All My Bloody Notes Bloody Maths Statistics Stochastic Dominance

Stochastic Dominance

The notion of a distribution being bigger than another

First order stochastic dominance

Key idea

Higher expected value from the dominant distribution


We say F(x)F(x) F.O.S.D G(x)G(x) if F(x)G(x),xF(x) \leq G(x), \forall x

Higher expected value (1)

Suppose the support of the distribution is some interval [α,β][\alpha, \beta]

αβxf(x)dx=αβxdF(x)=[xF(x)]αβαβF(x)dx\int^{\beta}_{\alpha} x f(x) dx = \int^{\beta}_{\alpha} x dF(x) = \left[ xF(x) \right]^{\beta}_{\alpha} - \int^{\beta}_{\alpha} F(x) dx

As F(x)G(x)xF(x) \leq G(x) \forall x, then αβF(x)dxαβG(x)dx\int^{\beta}_{\alpha} F(x) dx \leq \int^{\beta}_{\alpha} G(x) dx, minus a smaller quantity in turn implies that E(F(x))E(G(x))E(F(x)) \geq E(G(x)).

Higher expected value (2)

Consider some non-decreasing function u(x)u(x):

αβu(x)[f(x)g(x)]dx=αβu(x)d[F(x)G(x)]\int^{\beta}_{\alpha} u(x) [f(x) - g(x)] dx = \int^{\beta}_{\alpha} u(x) d[F(x) - G(x)]

Using integration by parts again, this equals to:

[u(x)[F(x)G(x)]]αβαβ[F(x)G(x)]u(x)dx=αβ[F(x)G(x)]u(x)dx\left[ u(x)[F(x) - G(x)] \right]^{\beta}_{\alpha} - \int^{\beta}_{\alpha} [F(x) - G(x)] u'(x) dx = - \int^{\beta}_{\alpha} [F(x) - G(x)] u'(x) dx

Again as F(x)G(x)xF(x) \leq G(x) \forall x and u(x)0u'(x) \geq 0, then we have:

EF(x)(u(x))EG(x)(u(x))E_{F(x)}(u(x)) \geq E_{G(x)}(u(x))

Conditional stochastic dominance

Key idea

Truncate from right, left-tail probability mass higher


Consider two differentiable distribution functions F(θ)F(\theta) and G(θ)G(\theta) with common support [α,β][\alpha, \beta].

Define Fy(θ)F_y(\theta) and Gy(θ)G_y(\theta) to be the two distribution functions right truncated at point yy. That is,

Fy(θ)=F(θ)F(y);θ[α,y]F_y(\theta) = \frac{F(\theta)}{F(y)}; \forall \theta \in [\alpha, y] Gy(θ)=G(θ)G(y);θ[α,y]G_y(\theta) = \frac{G(\theta)}{G(y)}; \forall \theta \in [\alpha, y]

Then we say F(.)F(.) conditionally stochastic dominate G(.)G(.) if

Fy(θ)=F(θ)F(y)<G(θ)G(y)=Gy(θ),y[α,β];θ<yF_y(\theta) = \frac{F(\theta)}{F(y)} < \frac{G(\theta)}{G(y)} = G_y(\theta), \forall y \in [\alpha, \beta]; \forall \theta < y

In words, we say F(.)F(.) conditionally stochastic dominate G(.)G(.) if for every right truncation yy, the resultant probability mass in the left-tail is less in F(.)F(.) than in G(.)G(.).

Useful links

Conditional stochastic dominance:

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