<p>If <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are independent positive random variables then the following are true by Jensen's Inequality:
<span class="math-container">\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)} &amp; \ldots (1)\end{eqnarray*}</span>
<span class="math-container">\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*}</span></p>
<p>However, if you are interested in comparing the ratios <span class="math-container">$\dfrac{\mathbb{E}\left(X^2\right)}{\mathbb{E}\left(Y^2\right)}$</span> and <span class="math-container">$\dfrac{\left(\mathbb{E}\left(X\right)\right)^2}{\left(\mathbb{E}\left(Y\right)\right)^2}$</span>, then the inequality depends on the choice of the distribution of random variables. When <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are identically distributed then
<span class="math-container">$\dfrac{\mathbb{E}\left(X^2\right)}{\mathbb{E}\left(Y^2\right)}=\dfrac{\left(\mathbb{E}\left(X\right)\right)^2}{\left(\mathbb{E}\left(Y\right)\right)^2}$</span>. When <span class="math-container">$X \sim \text{Unif}(0,1)$</span> and <span class="math-container">$Y=X+1\sim\text{Unif}(1,2)$</span>, then <span class="math-container">$\dfrac{\mathbb{E}\left(X^2\right)}{\mathbb{E}\left(Y^2\right)}=\dfrac{\mathbb{E}\left(X^2\right)}{\mathbb{E}\left(X^2+1+2X\right)}=\dfrac{1}{7}&gt;\dfrac{1}{9}=\dfrac{\left(\mathbb{E}\left(X\right)\right)^2}{\left(\mathbb{E}\left(Y\right)\right)^2}$</span>. Reversing the roles of <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> i.e. when <span class="math-container">$X \sim \text{Unif}(1,2)$</span> and <span class="math-container">$Y=X-1\sim\text{Unif}(0,1)$</span>, then <span class="math-container">$\dfrac{\mathbb{E}\left(X^2\right)}{\mathbb{E}\left(Y^2\right)}=\dfrac{\mathbb{E}\left(X^2\right)}{\mathbb{E}\left(X^2+1-2X\right)}=7 &lt; 9=\dfrac{\left(\mathbb{E}\left(X\right)\right)^2}{\left(\mathbb{E}\left(Y\right)\right)^2}$</span>.</p>

CC BY-SA 4.0
2023-09-22T21:42:06.920Z
Thank you! This was what I was looking for!
Elina Gilbert
CC BY-SA 4.0
2023-09-22T05:01:38.163Z
2023-09-22T05:31:37.823Z
2023-09-22T05:31:37.823Z
Amit
<p>A bit of a &quot;non answer&quot; but it appears that you cant say anything about how the a ratio of expectations is related to an expectation of ratios.</p>
<p>To illustrate, let <span class="math-container">$f(X)$</span> and <span class="math-container">$g(X)$</span> be concave functions and <span class="math-container">$X$</span> a random variable. By Jensens inequality we can say</p>
<p><span class="math-container">$$\mathbb{E}[f(X)]\leq f\left(\mathbb{E}[X]\right)$$</span>
and
<span class="math-container">$$\mathbb{E}[g(X)]\leq g\left(\mathbb{E}[X]\right)$$</span></p>
<p>Note that the reciprocal of the second inequality gives us:</p>
<p><span class="math-container">$$\frac{1}{\mathbb{E}[g(X)]}\geq \frac{1}{g\left(\mathbb{E}[X]\right)}$$</span></p>
<p>If we multiply this equation by the first equation we can say nothing about this relationship.</p>
<p><span class="math-container">$$\frac{\mathbb{E}[f(X)]}{\mathbb{E}[g(X)]} \lesseqqgtr \frac{f\left(\mathbb{E}[X]\right)}{g\left(\mathbb{E}[X]\right)} $$</span>.</p>
<p>Alternatively if <span class="math-container">$f(X)$</span> is concave and <span class="math-container">$g(X)$</span> is convex we can state:</p>
<p><span class="math-container">$$\frac{\mathbb{E}[f(X)]}{\mathbb{E}[g(X)]} \leq\frac{f\left(\mathbb{E}[X]\right)}{g\left(\mathbb{E}[X]\right)} $$</span>.</p>
<p>and if <span class="math-container">$f(X)$</span> is convex and <span class="math-container">$g(X)$</span> is concave we can say:</p>
<p><span class="math-container">$$\frac{\mathbb{E}[f(X)]}{\mathbb{E}[g(X)]} \geq \frac{f\left(\mathbb{E}[X]\right)}{g\left(\mathbb{E}[X]\right)} $$</span>.</p>
<p>but with <span class="math-container">$f(X)$</span> and <span class="math-container">$g(X)$</span> both being concave or convex we cannot say anything about this relationship, let alone about how <span class="math-container">$\frac{\mathbb{E}[f(X)]}{\mathbb{E}[g(X)]}$</span> relates to <span class="math-container">$\mathbb{E}\left[\frac{f(X)}{g(X)}\right]$</span>.</p>
<p>TL;DR: The ratio of two Jensen's inequalities cannot tell us much if both functions are concave or convex.</p>

