<p>Let $u(x) = x$ which is increasing and concave. Then the <em>defining condition</em> of SOSD reads</p>

<p>$$\int x\mathrm dF(x)\ge \int x\mathrm dG(x) \implies E_F(X) \geq E_G(X) \tag{1}$$</p>

<p>..which would contradict the case  $E_F(X) &lt; E_G(X)$ that would be permissible under a general "different means" postulate. On the other hand, we see that the "same mean" condition can be violated by the defining condition for SOSD itself. What does that tell us?</p>

<p>1) That $E_F(X) \geq E_G(X)$ is a <em>necessary condition</em> for $F$ to SOSD $G$.  </p>

<p>2) ...And so that the requirement "$F$ and $G$ have the same mean" wrongly restricts the application of the concept of SOSD.</p>

CC BY-SA 3.0
2016-10-20T14:54:35.727Z
2016-10-20T14:54:35.727Z
Alecos Papadopoulos
<p>Let $F$ and $G$ be two distributions with the same mean. $F$ is said to <strong>second order stochastically dominate</strong> (<strong>SOSD</strong>) $G$ if 
$$\int u(x)\mathrm dF(x)\ge \int u(x)\mathrm dG(x)\tag{1}$$
for all increasing and concave $u(\cdot)$. </p>

<p>This above definition is equivalent to<br>
$$\int_{-\infty}^x F(t)\mathrm dt\le \int_{-\infty}^xG(t)\mathrm dt,\qquad\forall x\in\mathbb R.\tag{2}$$</p>

<p>I was told that the requirement for $F$ and $G$ to have the same mean is not really necessary. Suppose $F$ and $G$ do <strong>not</strong> have the same mean. Can we then still have the equivalence between $(1)$ and $(2)$? </p>

<p>N.B. I was able to show $(2)\Rightarrow (1)$ without the same mean condition, but not the other way around.</p>

<microeconomics><probability>
second order stochastic dominance without the same mean
Herr K.
# second order stochastic dominance without the same mean

(Asked by: username_0 on 2016-10-19)

Let $F$ and $G$ be two distributions with the same mean. $F$ is said to **second order stochastically dominate** (**SOSD**) $G$ if
$$\int u(x)\mathrm dF(x)\ge \int u(x)\mathrm dG(x)\tag{1}$$
for all increasing and concave $u(\cdot)$.

This above definition is equivalent to  
$$\int\_{-\infty}^x F(t)\mathrm dt\le \int\_{-\infty}^xG(t)\mathrm dt,\qquad\forall x\in\mathbb R.\tag{2}$$

I was told that the requirement for $F$ and $G$ to have the same mean is not really necessary. Suppose $F$ and $G$ do **not** have the same mean. Can we then still have the equivalence between $(1)$ and $(2)$?

N.B. I was able to show $(2)\Rightarrow (1)$ without the same mean condition, but not the other way around.

**Tags:** microeconomics, probability

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## 1 answers

### Answer by username_1 on 2016-10-20. Score: 6 | Accepted answer

Let $u(x) = x$ which is increasing and concave. Then the *defining condition* of SOSD reads

$$\int x\mathrm dF(x)\ge \int x\mathrm dG(x) \implies E\_F(X) \geq E\_G(X) \tag{1}$$

..which would contradict the case $E\_F(X) < E\_G(X)$ that would be permissible under a general "different means" postulate. On the other hand, we see that the "same mean" condition can be violated by the defining condition for SOSD itself. What does that tell us?

1) That $E\_F(X) \geq E\_G(X)$ is a *necessary condition* for $F$ to SOSD $G$.

2) ...And so that the requirement "$F$ and $G$ have the same mean" wrongly restricts the application of the concept of SOSD.


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