<p>The motivation is I believe clearly stated in p. 197 beginning of the section, where they write that they want to use Second-order stochastic dominance to reflect comparisons related to "riskiness/dispersion". </p>

<p>To make this prominent, they use distributions with the same mean, so that it is <em>obvious</em> that we compare them <strong>not</strong> w.r.t their means, but w.r.t. riskiness/dispersion. In fact, the authors use the concept as equivalent to the stochastic dominance over mean-preserving spreads (see proposition 6.D.2 p. 199).</p>

<p>Using the general definition (which is defined by what they provide as property 6.D.2 p. 198) would not have offered them something more, in view of the reasons they want to use the concept.</p>

CC BY-SA 3.0
2017-10-20T23:10:09.233Z
Thank you. A natural question that follows is why they did not do the same thing for first-order stochastic dominance??
Yes
CC BY-SA 3.0
2017-10-20T23:47:57.623Z
@GaryMoore Because first-order SD is _all_ about expected (mean) returns, in fact it is _equivalent_ to say that the dominant has higher mean that the dominated.
Alecos Papadopoulos
CC BY-SA 3.0
2017-10-21T13:33:48.170Z
@AlecosPapadopoulos If lottery $F$ fosds lottery $G$, then $E[F] \ge E[G]$. However, it is not true that whenever $E[F] \ge E[G]$, $F$ fosds $G$. For a counter-example, see [Wikipedia](https://en.wikipedia.org/wiki/Stochastic_dominance). Hence, first-order stochastic dominance is more than just about the mean, and it is not equivalent to say that the dominant has a higher mean than the dominated.
Theoretical Economist
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2017-10-21T16:12:30.907Z
@TheoreticalEconomist Thanks for spotting this. Indeed I had in mind expected _utility_ "returns", $E[u(F)] \ge E[u(G)]$ for non-decreasing $u$, but I forgot the "utility" part and resulted in a mistaken assertion.
Alecos Papadopoulos
CC BY-SA 3.0
2017-10-20T21:19:49.050Z
2017-10-20T21:19:49.050Z
Alecos Papadopoulos
<p>Is not Mascolell's definition in his microeconomic theory of the second-order stochastic dominance narrower? He defines for distribution functions with the same mean only. Although he gives some motivation for doing this, somehow I do not get his motivation.</p>

<microeconomics><reference-request><risk>
Why does Mascolell define second-order stochastic dominance as such?
Yes
# Why does Mascolell define second-order stochastic dominance as such?

(Asked by: username_0 on 2017-10-20)

Is not Mascolell's definition in his microeconomic theory of the second-order stochastic dominance narrower? He defines for distribution functions with the same mean only. Although he gives some motivation for doing this, somehow I do not get his motivation.

**Tags:** microeconomics, reference-request, risk

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## 1 answers

### Answer by username_1 on 2017-10-20. Score: 3

The motivation is I believe clearly stated in p. 197 beginning of the section, where they write that they want to use Second-order stochastic dominance to reflect comparisons related to "riskiness/dispersion".

To make this prominent, they use distributions with the same mean, so that it is *obvious* that we compare them **not** w.r.t their means, but w.r.t. riskiness/dispersion. In fact, the authors use the concept as equivalent to the stochastic dominance over mean-preserving spreads (see proposition 6.D.2 p. 199).

Using the general definition (which is defined by what they provide as property 6.D.2 p. 198) would not have offered them something more, in view of the reasons they want to use the concept.

> **Comments:**
> **username_0**: Thank you. A natural question that follows is why they did not do the same thing for first-order stochastic dominance??  
> **username_1**: @GaryMoore Because first-order SD is _all_ about expected (mean) returns, in fact it is _equivalent_ to say that the dominant has higher mean that the dominated.  
> **username_2**: @AlecosPapadopoulos If lottery $F$ fosds lottery $G$, then $E[F] \ge E[G]$. However, it is not true that whenever $E[F] \ge E[G]$, $F$ fosds $G$. For a counter-example, see [Wikipedia](https://en.wikipedia.org/wiki/Stochastic_dominance). Hence, first-order stochastic dominance is more than just about the mean, and it is not equivalent to say that the dominant has a higher mean than the dominated.  
> **username_1**: @TheoreticalEconomist Thanks for spotting this. Indeed I had in mind expected _utility_ "returns", $E[u(F)] \ge E[u(G)]$ for non-decreasing $u$, but I forgot the "utility" part and resulted in a mistaken assertion.  

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