Mean preserving spread for normal distribution

Question

Is it true that any two normal distributions with the same mean can be ordered w.r.t. the relation of a mean preserving spread? My intuition would be that this is true but I cannot come up with a formal proof.

Thanks in advance

Answer 1

Hint: If both normal distributions have the same mean $\mu$, there is a positive constant $c$ such that $F_1(x + \mu) = F_2(c x + \mu)$ where $F_1$ and $F_2$ are the CDF's.

Answer 2

Using the first definition from Wiki,

Let $x_{A}$ and $x_{B}$ be the random variables associated with gambles A and B. Then B is a mean-preserving spread of A if and only if $x_{B} \overset{d}{=}(x_{A}+z)$ for some random variable $z$ having $E(z\mid x_{A})=0$ for all values of $x_{A}$. Here $\overset{d}{=}$ means "is equal in distribution to" (that is, "has the same distribution as").

For $\sigma^2_2>\sigma^2_1$ we can heuristically write $N(\mu,\sigma^2_1) \overset{d}{=} N(\mu,\sigma^2_2) +N(\mu,\sigma^2_2 - \sigma^2_1)$.

The proof of the equivalence of the definition can be found in Rothschild and Stiglitz 1970 or Machina and Pratt 1997