Stochastic superiority

Abstract

This paper introduces a definition of stochastic superiority. One random variable is stochastically superior to another whenever it stochastically dominates the other after the risk in each random variable has been optimally reduced. Stochastic superiority is implied by stochastic dominance, but the reverse is not true. Stochastic superiority allows more pairs of random alternatives to be ranked, and efficient sets to be smaller. A very strong sufficient condition for stochastic superiority is demonstrated to also be necessary when preferences are risk averse. This condition provides a relatively easy way to conduct stochastic superiority tests. As an alternative to “almost stochastic dominance,” stochastic superiority also provides a natural solution to the “left tail problem” that arises often when comparing random alternatives.

Appendix

1.1 Proof of Theorem 1

Assume $\alpha \stackrel{{\sim }}{\text{x}}+\left(1-\alpha \right)\rho$ stochastically dominates $\tilde{y}$ for all u(x) in U. Then from Lemma 1, for any λ in [0, 1] and $\gamma = \left( {1 - \lambda } \right)\rho ,\,\lambda \left[ {\alpha \tilde{x} + \left( {1 - \alpha } \right)\rho } \right] + \left( {1 - \lambda } \right)\rho = \lambda \alpha \tilde{x} + \left( {1 - \lambda \alpha } \right)\rho$ stochastically dominates $\lambda \tilde{y} + \left( {1 - \lambda } \right)\rho$ for all u(x) in U. Put differently, for every element in the set $\left\{ {\lambda \tilde{y} + \left( {1 - \lambda } \right)\rho } \right\}$, there is an element in the set $\left\{ {\lambda \alpha \tilde{x} + \left( {1 - \lambda \alpha } \right)\rho } \right\}$ that dominates the former for all u(x) in U. This is of course sufficient for the optimal choice from the set $\left\{ {\alpha \tilde{x} + \left( {1 - \alpha } \right)\rho } \right\}$ to be preferred or indifferent to the optimal choice from the set $\left\{ {\beta \tilde{y} + \left( {1 - \beta } \right)\rho } \right\}$ for every utility function in the set U. Therefore, $\stackrel{{\sim }}{\text{x}}$ is stochastically superior to $\tilde{y}$ for all u(x) in U for the given ρ. Q.E.D.

1.2 Proof of Theorem 2

Let ${\text{U}}\left( \alpha \right) = {\text{Eu}}\left( {\alpha \tilde{x} + \left( {1 - \alpha } \right)\rho } \right)$ for α in [0, 1] and a given ρ. What must be shown is that if for any u(x) in U there exists an α such that ${\text{U}}\left( \alpha \right) \ge {\text{Eu}}\left( {\tilde{y}} \right)$, then there exists an α₀ in [0, 1] such that ${\text{U}}\left( {\alpha_{0} } \right) \ge {\text{Eu}}\left( {\tilde{y}} \right)$ for all u(x) in U. Define A_u to be the set of all α in [0, 1] such that ${\text{U}}\left( \alpha \right) \ge {\text{Eu}}\left( {\tilde{y}} \right)$. For concave utility functions, A_u is a closed line segment in [0, 1]. Let A^* denote the intersection of these line segments. If A^* is not empty, then any α₀ in A^* is such that ${\text{U}}\left( {\alpha_{0} } \right) \ge {\text{Eu}}\left( {\tilde{y}} \right)$ for all u(x) in U and we are done.

Assume instead that A^* is empty. We must show that this is not possible when $\stackrel{{\sim }}{\text{x}}$ is stochastically superior to $\tilde{y}$. Let α_u denote the maximum value for α such that ${\text{U}}\left( \alpha \right) \ge {\text{Eu}}\left( {\tilde{y}} \right)$ for utility function u(x). That is, α_u is the right end point of the line segment A_u. Denote this set of real numbers, one for each utility function, as $\left\{ {\alpha_{{\text{u}}} } \right\}$. Let α_L denote the infimum or greatest lower bound of this set $\left\{ {\alpha_{{\text{u}}} } \right\}$. α_L exists because $\left\{ {\alpha_{{\text{u}}} } \right\}$ is a bounded set of real numbers. Similarly, let $\alpha_{{\text{u}}}^{ * }$ denote the minimum value for α such that ${\text{U}}\left( \alpha \right) \ge {\text{Eu}}\left( {\tilde{y}} \right)$. $\alpha_{{\text{u}}}^{ * }$ is the left end point of the line segment which is A_u. Denote this set of real numbers $\left\{ {\alpha_{{\text{u}}}^{ * } } \right\}$. Let α_R denote the supremum or least upper bound of this set $\left\{ {\alpha_{{\text{u}}}^{ * } } \right\}$.

