Higher Order Stochastic Dominance Verification

This tutorial demonstrates the function verify_dominance, which validates the dominance relationship between $Y$ and $X$.

Example 1

To verify if $Y$ stochastically dominates $X$ under the specified order SDorder = 2, we use the function verify_dominance

Defining Random Variables $X$ and $Y$

Consider two discrete random variables, $X$ and $Y$. Their values and associated probabilities are:

• \[Y = [3, 5, 7]\]

  with probabilities $p_Y = [0.3, 0.4, 0.3]$.

• \[X = [2, 4, 6]\]

  with probabilities $p_X = [0.2, 0.5, 0.3]$.

These represent two probability distributions that we compare using stochastic dominance.

Y = [3, 5, 7]
X = [2, 4, 6]
p_Y = [0.3, 0.4, 0.3]
p_X = [0.2, 0.5, 0.3]
SDorder = 2
verify_dominance(Y, X,SDorder;p_Y, p_X)

Output:

Y dominates X in stochastic order 2
true

This function checks whether $Y$ stochastically dominates $X$ of order SDorder. The verbose=true option ensures detailed output, providing insights into the dominance verification process.

Example 2

Consider two discrete random variables, $X$ and $Y$, with swapped roles. Their values and associated probabilities are:

• \[X = [3, 5, 7]\]

  with probabilities $p_X = [0.3, 0.4, 0.3]$.

• \[Y = [2, 4, 6]\]

  with probabilities $p_Y = [0.2, 0.5, 0.3]$.

These represent two probability distributions that we compare using stochastic dominance.

X = [3, 5, 7]
Y = [2, 4, 6]
p_X = [0.3, 0.4, 0.3]
p_Y = [0.2, 0.5, 0.3]
SDorder = 2
verify_dominance(Y,X,SDorder;p_Y,p_X,verbose=true)

Output:

Y doesn't dominates X in stochastic order 2
Y doesn't dominates X in stochastic order 2 with residual 1.6970562748477143
false

The verbose output indicates that with a residual value of 1.70, we can confirm that Y does not dominate X in the second-order stochastic dominance. However, in general, the user has control over the choice of residual (ε), and it can be selected as needed.