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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.

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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 $u$.

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    was my attempted solution incorrect? – 2017-01-23
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    Also, I do not see how $Y$ and $Z$ are uniform? – 2017-01-23
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    You state the $\mathbb{E}[Y] = 1.5$ but I don't see how you got that, please provide more information – 2017-01-23
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    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$0$ elsewhere. Likewise, the density of $Z$ is $f_z(z) = \frac{1}{3}$ over $0$0$ 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 $\mathbb{E}(Y)=\mathbb{E}(Z) = 1.5$. – 2017-01-24
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    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? – 2017-01-24
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    Take a look here http://math.stackexchange.com/questions/2140129/first-and-second-order-stochastic-dominance-given-two-asset-payoffs – 2017-02-13