# Example of Second-order Stochastic Dominance and Mean Preserving Spreads

Let $X=\{1,2,3,4\}$. Consider the following lotteries $F$ and $G$ as defined below: for all $x \in \{1,2,3,4\}$,

$$F(x) = \begin{cases} 0 \mbox{ if } x \leq 1, \\
\frac{1}{2} \mbox{ if } x=2, \\
1 \mbox{ if } x \geq 3.
\end{cases}$$ and for all $y \in \{1,2,3,4,5\}$, $$G(y) = \begin{cases} 0 \mbox{ if } x < 1, \\
\frac{1}{4} \mbox{ if } x=1, \\
\frac{1}{2} \mbox{ if } x=2, \\
\frac{3}{4} \mbox{ if } x=3, \\
1 \mbox{ if } x\geq 4.
\end{cases}$$

Clearly, we cannot compare $F$ and $G$ using first-order stochastic dominance (why?). We can establish that $F$ second-order stochastic dominates $G$ in two ways. First, observe that the expected value under $F$ and $G$ are the same $\left(2.5\right)$. Next, we compute that area under the cdfs.

|$X$ | Area under $F$ | Area under $G$ |
|----------|----------|----------|
| 1 | $0$ | $0$ |
| 2 | $0$ | $\frac{1}{4}$ |
| 3 | $\frac{1}{2}$ | $\frac{3}{4}$ |
| 4 | $\frac{3}{2}$ | $\frac{7}{2}$ |

From the above table, it is clear that $\displaystyle\sum_{i=1}^{x} G(i) \geq \displaystyle\sum_{i=1}^{x} F(i)$ for all $x \in \{1,2,3,4\}$. Therefore, as per the area under the CDF criterion, $F$ second-order stochastically dominates $G$.

Note that the lottery $F$'s support is $\{2,3\}$. Let us denote the random variable under $F$ as $x$. We formulate the random variable $y$ that follows $G$ where each $y$ can be thought of as either a downward/upward spread of the outcomes $x=2$ and $x=3$ where the magnitude of the spread is $1$. $y=1$ can be thought of as the downward spread of $x=2$; $y=2$ can be thought of as the downward spread of $x=3$; $y=3$ can be thought of as the upward spread of $x=2$; and $y=4$ can be thought of as the upward spread of $x=3$. Define a probability distribution function, $h_{2}(y)$, that has support of $1$ and $3$ as follows: $$h_{2}(y) = \begin{cases} 0 \mbox{ if } y < 1, \\
\frac{1}{2} \mbox{ if } y=1, \\
0 \mbox{ if } y=2, \\
\frac{1}{2} \mbox{ if } y=3, \\
0 \mbox{ if } y \geq 4.
\end{cases}$$ and a probability distribution function, $h_{3}(y)$, that has support of $2$ and $4$ as follows: $$h_{3}(y) = \begin{cases} 0 \mbox{ if } y < 1, \\
0 \mbox{ if } x=1, \\
\frac{1}{2} \mbox{ if } y=2, \\
0 \mbox{ if } y=3, \\
\frac{1}{2} \mbox{ if } y \geq 4.
\end{cases}$$ Since each $y$ is either a downward spread or an upward spread of a unique $x$ in the support of the CDF $F$, observe that probability distribution function of $g$ can therefore be written a compound lottery as follows:
$$g(y) = \begin{cases} f(2)h_{2}(1) = \frac{1}{4} \mbox{ if } y \leq 1, \\
f(3)h_{3}(2) = \frac{1}{4} \mbox{ if } y=1, \\
f(2)h_{2}(3) = \frac{1}{4} \mbox{ if } y=2, \\
f(3)h_{3}(4) = \frac{1}{4} \mbox{ if } y=3, \\
0 \mbox{ if } x \geq 4.
\end{cases}$$

> **Exercise**: Let $X = \{1,2,3,4,5\}$. Consider the following lotteries $F$ and $G$ as defined below: for all $x \in \{1,2,3,4,5\}$,$$F(x) = \begin{cases} 0 \mbox{ if } x \leq 1, \\
\frac{1}{3} \mbox{ if } x=2, \\
\frac{2}{3} \mbox{ if } x=3, \\
1 \mbox{ if } x \geq 4.
\end{cases}$$ and for all $y \in \{1,2,3,4,5\}$, $$G(y) = \begin{cases} 0 \mbox{ if } x < 1, \\
\frac{1}{12} \mbox{ if } x=1, \\
\frac{3}{12} \mbox{ if } x=2, \\
\frac{6}{12} \mbox{ if } x=3, \\
\frac{6}{12} \mbox{ if } x=4, \\
1 \mbox{ if } x \geq 5. \\
\end{cases}$$ Is $G$ a mean-preserving spread of $F$? If yes, why? If not, can you *fix* $G$ so that $G$ is a mean-preserving spread of $F$?