# Second Order Stochastic Dominance

#### santicanel

##### New Member
how can I test for second order stochastic dominance? is easy to understand the idea but when I have to normal distributions with different means and different standard deviations I don't know how to calculate if the integrals of the CDF intersect

#### BGM

##### TS Contributor
Suppose you want to check whether $$X_1$$ has a second order stochastic dominance over $$X_2$$. So you would like to consider the following function:

$$D(x) = \int_{-\infty}^{x} [F_{X_2}(t) - F_{X_1}(t)] dt$$

and check whether $$D(x) \geq 0 ~~ \forall x \in \mathbb{R}$$

If their CDF is nice enough, $$D$$ will be differentiable and you can try to locate the local minimum points of $$D$$, and check whether they are all non-negative. By differentiation, it is easy to see that all critical point(s) $$x$$ need to satisfy

$$F_{X_1}(x) = F_{X_2}(x)$$

and for those critical points will be either local maximum or local minimum.

In particular, if you assume that $$X_1 \sim (\mu_1, \sigma^2_1)$$ and $$X_2 \sim (\mu_2, \sigma^2_2)$$,

$$F_{X_1}(x) = F_{X_2}(x) \iff \Pr\{X_1 \leq x\} = \Pr\{X_2 \leq x\}$$

$$\iff \Pr\left\{Z \leq \frac {x - \mu_1} {\sigma_1} \right\}= \Pr\left\{Z \leq \frac {x - \mu_2} {\sigma_2} \right\}$$

$$\iff \frac {x - \mu_1} {\sigma_1} = \frac {x - \mu_2} {\sigma_2}$$

$$\iff (\sigma_2 - \sigma_1)x = \sigma_2\mu_1 - \sigma_1\mu_2$$

Now we see that if their variances are equal $$\sigma_1 = \sigma_2$$, their CDFs will not intersect each other (trivial case: CDFs are overlapping if their mean also equal). So you would like to check whether

$$F_{X_2}(x) > F_{X_1}(x) \iff \frac {x - \mu_1} {\sigma} < \frac {x - \mu_2} {\sigma} \iff \mu_1 > \mu_2$$

as expected.

When the variances are not equal $$\sigma_1 \neq \sigma_2$$, the above equation will give one and only one root. That means the function $$D$$ have one critical point only:

$$x^* = \frac {\sigma_2\mu_1 - \sigma_1\mu_2} {\sigma_2 - \sigma_1}$$

So now you need to verify the following:

1. $$F_{X_2}(x) - F_{X_1}(x) > 0 ~~ \forall x < x^*$$

and this ensure $$D(x) > 0 ~~ \forall x < x^*$$. From the definition it ensure that $$D$$ will be strictly increasing up to $$x^*$$ and strictly decreasing until $$+\infty$$. And thus we would also need to verify

2. $$\lim_{x\to+\infty}D(x) \geq 0$$

For the first condition it can be similarly translated as

$$(\sigma_2 - \sigma_1)x < \sigma_2\mu_1 - \sigma_1\mu_2 ~~ \forall x < x^*$$

and this is equivalent to $$\sigma_2 > \sigma_1$$
(as expected; a larger variance ensure a thicker tail)

For the second condition, note that

$$\lim_{x\to+\infty}D(x) = \int_{-\infty}^{+\infty} [F_{X_2}(t) - F_{X_1}(t)] dt$$

Therefore

$$\lim_{x\to+\infty}D(x) \geq 0 \iff \int_{-\infty}^{+\infty} [F_{X_2}(t) - F_{X_1}(t)]dt \geq 0$$

$$\iff \int_{-\infty}^{0} \{-F_{X_1}(t) - [-F_{X_2}(t)]\}dt + \int_{0}^{+\infty} \{[1 - F_{X_1}(t)] - [1 - F_{X_2}(t)] \}dt \geq 0$$

$$\iff \int_{-\infty}^{0} [-F_{X_1}(t)]dt + \int_{0}^{+\infty} [1 - F_{X_1}(t)]dt \geq \int_{-\infty}^{0} [-F_{X_2}(t)]dt + \int_{0}^{+\infty} [1 - F_{X_2}(t)]dt$$

$$\iff E[X_1] \geq E[X_2]$$

As a result, the conditions will be $$mu_1 \geq \mu_2$$ and $$\sigma_1 \leq \sigma_2$$ and this should be enough for normal distribution.

The issue becomes interesting if you have the two independent random sample and want to test the stochastic dominance. Again if you have the normality assumption, you may have the above condition as your null hypothesis.

