<p>The answer to 1.</p>
<p>Your conjecture is correct. Consider lotteries <span class="math-container">$A,B$</span> where <span class="math-container">$A$</span> guarantues a payoff of 1 while <span class="math-container">$B$</span> yields 0 or 4, each with 50% probability.</p>
<p><span class="math-container">$B$</span> does not SOSD <span class="math-container">$A$</span>, as you can easily find an agent risk averse enough that they will prefer <span class="math-container">$A$</span>, e.g. an agent whose preferences are described by <span class="math-container">$u(x) = \ln(x)$</span>.</p>
<p><span class="math-container">$A$</span> does not SOSD <span class="math-container">$B$</span> either, as <span class="math-container">$E(B) &gt; E(A)$</span>, meaning an agent with <span class="math-container">$u(x) = x$</span> would prefer <span class="math-container">$B$</span>.</p>

CC BY-SA 4.0
2021-04-30T10:04:56.857Z
Yes this seems right. I guess one should be able to construct similar examples (e.g. with asymmetric distributions?) even when means are equal?
afreelunch
CC BY-SA 4.0
2021-04-30T10:32:11.917Z
@afreelunch I don't think so, the trade off here is definitely between expected value and risk. I recommend taking the "accept tick" back, and waiting for an answer to 2. :)
Giskard
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2021-04-30T11:29:39.123Z
Yes good point. For what it is worth, I have tried and failed to come up with counterexamples to 2
afreelunch
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2021-04-30T14:02:39.533Z
@Bayesian I did, perhaps I am missing something. Using notation from there, assuming $E(x) = E(y)$ and $\epsilon = y-x$, is $E(\epsilon|x) = 0$ trivial? Without this, $E(x) = E(y)$ does not imply that either variable is a mean-preserving spread of the other.
Giskard
CC BY-SA 4.0
2021-04-30T14:21:14.987Z
The $E[\epsilon|x]=0$ is part of the definition of a mean-preserving spread, no?
My arguments was based on statement 1 & 3 of Thm 4.2 being equivalent.
Bayesian
CC BY-SA 4.0
2021-04-30T14:25:47.887Z
@Bayesian All these only hold if $G$ is a mean preserving spread of $F$. Seems like this does not follow from equal expected value. Will try to construct a counterexample.
Giskard
CC BY-SA 4.0
2021-04-30T14:57:36.663Z
Could not get one. Seems like $G$ and $F$ would need to have several intersection points in the interior of the domain, so the functions will not have simple formulas, or I may be wrong and a counterexample could be impossible.
Giskard
CC BY-SA 4.0
2021-04-30T15:00:07.927Z
I believe this is a standard result, and I just posted the first google hit. Before you put too much effort into constructing a counterexample, we can look for a more convincing proof.
Bayesian
CC BY-SA 4.0
2021-05-01T06:29:37.860Z
@afreelunch I recommend accepting the [Herr K.'s answer](https://economics.stackexchange.com/a/43680/1601), because it is complete in itself: the negative answer to 2. is also a negative answer to 1.
Giskard
CC BY-SA 4.0
2021-04-30T09:50:58.537Z
2021-04-30T09:50:58.537Z
Giskard
<p>The answer to <strong>2.</strong> is <strong>no</strong>.</p>
<p>One way to see this is from MWG's Property 6.D.2: <span class="math-container">$F$</span> SOSD <span class="math-container">$G$</span> if and only if
<span class="math-container">\begin{equation}
\int_0^xF(t)\mathrm dt \le \int_0^xG(t)\mathrm dt \quad\text{for all }x.
\end{equation}</span>
<a href="https://www.princeton.edu/%7Edixitak/Teaching/EconomicsOfUncertainty/Slides&amp;Notes/Notes04.pdf" rel="nofollow noreferrer">Dixit</a> calls the two integrals <em>super-cumulative</em> functions of <span class="math-container">$F$</span> and <span class="math-container">$G$</span>, respectively. Hence, a characterization of SOSD is that the super-cumulative of the dominating distribution always lies below the super-cumulative of the dominated distribution. (This is reminiscent of the characterization of FOSD as the <em>cdf</em> of the dominating distribution always lying below the <em>cdf</em> of the dominated one.)</p>
<p>For a counterexample, we only need to come up with distributions <span class="math-container">$F$</span> and <span class="math-container">$G$</span> such that their super-cumulatives cross. Here's one:
<span class="math-container">\begin{align}
f(x)&amp;=\begin{cases}
0.1 &amp; x\in\{0,6\}\\
0.4 &amp; x\in\{2,4\}\\
0 &amp; \text{elsewhere}
\end{cases}\\
g(x)&amp;=\begin{cases}
\frac13 &amp; x\in\{1,3,5\}\\
0 &amp; \text{elsewhere}
\end{cases}
\end{align}</span>
Both distributions have the same mean of <span class="math-container">$3$</span>, but as the following figure shows, their super-cumulatives (<span class="math-container">$S_F$</span> and <span class="math-container">$S_G$</span>) cross at several points, and so neither distribution SOSD the other.</p>
<p><a href="https://i.sstatic.net/sGSck.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/sGSck.png" alt="enter image description here" /></a></p>
<p>Here, <span class="math-container">$S_F(x)=\int_0^x F(t)\mathrm dt = \int_0^x\int_0^tf(s)\mathrm ds\mathrm dt$</span>, and <span class="math-container">$S_G$</span> is similarly defined. Since <span class="math-container">$f$</span> and <span class="math-container">$g$</span> are probability mass functions, the cdf's <span class="math-container">$F$</span> and <span class="math-container">$G$</span> are step-functions (shown below). Integrating the step-functions yields the continuous and piece-wise linear <span class="math-container">$S_F$</span> and <span class="math-container">$S_G$</span> above.</p>
<p><a href="https://i.sstatic.net/g6UHL.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/g6UHL.png" alt="enter image description here" /></a></p>
<hr />
<p><strong>Edit</strong></p>
<p>As OP noted in a comment, &quot;ALL risk averse people with iso-elastic utility (<span class="math-container">$u=x^\alpha$</span>, <span class="math-container">$\alpha\in(0,1)$</span>) prefer gamble <span class="math-container">$G$</span> to gamble <span class="math-container">$F$</span>&quot;. The negative answer above suggests that there must be a concave function with which <span class="math-container">$F$</span> is preferred to <span class="math-container">$G$</span>. Here is an example:
<span class="math-container">\begin{equation} 
u(x)=\begin{cases} 2x&amp; x\le 2\\ 4&amp; x&gt;2 \end{cases} 
\end{equation}</span>
This function is concave, and <span class="math-container">$\mathbb E_F(u)=3.6&gt;3.\overline{33}=\mathbb E_G(u)$</span>.</p>

