-1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What does preservation of linearity has to do with any of these...? $\endgroup$ – Giskard Jan 23 '17 at 16:30
4
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.