If α_L > α_R, then for all α in [α_R, α_L], ${\text{U}}\left( \alpha \right) \ge {\text{Eu}}\left( {\tilde{y}} \right)$ for all u(x) in U and hence A^* is not empty. When α_L = α_R, and α_L is in $\left\{ {\alpha_{{\text{u}}} } \right\}$ and α_R is in $\left\{ {\alpha_{{\text{u}}}^{ * } } \right\}$, the same thing is true; that is ${\text{U}}\left( {\alpha_{L} } \right) \ge {\text{Eu}}\left( {\tilde{y}} \right)$ for all u(x) in U. Thus, the hypothesis A^* empty rules this out.

If α_L < α_R, for any α₀ in [α_L, α_R], there exists a utility function u₁(x) such that ${\text{U}}_{1} \left( {\alpha_{0} } \right) < {\text{Eu}}\left( {\tilde{y}} \right)$ and the α^* which maximizes U₁(α) occurs at a point α^* < α₀ (because α_L < α₀). Similarly, there also exists a utility function u₂(x) such that ${\text{U}}_{2} \left( {\alpha_{0} } \right) < {\text{Eu}}\left( {\tilde{y}} \right)$ and the α^* which maximizes U₂(α) occurs at a point α^* > α₀ (because α₀ < α_R). This same thing is true when α_L = α_R, and α_L is not in $\left\{ {\alpha_{{\text{u}}} } \right\}$ and α_R is not in $\left\{ {\alpha_{{\text{u}}}^{ * } } \right\}$. Either of these two cases allows construction of a utility function of the form ${\text{u}}_{\lambda } \left( x \right) = \lambda {\text{u}}_{1} \left( x \right) + \left( {1 - \lambda } \right){\text{u}}_{2} \left( x \right)$ which satisfies ${\text{U}}_{\lambda } \left( {\alpha_{0} } \right) < {\text{Eu}}\left( {\tilde{y}} \right)$ and the maximum α^* for this utility function occurs at α₀. Hence stochastic superiority is violated so this case is also not possible.

The final two cases to discuss are when α_L = α_R, and α_L is in $\left\{ {\alpha_{{\text{u}}} } \right\}$ and α_R is not in $\left\{ {\alpha_{{\text{u}}}^{ * } } \right\}$, or vice versa. The discussion is for the first of these cases and the second is resolved in a similar fashion. When α_L = α_R, and α_L is in $\left\{ {\alpha_{{\text{u}}} } \right\}$ and α_R is not in $\left\{ {\alpha_{{\text{u}}}^{ * } } \right\}$, there exists a utility function u₁(x) such that ${\text{U}}_{1} \left( {\alpha_{{\text{L}}} } \right) = {\text{Eu}}\left( {\tilde{y}} \right)$ and the α^* which maximizes U₁(α) occurs at a point α^* < α_L. There also exists a utility function u₂(x) such that U₂(α_L) < Eu($\stackrel{{\sim }}{\text{y}}$) and the α^* which maximizes U₂(α) occurs at a point α^* > α_L. This allows construction of a utility function of the form ${\text{u}}_{\lambda } \left( x \right) = \lambda {\text{u}}_{1} \left( x \right) + \left( {1 - \lambda } \right){\text{u}}_{2} \left( x \right)$ which satisfies ${\text{U}}_{\lambda } \left( {\alpha_{{\text{L}}} } \right) < {\text{Eu}}\left( {\tilde{y}} \right)$ and the maximum α^* for this utility function occurs at α_L. Thus, this case is not possible either and it can be concluded that A^* is not empty when stochastic superiority of $\stackrel{{\sim }}{\text{x}}$ over $\tilde{y}$ holds. Q.E.D.

1.3 Proof of Theorem 3

Suppose that $\tilde{x}$ is stochastically superior to $\tilde{y}$ on some convex set of concave utility functions U for a given ρ. Then, from Theorem 2, there exists an α₀ in [0, 1] such that $\alpha_{0} \tilde{x} + \left( {1 - \alpha_{0} } \right)\rho$ stochastically dominates $\tilde{y}$ on U. Since for $\rho^{\prime} \ge \rho ,\,\alpha_{0} \tilde{x} + \left( {1 - \alpha_{0} } \right)\rho^{\prime}$ stochastically dominates $\alpha_{0} \tilde{x} + \left( {1 - \alpha_{0} } \right)\rho$, by transitivity it also stochastically dominates $\tilde{y}$ on U. Therefore, according to Theorem 1, $\stackrel{{\sim }}{\text{x}}$ is stochastically superior to $\tilde{y}$ on U for ρ′. Q.E.D.

1.4 Proof of Theorem 4

Sufficiency is obvious. To show necessity, suppose that $\left(\stackrel{{\sim }}{\text{x}}, {\rho }_{x}\right)$ is stochastically superior to $\left(\stackrel{{\sim }}{\text{y}}, {\rho }_{y}\right)$ on U for some ρ_y such that ρ_x ≥ ρ_y. Then for every u(x) in U there exists an α ∈ [0, 1] such that $\alpha \tilde{x} + \left( {1 - \alpha } \right)\rho_{x}$ is preferred or indifferent to $\tilde{y}$. According to Theorem 2, there exists an α₀ ∈ [0, 1] such that $\alpha_{0} \tilde{x} + \left( {1 - \alpha_{0} } \right)\rho_{x}$ stochastically dominates $\tilde{y}$ on U. From Theorem 1, this implies that $\left(\stackrel{{\sim }}{\text{x}}, {\rho }_{x}\right)$ is stochastically superior to $\left(\stackrel{{\sim }}{\text{y}}, {\rho }_{x}\right)$ on U. Q.E.D.

