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In the Wikipedia article for mean-preserving spread, the following is claimed without citation:

If B is a mean-preserving spread of A, then B has a higher variance than A; but the converse is not in general true, because the variance is a complete ordering while ordering by mean-preserving spreads is only partial.

My intuition about the very meaning of "B is a mean-preserving spread of A" has been that the B distribution enjoys the same mean but a higher variance than A. I usually think of this as a risk-averse individual preferring the certainty of winning a dollar to a $.001$ chance of winning $1,000. In my mind, the individual would rank lotteries A and B as follows:

  1. Comparing the means of A and B
  2. Comparing the variances of A and B

And if 1 and 2 turn out to be equal, then the risk-averse individual is indifferent between A and B. But according to this article, that is false. What am I missing?

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up vote 1 down vote accepted

An example of two distributions with the same mean that are incomparable with respect to the partial order "is a mean-preserving spread of" but have different variances is furnished by a first distribution that has discrete probabilities $1/2$ at $-1$ and $1/4$ each at $+1\pm\epsilon_1$ and a second distribution that has discrete probabilities $1/2$ at $+1$ and $1/4$ each at $-1\pm\epsilon_2$, with $\epsilon_1\lt\epsilon_2\lt1$. The mean in both cases is $0$, the second distribution has a higher variance than the first distribution, but you cannot obtain the second distribution from the first distribution by spreading, since there's no way to get rid of the probability at $+1+\epsilon_1$ by spreading without spreading some of it even further away from $+1$.

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Edit: Can you clarify why you chose these particular distributions to illustrate your example? Do you mean there is no way to transform or produce a map from the first distribution into the second? –  tacos_tacos_tacos Jan 27 '13 at 3:03
Alternatively, the significance of choosing the same example as you provided, except with both of the $\frac{1}{2}$ probabilities at $+1$? –  tacos_tacos_tacos Jan 27 '13 at 3:08
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