0
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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

1
$\begingroup$

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

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.