Stochastic Dominance for convex sum of two random variables with same distribution

Question

This is a very widely used result in finance and economics, and seems fairly intuitive as well (riskier asset is less preferred by risk-averse (concave utility) investors). However, I have not been able to find a proof for this.

I see a very similar question here but it doesn't require the function to be increasing (the accepted answer gives a counter example but with decreasing function).

Formally, let $X$ and $Y$ be two r.v. such that $E(X) = E(Y)$ but $Var(Y) > Var(X)$. Let $g(x)$ be a continuous increasing concave function.

We want to show: $$E[g(Y)] \leq E[g(X)]$$

EDIT: I found from here that, the above is not actually true. In fact, whenever for every increasing concave function, $E[g(Y)] \leq E[g(X)]$ then $X$ is defined to second order stochastically dominate(SOSD) $Y$. It is also easy to show that $X$ SOSD $Y \implies Var(X) < Var(Y)$.

But now I have another doubt due to which I am changing the question (not deleting the above part but changing the title):

(Further edit) Let $X$ and $Y$ have same cdf $F(.)$. Now let $Z_1, Z_2$ be two convex combinations of $X, Y$, i.e., $Z_i=a_iX+(1-a_i)Y$ and $0.5>a_1>a_2>0$.

Does $Z_1$ SSOD $Z_2$? (Intuitively it should because variance of $Z_1 < Z_2$.

Answer 1

Your source has Theorem 2', which says that $Z$ SOSD $X$ if for every concave function $u$

$$ \mathbb{E}u(Z) \geq \mathbb{E}u(X), $$

whatever the sign of the first derivative, $u'$.

Now, \begin{align} \mathbb{E}u(Z) &= \mathbb{E}u(aX + (1-a)Y) \\ \text{[Jensen inequality]}\quad \quad &\geq \mathbb{E}\left(au(X)+(1-a)u(Y)\right)\\ \text{[linearity of $\mathbb{E}$]}\quad \quad & = a\mathbb{E}u(X)+(1-a)\mathbb{E}u(Y)\\ \text{[$X,Y$ iid]}\quad \quad & = \mathbb{E}u(X), \end{align}

or in short

$$\mathbb{E}u(Z) \geq \mathbb{E}u(X),$$

which holds iff $Z$ SOSD $X$ ($Y$) by the Theorem 2' mentioned in the beginning.