# Ratio of two Jensen inequality

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.

• What is the context? What are $e_1$ and $e_2$? How do you define "superior"? Why are you surprised that $\frac{a}{b}>\frac{c}{d}$? Sep 20, 2023 at 17:56
• 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? Sep 20, 2023 at 23:25
• 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? Sep 21, 2023 at 6:37
• 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.
– EconJohn
Sep 22, 2023 at 5:35

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 $$\dfrac{\mathbb{E}\left(X^2\right)}{\mathbb{E}\left(Y^2\right)}=\dfrac{\mathbb{E}\left(X^2\right)}{\mathbb{E}\left(X^2+1-2X\right)}=7 < 9=\dfrac{\left(\mathbb{E}\left(X\right)\right)^2}{\left(\mathbb{E}\left(Y\right)\right)^2}$$.

• Thank you! This was what I was looking for! 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:

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

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

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.