CC BY-SA 4.0
2023-09-22T16:05:06.310Z
2023-09-22T20:43:09.870Z
2023-09-22T20:43:09.870Z
EconJohn
<p>I have these pair of numbers
<span class="math-container">$ (a, b) = (\frac{4}{9}, \frac{1}{9}) $</span> and <span class="math-container">$(c, d) = (\frac{1}{2}, \frac{1}{6}) $</span>. (Number mean nothing, just for illustration and simplification)</p>
<p>Note that - (a, b) are pair of numbers which represent <span class="math-container">$((E(e_1))^2, (E(e_2))^2) $</span> and (c, d) are pair for <span class="math-container">$((E(e_1^2)), (E(e_2^2))) $</span></p>
<p>Where, E is the expectation.</p>
<p>Clearly, c&gt;a and d&gt;b (by Jensen inequality)   But when I divide <span class="math-container">$\frac{a}{b}$</span> and <span class="math-container">$\frac{c}{d}$</span>, I get <span class="math-container">$\frac{a}{b}$</span> &gt; <span class="math-container">$\frac{c}{d}$</span>.</p>
<p>So, does Jensen's inequality flips when two Jensen inequality are divided?</p>
<p>Is there a property which ensure that this will be true always.</p>
<p>Please help</p>

CC BY-SA 4.0
2023-09-20T17:56:03.443Z
What is the context? What are $e_1$ and $e_2$? How do you define "superior"? Why are you surprised that $\frac{a}{b}>\frac{c}{d}$?
Richard Hardy
CC BY-SA 4.0
2023-09-20T23:25:16.927Z
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?
Elina Gilbert
CC BY-SA 4.0
2023-09-21T06:37:46.930Z
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?
Richard Hardy
CC BY-SA 4.0
2023-09-22T05:35:26.550Z
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
<statistics><expected-utility><inequality><comparative-statics><functional-form>
Ratio of two Jensen inequality
Elina Gilbert
# Ratio of two Jensen inequality

(Asked by: username_0 on 2023-09-20)

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

**Tags:** statistics, expected-utility, inequality, comparative-statics, functional-form

> **Comments:**
> **username_1**: What is the context? What are $e_1$ and $e_2$? How do you define "superior"? Why are you surprised that $\frac{a}{b}>\frac{c}{d}$?  
> **username_0**: 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?  
> **username_1**: 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?  
> **username_2**: 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.  

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## 2 answers

### Answer by username_3 on 2023-09-22. Score: 1 | Accepted answer

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

> **Comments:**
> **username_0**: Thank you! This was what I was looking for!  

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### Answer by username_2 on 2023-09-22. Score: 1

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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