1
$\begingroup$

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 $$F_z(w) = \begin{cases} 0 \ &\text{if} \ w < 0\\ \frac{1}{3}w \ &\text{if} \ 0\leq w < 3\\ 1 \ &\text{if} \ 3\leq w \end{cases}$$ Determine whether or not $F_y$ or $F_z$ is first order or second order stochastically dominates 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 dominance applies, any suggestions are greatly appreciated.

$\endgroup$

1 Answer 1

Reset to default
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? $\endgroup$ Commented Jan 23, 2017 at 23:47
  • $\begingroup$ Also, I do not see how $Y$ and $Z$ are uniform? $\endgroup$ Commented Jan 23, 2017 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 $\endgroup$ Commented Jan 23, 2017 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 $\mathbb{E}(Y)=\mathbb{E}(Z) = 1.5$. $\endgroup$ Commented Jan 24, 2017 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? $\endgroup$ Commented Jan 24, 2017 at 1:16

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.