CC BY-SA 4.0
2021-04-30T22:53:05.887Z
@Giskard: There was indeed a mistake in the original graph (I plotted the wrong rows of the functional values). Thanks for catching that. Your other concerns are addressed in the edit.
Herr K.
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2021-04-30T23:30:49.667Z
+1 I now see where I went wrong and deleted my comment where I was misled.
Bayesian
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2021-05-02T10:56:39.420Z
@HerrK. thanks this looks persuasive. Though one thing I find strange is that ALL risk averse people with iso-elastic utility ($u = x^\alpha$, $\alpha \in (0, 1)$) prefer gamble $G$ to gamble $F$. Can you construct a (concave) utility function which leads to a preference for $F$?
afreelunch
CC BY-SA 4.0
2021-05-02T14:32:05.040Z
Great example! (Though you have a typo and have written $x > 4$ not $x > 2$)
afreelunch
CC BY-SA 4.0
2021-04-30T20:01:53.950Z
2021-05-02T14:53:13.847Z
2021-05-02T14:53:13.847Z
Herr K.
<p>I have two very basic questions about second order stochastic dominance (SOSD):</p>
<ol>
<li>Am I right in thinking that this is only a partial order, i.e. you can find a pair of lotteries such that neither SOSD the other?</li>
<li>Could it be that, if we only consider lotteries with equal means, SOSD is then a complete order? [I suspect not...]</li>
</ol>

CC BY-SA 4.0
2021-04-30T09:47:19.740Z
See Theorem 4.2 here: https://ocw.mit.edu/courses/economics/14-123-microeconomic-theory-iii-spring-2015/lecture-notes-and-slides/MIT14_123S15_Chap4.pdf
Bayesian
<risk>
Second order stochastic dominance
afreelunch
# Second order stochastic dominance

(Asked by: username_0 on 2021-04-30)

I have two very basic questions about second order stochastic dominance (SOSD):

1. Am I right in thinking that this is only a partial order, i.e. you can find a pair of lotteries such that neither SOSD the other?
2. Could it be that, if we only consider lotteries with equal means, SOSD is then a complete order? [I suspect not...]

**Tags:** risk

> **Comments:**
> **username_1**: See Theorem 4.2 here: https://ocw.mit.edu/courses/economics/14-123-microeconomic-theory-iii-spring-2015/lecture-notes-and-slides/MIT14_123S15_Chap4.pdf  

---

## 2 answers

### Answer by username_3 on 2021-04-30. Score: 8 | Accepted answer

The answer to **2.** is **no**.

One way to see this is from MWG's Property 6.D.2: $F$ SOSD $G$ if and only if
\begin{equation}
\int\_0^xF(t)\mathrm dt \le \int\_0^xG(t)\mathrm dt \quad\text{for all }x.
\end{equation}
[Dixit](https://www.princeton.edu/%7Edixitak/Teaching/EconomicsOfUncertainty/Slides&Notes/Notes04.pdf) calls the two integrals *super-cumulative* functions of $F$ and $G$, respectively. Hence, a characterization of SOSD is that the super-cumulative of the dominating distribution always lies below the super-cumulative of the dominated distribution. (This is reminiscent of the characterization of FOSD as the *cdf* of the dominating distribution always lying below the *cdf* of the dominated one.)

