First and second order stochastic dominance 
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.
 A: 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 $u$.
