0
$\begingroup$

Consider the two distributions $$F_y(w) = \begin{cases} 0 \ &\text{if} \ w < 1\\ w-1 \ &\text{if} \ 1\leq w < 2\\ 1 \ &\text{if} \ 2\leq w \end{cases}$$ and $$F_z(w) = \begin{cases} 0 \ &\text{if} \ w < 1\\ \frac{1}{3}w \ &\text{if} \ 0\leq w < 3\\ 1 \ &\text{if} \ 3\leq w \end{cases}$$ Determine whether or not $F_y$ or $F_z$ is first order or second order stochastically dominantes the other.

Attempted solution - Suppose $w = 1.6$ then $F_y(w) = .6$ and $F_z(w) = .533$ so $F_y(w) > F_z(w)$ when $w = 1.6$. But if $w = 1.5$ then $F_y(w) = F_z(w)$, so $F_y$ nor $F_z$ is first order stochastically dominates the other.

I am not sure how to show if second order stochastically dominante applies, any suggestions are greatly appreciated.

$\endgroup$
1
$\begingroup$

Since $F_y(1) = 0 < \frac{1}{3} = F_z(1)$ and $F_y(2) = 1 > \frac{2}{3} = F_z(2)$. Therefore, neither $F_y$ nor $F_z$ FOSD the other one.

Suppose $Y\sim F_y$ and $Z\sim F_z$. Notice that $Y$ is a uniform random variable over the interval $(1,2)$ and $Z$ is the uniform random variable over the interval $(0, 3)$. Both have the same mean: $\mathbb{E}(Y) = \mathbb{E}(Z) = 1.5$. We will show that $F_y$ second order stochastically dominates $F_z$ i.e. for every concave function $u$ the following holds: $\mathbb{E}(u(Y))\geq \mathbb{E}(u(Z))$.

\begin{eqnarray*} \mathbb{E}(u(Z)) & = & \displaystyle\int_0^3 u(z)\frac{1}{3}dz\\ & = & \int_0^1 u(z)\frac{1}{3}dz+\int_1^2 u(z)\frac{1}{3}dz+\int_2^3 u(z)\frac{1}{3}dz \\ & = & \int_0^1 u(z)\frac{1}{3}dz+\int_1^2 u(z)\frac{1}{3}dz+\int_0^1 u(z+2)\frac{1} {3}dz \\ & = & \int_0^1 (u(z)+u(z+2))\frac{1}{3}dz+\int_1^2 u(z)\frac{1}{3}dz \\ & = & \int_0^1 \left(\frac{1}{2}u(z)+\frac{1}{2}u(z+2)\right)\frac{2}{3}dz+\int_1^2 u(z)\frac{1}{3}dz \\ & \leq & \int_0^1 u(z+1)\frac{2}{3}dz+\int_1^2 u(z)\frac{1}{3}dz \ \ \ldots \ [\text{By Concavity of $u$}] \\ & = & \int_1^2 u(z)\frac{2}{3}dz+\int_1^2 u(z)\frac{1}{3}dz \\ & = & \int_1^2 u(z)dz \\ & = & \mathbb{E}(u(Y))\end{eqnarray*}

Therefore, $\mathbb{E}(u(Y))\geq \mathbb{E}(u(Z))$ for every concave $u$.

$\endgroup$
6
  • $\begingroup$ was my attempted solution incorrect? $\endgroup$ – Wolfy Jan 23 '17 at 23:47
  • $\begingroup$ Also, I do not see how $Y$ and $Z$ are uniform? $\endgroup$ – Wolfy Jan 23 '17 at 23:48
  • $\begingroup$ You state the $\mathbb{E}[Y] = 1.5$ but I don't see how you got that, please provide more information $\endgroup$ – Wolfy Jan 23 '17 at 23:52
  • $\begingroup$ Given the CDFs of $Y$, just differentiate it with respect to y and you will get its density as: $f_y(y) = 1$ over $1<y<2$ and $0$ elsewhere. Likewise, the density of $Z$ is $f_z(z) = \frac{1}{3}$ over $0<z<3$ and $0$ elsewhere. Therefore, $Y$ and $Z$ are uniform. To find the means, simply find $\mathbb{E}(Y)$ and $\mathbb{E}(Z)$. Given their densities, you will get $\mathbb{E}(Y)=\mathbb{E}(Z) = 1.5$. $\endgroup$ – Amit Jan 24 '17 at 0:24
  • $\begingroup$ The second line here: $$\int_0^1 u(z)\frac{1}{3}dz+\int_1^2 u(z)\frac{1}{3}dz+\int_2^3 u(z)\frac{1}{3}dz$$ where are these limits coming from? Is it the CDF? $\endgroup$ – Wolfy Jan 24 '17 at 1:16

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.