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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...]
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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 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

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


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 $\mathbb E_F(u)=3.6>3.\overline{33}=\mathbb E_G(u)$.

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    $\begingroup$ @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. $\endgroup$
    – Herr K.
    Apr 30 at 22:53
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    $\begingroup$ +1 I now see where I went wrong and deleted my comment where I was misled. $\endgroup$
    – Bayesian
    Apr 30 at 23:30
  • $\begingroup$ @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 $F$? $\endgroup$
    – afreelunch
    May 2 at 10:56
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    $\begingroup$ Great example! (Though you have a typo and have written $x > 4$ not $x > 2$) $\endgroup$
    – afreelunch
    May 2 at 14:32
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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 $B$.

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  • $\begingroup$ Yes this seems right. I guess one should be able to construct similar examples (e.g. with asymmetric distributions?) even when means are equal? $\endgroup$
    – afreelunch
    Apr 30 at 10:04
  • $\begingroup$ @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. :) $\endgroup$
    – Giskard
    Apr 30 at 10:32
  • $\begingroup$ Yes good point. For what it is worth, I have tried and failed to come up with counterexamples to 2 $\endgroup$
    – afreelunch
    Apr 30 at 11:29
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    $\begingroup$ 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. $\endgroup$
    – Bayesian
    Apr 30 at 15:00
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    $\begingroup$ @afreelunch I recommend accepting the Herr K.'s answer, because it is complete in itself: the negative answer to 2. is also a negative answer to 1. $\endgroup$
    – Giskard
    May 1 at 6:29

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