0
$\begingroup$

The mean preserving spread is defined as follows:

Consider two lotteries g and h. Let $x_g$ und $x_h$ denote the corresponding random variables. Then h is a mean preserving spread (MPS) of g, if: $x_h = x_g + z$ for some random variable z with $E[z|x_g] = 0\ \forall\ x_g$. (This definition is supposed to be complete and exhaustive and I have found it in multiple sources)

An example for an MPS is the following: enter image description here Lottery g is replaced with a new lottery (lets call it h) by adding some noise, which makes the payoffs more extreme but preserves the mean.

My problem is: As far as I understand it, g (the old lottery) is also a mean preserving spread of h (the new lottery). At least the definition from the beginning allows for this: If I set z in a manner that the first outcome becomes larger and the second one smaller, I can still have $E[z|x_g] = 0\ \forall\ x_g$, but the total risk is now smaller. I have basically spread the lottery negatively. Just that MPS is now an entirely useless concept, since it does not allow for judgements with regards to the value of utility for an individual with concave utility functions (or any other shape for that matter.

$\endgroup$
0
$\begingroup$

I realized my mistake whilst writing this, so I will just share the answer as well.

Essentially my mistake was, that I assumed this aspect of the definition ($E[z|x_g] = 0\ \forall\ x_g$) to mean that the expected value of z has to be 0 given, that we are in the lottery that is given by the random variable $x_g$. This reading was however wrong. It actually means, that the expected value of z has to be 0, given that we are at a particular value of $x_g$. This has to be individually true for all values of $x_g$.

Thus, using this definition correctly a negative spreading is not possible anymore, the MPS of the original lottery is therefore always at least as risky as the original lottery. A risk averse decision maker will thus never want to choose an MPS.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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