# Implication of first-order stochastic dominance

Use the utility index $U(x) = x$ to prove that if the distribution of $F$ first-order stochastically dominates distribution $G$, then the mean of $x$ under $G$ cannot exceed the mean of $x$ under $F$.

Attempted proof - Suppose the $F$ is first-order stochastically dominates $G$ then $$F(x) \leq G(x) \ \ \forall x$$ Since the expectation preserves linearity then it follows that $$\mathbb{E}\left[F(x)\right] \leq \mathbb{E}\left[G(x)\right] \ \ \forall x$$

I am not sure if this is correct or rigorous enough. Any suggestions are greatly appreciated.

• What does preservation of linearity has to do with any of these...? – Giskard Jan 23 '17 at 16:30

We are given two CDFs $F$ and $G$, such that $F$ FOSD $G$ i.e. $F(x) \leq G(x)$ $\forall x$. Consider the random variables $X\sim F$ and $Y\sim G$. Also, suppose $X$ and $Y$ take non-negative values.
We want to show that $\mathbb{E}(X) \geq \mathbb{E}(Y)$.
Here is the intuition: $F(x) \leq G(x)$ $\forall x$ means that the probability that the random variable $X$ take values smaller than $x$ is smaller than the probability that $Y$ take values smaller than $x$, and this is true for every $x$. Therefore, $X$ take higher values than $x$ more often than $Y$ takes indicating that $X$ will have the higher mean than $Y$.