(a) Is it possible to have two r.v.s X and Y such that X and...

(a) Is it possible to have two r.v.s $X$ and $Y$ such that $X$ and $Y$ have the same distribution but $P(X<Y) \geq p$, where:
- $p=0.9$ ?
- $p=0.99$ ?
- $p=0.9999999999999$ ?
- $p=1$ ?
For each, give an example showing it is possible, or prove it is impossible.

Key Concepts

Copulas and the Fréchet–Hoeffding bounds

A copula is a function that links together marginal distributions to form a joint distribution. The Fréchet–Hoeffding bounds give the maximal and minimal possible values for probabilities of events (like {X < Y}) over all couplings with fixed marginals. In particular, these bounds are used to understand the best?and?worst case scenarios for probabilities of ordering events when the marginal laws are fixed.

Discrete vs. Continuous Distributions

A key point is that when the common law of X and Y is continuous (i.e. has no atoms) then one always has P(X < Y) = P(Y < X) = 1/2 for any joint law having those marginals. However, if the marginal law has atoms (is at least partly discrete) then by choosing a non?symmetric (extremal) coupling one may arrange that P(X < Y) is as close to 1 as one wishes (though it can never equal 1). This distinction is crucial in answering the different parts of the problem.

Coupling of Random Variables

This is the idea of choosing a joint law for two random variables that are required to have given marginal distributions. Although one can choose many different joint distributions (couplings) for two fixed marginals, some choices maximize or minimize certain joint probabilities (such as the event that one variable is less than the other). In this problem one is allowed to choose any joint law for X and Y, provided that both have the same distribution.

Identical Marginal Distributions

Two random variables having the same distribution means that they have the same cumulative distribution function (cdf) (or probability mass function if discrete) even though their joint distribution may be very different. This requirement restricts the marginals but leaves open the possibility of non?symmetric couplings, which are used in questions asking how large P(X < Y) can be made.

Maximal Probability Ordering (and its Limitations)

The problem is really about how high one can make P(X < Y) when X and Y share the same law. With continuous distributions, symmetry forces P(X < Y) to be exactly 1/2, but when the law has atoms one may ‘separate’ the atoms in a careful joint construction to make P(X < Y) arbitrarily close to 1 – though not equal to 1. In particular, one may obtain examples with P(X < Y) ? 0.9, 0.99, or even 0.9999999999999, but one cannot have P(X < Y) = 1 unless the distribution is degenerate.

Transcript

Please give Ace some feedback