Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $+1$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.