1.5 Proof of Theorem 5

Equation 1 implies that ρ_x > ρ_y, and therefore according to Theorems 1, 2 and 4, $\left(\stackrel{{\sim }}{\text{x}}, {\rho }_{x}\right)$ is stochastically superior to $\left(\stackrel{{\sim }}{\text{y}}, {\rho }_{y}\right)$ in the second degree if and only if there is an α in [0, 1] such that $\alpha \tilde{x} + \left( {1 - \alpha } \right)\rho_{x}$ stochastically dominates $\tilde{y}$ in the second degree, where $\rho_{x} = {\text{w}} - {\text{e}}_{2} - \left( {1 + \pi } \right){\text{p}}_{2} {\text{L}}$.

To establish the condition under which there exists an α in [0, 1] such that $\alpha \tilde{x} + \left( {1 - \alpha } \right)\rho_{x}$ stochastically dominates $\tilde{y}$ in the second degree, note that $\alpha \tilde{x} + \left( {1 - \alpha } \right)\rho_{x}$ is a binary distribution yielding a low outcome $\alpha \left( {{\text{w}} - {\text{e}}_{2} - {\text{L}}} \right) + \left( {1 - \alpha } \right)\left( {{\text{w}} - {\text{e}}_{2} - \left( {1 + \pi } \right){\text{p}}_{2} {\text{L}}} \right)$ and a high outcome $\alpha \left( {{\text{w}} - {\text{e}}_{2} } \right) + \left( {1 - \alpha } \right)\left( {{\text{w}} - {\text{e}}_{2} - \left( {1 + \pi } \right){\text{p}}_{2} {\text{L}}} \right)$ with probabilities p₂ and (1 − p₂) respectively. Similarly, $\tilde{y}$ is a binary distribution yielding a low outcome ${\text{w}} - {\text{e}}_{1} - {\text{L}}$ and a high outcome w − e₁ with probabilities p₁ and (1 − p₁) respectively. Because p₂ < p₁, $\alpha \tilde{x} + \left( {1 - \alpha } \right)\rho_{x}$ stochastically dominates $\tilde{y}$ in the second degree if and only if the low outcome of the former is no less than the low outcome of the latter, and the mean of the former is no less than the mean of the latter. These conditions are expressed as eqs. (3) and (4) given below:

$$\alpha \left( {{\text{w}} - {\text{e}}_{2} - {\text{L}}} \right) + \left( {1 - \alpha } \right)\left( {{\text{w}} - {\text{e}}_{2} - \left( {1 + \pi } \right){\text{p}}_{2} {\text{L}}} \right) \ge {\text{w}} - {\text{e}}_{1} - {\text{L}}$$

(3)

and

$$\alpha \left( {{\text{w}} - {\text{e}}_{2} } \right) - \alpha {\text{p}}_{2} {\text{L}} + \left( {1 - \alpha } \right)\left( {{\text{w}} - {\text{e}}_{2} - \left( {1 + \pi } \right){\text{p}}_{2} {\text{L}}} \right) \ge {\text{w}} - {\text{e}}_{1} - {\text{p}}_{1} {\text{L}}$$

(4)

Now Eqs. (3) and (4) can be equivalently expressed as:

$$\alpha \le 1-\frac{{\text{e}}_{2}-{\text{e}}_{1}}{\left[1-\left(1+\pi \right){\text{p}}_{2}\right]{\text{L}}}$$

(5)

and

$$\alpha \ge 1-\frac{\left({\text{p}}_{1}-{\text{p}}_{2}\right){\text{L}}-\left({\text{e}}_{2}-{\text{e}}_{1}\right)}{\pi {\text{p}}_{2}{\text{L}}}$$

(6)

respectively.

There exists such an α that satisfies both Eqs. (5) and (6) if and only if

$$\frac{{\left( {{\text{p}}_{1} - {\text{p}}_{2} } \right){\text{L}} - \left( {{\text{e}}_{2} - {\text{e}}_{1} } \right)}}{{\pi {\text{p}}_{2} {\text{L}}}} \ge \frac{{{\text{e}}_{2} - {\text{e}}_{1} }}{{\left[ {1 - \left( {1 + \pi } \right){\text{p}}_{2} } \right]{\text{L}}}}$$

or equivalently

$$\pi \le \frac{{\left( {1 - {\text{p}}_{2} } \right)}}{{{\text{p}}_{2} }}\left[ {1 - \frac{{\left( {{\text{e}}_{2} - {\text{e}}_{1} } \right)}}{{{\text{L}}\left( {{\text{p}}_{1} - {\text{p}}_{2} } \right)}}} \right]$$