#### santicanel

##### New Member
Sorry took me a while to answer. Thanks you very much. Really helped!

Was reading Levy's book on stochastic dominance and remember my question

Regards!

Suppose you want to check whether $$X_1$$ has a second order stochastic dominance over $$X_2$$. So you would like to consider the following function:

$$D(x) = \int_{-\infty}^{x} [F_{X_2}(t) - F_{X_1}(t)] dt$$

and check whether $$D(x) \geq 0 ~~ \forall x \in \mathbb{R}$$

If their CDF is nice enough, $$D$$ will be differentiable and you can try to locate the local minimum points of $$D$$, and check whether they are all non-negative. By differentiation, it is easy to see that all critical point(s) $$x$$ need to satisfy

$$F_{X_1}(x) = F_{X_2}(x)$$

and for those critical points will be either local maximum or local minimum.

In particular, if you assume that $$X_1 \sim (\mu_1, \sigma^2_1)$$ and $$X_2 \sim (\mu_2, \sigma^2_2)$$,

$$F_{X_1}(x) = F_{X_2}(x) \iff \Pr\{X_1 \leq x\} = \Pr\{X_2 \leq x\}$$

$$\iff \Pr\left\{Z \leq \frac {x - \mu_1} {\sigma_1} \right\}= \Pr\left\{Z \leq \frac {x - \mu_2} {\sigma_2} \right\}$$

$$\iff \frac {x - \mu_1} {\sigma_1} = \frac {x - \mu_2} {\sigma_2}$$

$$\iff (\sigma_2 - \sigma_1)x = \sigma_2\mu_1 - \sigma_1\mu_2$$

Now we see that if their variances are equal $$\sigma_1 = \sigma_2$$, their CDFs will not intersect each other (trivial case: CDFs are overlapping if their mean also equal). So you would like to check whether

$$F_{X_2}(x) > F_{X_1}(x) \iff \frac {x - \mu_1} {\sigma} < \frac {x - \mu_2} {\sigma} \iff \mu_1 > \mu_2$$

as expected.

When the variances are not equal $$\sigma_1 \neq \sigma_2$$, the above equation will give one and only one root. That means the function $$D$$ have one critical point only:

$$x^* = \frac {\sigma_2\mu_1 - \sigma_1\mu_2} {\sigma_2 - \sigma_1}$$

So now you need to verify the following:

1. $$F_{X_2}(x) - F_{X_1}(x) > 0 ~~ \forall x < x^*$$

and this ensure $$D(x) > 0 ~~ \forall x < x^*$$. From the definition it ensure that $$D$$ will be strictly increasing up to $$x^*$$ and strictly decreasing until $$+\infty$$. And thus we would also need to verify

2. $$\lim_{x\to+\infty}D(x) \geq 0$$

For the first condition it can be similarly translated as

$$(\sigma_2 - \sigma_1)x < \sigma_2\mu_1 - \sigma_1\mu_2 ~~ \forall x < x^*$$

and this is equivalent to $$\sigma_2 > \sigma_1$$
(as expected; a larger variance ensure a thicker tail)

For the second condition, note that

$$\lim_{x\to+\infty}D(x) = \int_{-\infty}^{+\infty} [F_{X_2}(t) - F_{X_1}(t)] dt$$

Therefore

$$\lim_{x\to+\infty}D(x) \geq 0 \iff \int_{-\infty}^{+\infty} [F_{X_2}(t) - F_{X_1}(t)]dt \geq 0$$

$$\iff \int_{-\infty}^{0} \{-F_{X_1}(t) - [-F_{X_2}(t)]\}dt + \int_{0}^{+\infty} \{[1 - F_{X_1}(t)] - [1 - F_{X_2}(t)] \}dt \geq 0$$

$$\iff \int_{-\infty}^{0} [-F_{X_1}(t)]dt + \int_{0}^{+\infty} [1 - F_{X_1}(t)]dt \geq \int_{-\infty}^{0} [-F_{X_2}(t)]dt + \int_{0}^{+\infty} [1 - F_{X_2}(t)]dt$$

$$\iff E[X_1] \geq E[X_2]$$

As a result, the conditions will be $$mu_1 \geq \mu_2$$ and $$\sigma_1 \leq \sigma_2$$ and this should be enough for normal distribution.

The issue becomes interesting if you have the two independent random sample and want to test the stochastic dominance. Again if you have the normality assumption, you may have the above condition as your null hypothesis.