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Let $f(x)$ be concave, $X_1$ a random variable and $X_2$ a mean-preserving spread of $X_1$.

The entry on wikipedia defines mean-preserving spread as any $X_2$ such that

$$x_{2} \; {\overset {d}{=}}\; (x_{1}+z)$$

where $E[Z | x_1] = 0$ for all values of $x_1$. (Check the comment below for what seems to be an alternative definition)

My question: is the following statement always true?

$$ E[f(X_2)] \leq E[f(X_1)]$$

Not sure how to go about proving or disproving this. I was trying to use Jensen's inequality, which says that for a concave $f$:

$$ f(E[X]) \geq E[f(X)] $$

My intuition is that the higher variance means we're picking more of the extreme values with $f(X_2)$ when compared to $f(X_1)$ and the concavity of $f$ means $f(X_2)$ should tend to be lower than $f(X_1)$.

Sorry for being vague. Thanks!

cd98
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  • Do you provide a *definition* of "mean-preserving spread" or do you just quote one of its properties? From the name I would guess you are referring to any transformation in the family $$X_1\to E[X_1]+\sigma(X_1-E[X_1]),\ \sigma\ge 1.$$ However, that is far more restrictive than the equation you give. – whuber Mar 09 '18 at 15:17
  • @whuber Thanks, I was being sloppy. Wikipedia seems to be a less restrictive definition than yours. I'll edit my question to reflect this – cd98 Mar 10 '18 at 02:37
  • @cd98: perhaps you missed to mention that $f(X)$ is also increasing (for example, utility function is often assumed to be increasing and concave). – Dayne Dec 17 '20 at 05:56

2 Answers2

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Here is a counter example.

Suppose $X_1 \sim N(0,1)$, and $Z \sim N(0, c)$ for any $c > 0$. Then $X_2 = X_1 + Z$ is a mean preserving spread of $X_1$. Also note that $X_2 \sim N(0, c+1)$.

Now consider the concave function $f(x) = e^{-x}\,.$ Recall that for $X \sim N(\mu, \sigma^2)$ we have that $-X \sim N(-\mu, \sigma^2)$, and thus $\exp(-X) \sim $ LogNormal$(-\mu, \sigma^2)$. Finally, the expected value of this LogNormal distribution is $\exp(-\mu + \sigma^2/2)$.

So $\exp(-X_1) \sim LogNormal(0, 1)$ and $\exp(-X_2) \sim LogNormal(0, c+1)$. Looking at the expected values

$$E[\exp(-X_1) ] = \exp\left(\dfrac{1}{2}\right) < E[\exp(-X_2)] = \exp \left(\dfrac{c+1}{2} \right)$$

Which is the opposite of what you want.

Greenparker
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As far as I understand, your question is answered here https://math.stackexchange.com/questions/2131742/expectation-concave-function by applying Jensen's Inequality and the law of iterated expectations.

LyxUser12345
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