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.

  • $\begingroup$ What does preservation of linearity has to do with any of these...? $\endgroup$ – 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 $Y$.

Here is the proof:

\begin{eqnarray*} & F(x) \leq G(x) \ \ \ \forall x \\ \rightarrow & 1 - F(x) \geq 1- G(x) \ \ \ \forall x \\ \rightarrow & \int_{0}^{\infty}1 - F(x) dx \geq \int_{0}^{\infty} 1- G(x)dx \\ \rightarrow & \int_{0}^{\infty}\Pr(X> x) dx \geq \int_{0}^{\infty} \Pr(Y> x)dx \\ \rightarrow & \int_{0}^{\infty} \int_{x}^{\infty}f_X(a) da dx \geq \int_{0}^{\infty} \int_{x}^{\infty}f_Y(a) da dx \\ \rightarrow & \int_{0}^{\infty} \int_{0}^{a}f_X(a) dx da \geq \int_{0}^{\infty} \int_{0}^{a}f_Y(a) dx da \\ \rightarrow & \int_{0}^{\infty} \int_{0}^{a} dx \ f_X(a) \ da \geq \int_{0}^{\infty} \int_{0}^{a} dx \ f_Y(a) \ da \\ \rightarrow & \int_{0}^{\infty} a \ f_X(a) \ da \geq \int_{0}^{\infty} a \ f_Y(a) \ da \\ \rightarrow & \mathbb{E}(X) \geq \mathbb{E}(Y) \ \end{eqnarray*}


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