# Second order stochastic dominance

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...]

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 $$$$\int_0^xF(t)\mathrm dt \le \int_0^xG(t)\mathrm dt \quad\text{for all }x.$$$$ Dixit calls the two integrals super-cumulative functions of $$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.

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.

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: $$$$u(x)=\begin{cases} 2x& x\le 2\\ 4& x>2 \end{cases}$$$$ This function is concave, and $$\mathbb E_F(u)=3.6>3.\overline{33}=\mathbb E_G(u)$$.

• @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. Apr 30, 2021 at 22:53
• +1 I now see where I went wrong and deleted my comment where I was misled. Apr 30, 2021 at 23:30
• @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 $F$? May 2, 2021 at 10:56
• Great example! (Though you have a typo and have written $x > 4$ not $x > 2$) May 2, 2021 at 14:32

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 $$B$$.

• Yes this seems right. I guess one should be able to construct similar examples (e.g. with asymmetric distributions?) even when means are equal? Apr 30, 2021 at 10:04
• @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. :) Apr 30, 2021 at 10:32
• Yes good point. For what it is worth, I have tried and failed to come up with counterexamples to 2 Apr 30, 2021 at 11:29
• 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. Apr 30, 2021 at 15:00
• @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. May 1, 2021 at 6:29