Second-order stochastic dominance mean and variance

Question

Consider two random variables $X$ and $Y$. If $X$ dominates $Y$ by the second order, does that mean: (i) $\operatorname{E}(X)\geq \operatorname{E}(Y)$m and (ii) their variance $\operatorname{var}(X)\leq \operatorname{var}(Y)$?

If yes, can anyone prove it?

Answer 1

Let $X\sim F$ and $Y\sim G$. We say that $F$ second-order stochastically dominates (SOSD) $G$ if and only if \begin{equation} \int u(x)\mathrm dF(x)\ge \int u(y)\mathrm dG(y) \end{equation} for all increasing and concave $u(\cdot)$.

In particular, let $u(z)=z$, which is increasing and concave. Then it follows that \begin{equation} \int x\mathrm dF(x)\ge \int y\mathrm dG(y) \quad\Leftrightarrow\quad \mathrm E(X)\ge \mathrm E(Y). \end{equation} Thus $F$ SOSD $G$ implies that $X$ has a higher mean (though not necessarily the other way around).

Answer 2

I think part ii) is true when the distribution share the same mean or have common support, the proof is basically using Integration by part multiple times:

I restrict my discussion to the case of bounded continuous distributions, (denote $X \sim F$ while $Y \sim G$) and F,G have support contained in $[\underline{M},\bar{M}]$ and X and Y have the same mean. And I am using the definition $F \underset{SOSD}{\succsim} Y \Leftrightarrow \int^{z}_{-\infty}F(x)dx \leq \int^{z}_{-\infty}G(x)dx$ at all z.

Firstly notice that $E(X) - E(Y)=\int^{\bar{M}}_{\underline{M}}zd(F(z)-G(z)) = z(F(z)-G(z))|^{\bar{M}}_{\underline{M}} - \int^{\bar{M}}_{\underline{M}}F(z)-G(z)dz =0 \Rightarrow \int^{\bar{M}}_{\underline{M}}F(z)dz = \int^{\bar{M}}_{\underline{M}}G(z)dz$.

Then the formula $Var(Z)= E(Z^2)-E(Z)^2$ tells us we only need to compare $\int^{\bar{M}}_{\underline{M}}z^2dF(z)$ with $\int^{\bar{M}}_{\underline{M}}z^2dG(z)$.

And we have $Var(X)-Var(Y)=\int^{\bar{M}}_{\underline{M}}z^2d(F(z)-G(z))\\ =z^2(F(z)-G(z))|^{\bar{M}}_{\underline{M}} - \int^{\bar{M}}_{\underline{M}}2z(F(z)-G(z))dz\\ = -2z(\int^{z}_{\underline{M}}F(x)dx-\int^{z}_{\underline{M}}G(x)dx)|^{\bar{M}}_{\underline{M}} + \int^{\bar{M}}_{\underline{M}}(\int^{z}_{\underline{M}}F(x)dx-\int^{z}_{\underline{M}}G(x)dx)dz<0$

because $\int^{z}_{-\infty}F(z)dz < \int^{z}_{-\infty}G(z)dz$ at all z.

When the two distribution have different support, we can found a counter example: $X\sim Uni(1,3)$ and $Y \sim Uni(-1,0)$, we have $X \ \underset{SOSD}{\succsim} Y$ but $Var(X)>Var(Y)$.