I need to show that one distribution, $\Phi_1$, $\textbf{second-order}$ stochastically dominates the other, $\Phi_2$. These are discrete distributions. The distributions are:
$\Phi_1(x)$=
$\begin{cases}
0.25,& \text{if } 0 \leq x<10\\
0.6,& \text{if } 10 \leq x<20 \\
0.85 & \text{if } 20 \leq x<40 \\
0.95 & \text{if } 40 \leq x<100 \\
1 & \text{if } x \geq 100
\end{cases}$
and,
$\Phi_2(x)$=
$\begin{cases}
0.28, & \text{if } 0 \leq x<10\\
0.8, & \text{if } 10 \leq x<35 \\
0.86 & \text{if } 35 \leq x<100 \\
1 & \text{if } x \geq 100 \\
\end{cases}$
I am looking for SOSD relationships precisely. The way I'm going about this is as follows. I'm showing that $\int^{x}_{-\infty} \Phi_1 (x) dx \leq \int^{x}_{-\infty} \Phi_2 (x) dx$ for all $x$:
$\int^{10}_{0} 0.25 dx = 2.5 \leq \int^{10}_{0} 0.28 dx = 2.8$
$\int^{20}_{10} 0.6 dx = 6 \leq \int^{20}_{10}0.8 dx = 8 $
However, from now on:
$\int^{35}_{20} 0.85 dx = 12.75 \geq \int^{35}_{20} 0.8 dx = 12$
$\int^{35}_{40} 0.85 dx \leq \int^{35}_{40} 0.86 dx $
$\int^{40}_{100} 0.95 dx \geq \int^{40}_{100} 0.86 dx$
What I am getting wrong? $\int^{x}_{-\infty} \Phi_1 (x) dx \leq \int^{x}_{-\infty} \Phi_2 (x) dx$ for some $x$, but not all of them.
Best Answer
$$\int_{-\infty}^{10}\Phi_1(x)dx=2.5<\int_{-\infty}^{10}\Phi_2(x)dx=2.8$$
and the integral can never be larger for the first cdf because .25 is less than .28. Then for 10 to 20 we have
$$\int_{-\infty}^{20}\Phi_1(x)dx=2.5+6=8.5<\int_{-\infty}^{20}\Phi_2(x)dx=2.8+8=10.8$$
and it could not have been greater because .6 < .8. Then for 20 to 35 we have
$$\int_{-\infty}^{35}\Phi_1(x)dx=8.5+12.75=21.25<\int_{-\infty}^{35}\Phi_2(x)dx=10.8+12=22.8$$
For 35 to 40 we have
$$\int_{-\infty}^{40}\Phi_1(x)dx=21.25+4.25=25.5<\int_{-\infty}^{40}\Phi_2(x)dx=22.8+4.3=27.1$$
And lastly
$$\int_{-\infty}^{100}\Phi_1(x)dx=25.5+57=82.5<\int_{-\infty}^{100}\Phi_1(x)dx=27.1+55.9=83$$