Does the sum of two variables second order stochastic dominate the mixture of two variables

Question

hello I have two independent variable P and Q. They are both non-negative. Let $\alpha \in (0,1)$. Now I define two new variables on them:

The first variable is the sum of the two variables $$R_1:=\alpha P+(1-\alpha)Q.$$

The second variable R2 is the mixture of the two variables: there is a probability of $\alpha$ that we get P and $1-\alpha$ probability to get Q. Or formally, let 𝑍 be an independent, binary variable with $$𝑃(𝑍=1)=\alpha, 𝑃(𝑍=0)=1−\alpha,$$ then $$𝑅_2:=𝑍𝑃+(1−𝑍)𝑄 $$.

Here is an example. Let $P=Q=(10,0.5;0,0.5)$ which means there is 0.5 probability to get 10 and 0.5 probability to get 0. Let $\alpha=0.5$. Then $R_1=(10,0.25;5,0.5;0,0.25)$ and $R_2=(10,0.5;0,0.5)$.

From the a previous post, I understand that $$E(R_1)=E(R_2)\; and \;Var(R_1)<Var(R_2.)$$ However I wonder if we can get the stronger result that $R_1$ second order stochastic dominates $R_2$? That is for all $k \geq a$ $$\int_a^k cdf_1(t)\,dt \leq \int_a^k cdf_2(t)\,dt$$

Answer 1

We will use Theorem 4.2 in this MIT notes to prove that $R_1$ second order stochastic dominates $R_2$.

Let $F_p(\dot): \mathbb{R} \rightarrow [0,1]$ and $F_p(\dot): \mathbb{R} \rightarrow [0,1]$ denote the two cumulative distributions. Let $u(\cdot)$ denote an increasing and concave function. Denote

$$EU(R_1)=\int_a^b \int_a^b u(\alpha y+(1-\alpha)z ) dF_p(y) dF_q(z) $$ Since $u(\cdot)$ is concave, it has the property that for any $\alpha \in [0,1]$ and any $y,z$ $$ u(\alpha y+(1-\alpha)z) \geq \alpha u(y)+(1-\alpha)u(z). $$ So we have \begin{align*} EU(R_1)& \geq \int_a^b \int_a^b \alpha u(y)+ (1-\alpha)u(z)dF_p(y) dF_q(z) \\ &=\alpha \int_a^b \int_a^b u(y)dF_p(y) dF_q(z)+ (1-\alpha) \int_a^b \int_a^b u(z)dF_p(y) dF_q(z) \\ &=\alpha \int_a^b \left( \int_a^b u(y)dF_p(y) \right ) dF_q(z)+ (1-\alpha) \int_a^b \left( \int_a^b u(z) dF_q(z) \right ) dF_p(y) \\ &=\alpha \int_a^b E(u(P)) dF_q(z)+ (1-\alpha) \int_a^b E(u(Q)) dF_p(y) \\ &=\alpha E(u(P)) \int_a^b dF_q(z)+ (1-\alpha) E(u(Q))\int_a^b dF_p(y) \\ &=\alpha E(u(P))+ (1-\alpha)E(u(Q)) \end{align*}

Denote $$EU(R_2)=\alpha E(u(P))+ (1-\alpha)EU(u(Q))$$ so we have $$ EU(R_1) \geq EU(R_2).$$

Note we also have $$ E(R_1)=E(R_2).$$