For a counterexample, we only need to come up with distributions $F$ and $G$ such that their super-cumulatives cross. Here's one:
\begin{align}
f(x)&=\begin{cases}
0.1 & x\in\{0,6\}\\
0.4 & x\in\{2,4\}\\
0 & \text{elsewhere}
\end{cases}\\
g(x)&=\begin{cases}
\frac13 & x\in\{1,3,5\}\\
0 & \text{elsewhere}
\end{cases}
\end{align}
Both distributions have the same mean of $3$, but as the following figure shows, their super-cumulatives ($S\_F$ and $S\_G$) cross at several points, and so neither distribution SOSD the other.

[![enter image description here](https://i.sstatic.net/sGSck.png)](https://i.sstatic.net/sGSck.png)

Here, $S\_F(x)=\int\_0^x F(t)\mathrm dt = \int\_0^x\int\_0^tf(s)\mathrm ds\mathrm dt$, and $S\_G$ is similarly defined. Since $f$ and $g$ are probability mass functions, the cdf's $F$ and $G$ are step-functions (shown below). Integrating the step-functions yields the continuous and piece-wise linear $S\_F$ and $S\_G$ above.

[![enter image description here](https://i.sstatic.net/g6UHL.png)](https://i.sstatic.net/g6UHL.png)

---

**Edit**

As OP noted in a comment, "ALL risk averse people with iso-elastic utility ($u=x^\alpha$, $\alpha\in(0,1)$) prefer gamble $G$ to gamble $F$". The negative answer above suggests that there must be a concave function with which $F$ is preferred to $G$. Here is an example:
\begin{equation}
u(x)=\begin{cases} 2x& x\le 2\\ 4& x>2 \end{cases}
\end{equation}
This function is concave, and $\mathbb E\_F(u)=3.6>3.\overline{33}=\mathbb E\_G(u)$.

> **Comments:**
> **username_3**: @username_2: There was indeed a mistake in the original graph (I plotted the wrong rows of the functional values). Thanks for catching that. Your other concerns are addressed in the edit.  
> **username_1**: +1 I now see where I went wrong and deleted my comment where I was misled.  
> **username_0**: @HerrK. thanks this looks persuasive. Though one thing I find strange is that ALL risk averse people with iso-elastic utility ($u = x^\alpha$, $\alpha \in (0, 1)$) prefer gamble $G$ to gamble $F$. Can you construct a (concave) utility function which leads to a preference for $F$?  
> **username_0**: Great example! (Though you have a typo and have written $x > 4$ not $x > 2$)  

---

### Answer by username_2 on 2021-04-30. Score: 7

The answer to 1.

Your conjecture is correct. Consider lotteries $A,B$ where $A$ guarantues a payoff of 1 while $B$ yields 0 or 4, each with 50% probability.

$B$ does not SOSD $A$, as you can easily find an agent risk averse enough that they will prefer $A$, e.g. an agent whose preferences are described by $u(x) = \ln(x)$.

$A$ does not SOSD $B$ either, as $E(B) > E(A)$, meaning an agent with $u(x) = x$ would prefer $B$.

> **Comments:**
> **username_0**: Yes this seems right. I guess one should be able to construct similar examples (e.g. with asymmetric distributions?) even when means are equal?  
> **username_2**: @username_0 I don't think so, the trade off here is definitely between expected value and risk. I recommend taking the "accept tick" back, and waiting for an answer to 2. :)  
> **username_0**: Yes good point. For what it is worth, I have tried and failed to come up with counterexamples to 2  
> **username_2**: @username_1 I did, perhaps I am missing something. Using notation from there, assuming $E(x) = E(y)$ and $\epsilon = y-x$, is $E(\epsilon|x) = 0$ trivial? Without this, $E(x) = E(y)$ does not imply that either variable is a mean-preserving spread of the other.  
> **username_1**: The $E[\epsilon|x]=0$ is part of the definition of a mean-preserving spread, no? My arguments was based on statement 1 & 3 of Thm 4.2 being equivalent.  
> **username_2**: @username_1 All these only hold if $G$ is a mean preserving spread of $F$. Seems like this does not follow from equal expected value. Will try to construct a counterexample.  
> **username_2**: Could not get one. Seems like $G$ and $F$ would need to have several intersection points in the interior of the domain, so the functions will not have simple formulas, or I may be wrong and a counterexample could be impossible.  
> **username_1**: I believe this is a standard result, and I just posted the first google hit. Before you put too much effort into constructing a counterexample, we can look for a more convincing proof.  
> **username_2**: @username_0 I recommend accepting the [username_3.'s answer](https://economics.stackexchange.com/a/43680/1601), because it is complete in itself: the negative answer to 2. is also a negative answer to 1.  

---

