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

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?
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)&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}
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, &quot;ALL risk averse people with iso-elastic utility ($u=x^\alpha$, $\alpha\in(0,1)$) prefer gamble $G$ to gamble $F$&quot;. 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&amp; x\le 2\\ 4&amp; x&gt;2 \end{cases} 
\end{equation}
This function is concave, and $\mathbb E_F(u)=3.6&gt;3.\overline{33}=\mathbb E_G(u)$.