geq \frac{f\left(\mathbb{E}[X]\right)}{g\left(\mathbb{E}[X]\right)} $$.

but with $f(X)$ and $g(X)$ both being concave or convex we cannot say anything about this relationship, let alone about how $\frac{\mathbb{E}[f(X)]}{\mathbb{E}[g(X)]}$ relates to $\mathbb{E}\left[\frac{f(X)}{g(X)}\right]$.

TL;DR: The ratio of two Jensen's inequalities cannot tell us much if both functions are concave or convex.


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