# 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 $$\begin{equation} \int_0^xF(t)\mathrm dt \le \int_0^xG(t)\mathrm dt \quad\text{for all }x. \end{equation}$$ 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. 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: $$\begin{equation} u(x)=\begin{cases} 2x& x\le 2\\ 4& x>2 \end{cases} \end{equation}$$ This function is concave, and $$\mathbb E_F(u)=3.6>3.\overline{33}=\mathbb E_G(u)$$.

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

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 $$B$$.