In my previous post I gave a few examples of cross-embedded models with multiple ancillaries. In this post, I’ll present the Evans et al. proof that (C) entails (L), which involves constructing a cross-embedded model.

Evans et al. start with an arbitrary experimental outcome (E,x

^{0}) and construct by stipulation a hypothetical outcome of a hypothetical Bernoulli experiment (B,h) that has the same likelihood function. They then build a cross-embedded experiment out of E and B. To preview where this is going, they invoke (C) twice to establish that the outcome (x^{0},h) of the cross-embedded experiment has the same evidential meaning as (E,x^{0}) and as (B,h), and thus that (E,x^{0}) and (B,h) have the same likelihood function as one another. They then repeat the process with an arbitrary experimental outcome (E’,y^{0}) that has the same likelihood function as (E,x^{0}) to establish that (E’,y^{0}) has the same evidential meaning as (B,h), and thus that it has the same evidential meaning as (E,x^{0}). The likelihood principle follows immediately.Let f(X;θ) be the likelihood function for experiment E at sample point x. Evans et al. construct the following cross-embedded experiment:

V\X | x ^{0} | x ^{1} | … | … | x ^{i} | … |

h | ½ f(x ^{0};θ) | ½ f(x ^{1};θ) | … | … | ½ f(x ^{i};θ) | … |

t | ½ - ½f(x ^{0};θ) | ½ f(x ^{0};θ) | 0 | … | 0 | … |

The indicator variable for h is ancillary: its distribution is Bernoulli(1/2) independent of θ. The indicator variable for X=x

^{0}(as opposed to X itself) is ancillary: its distribution is also Bernoulli(1/2) independent of θ. Thus, (C) says that one can conditionalize on these variables without changing evidential meaning.

Call this hypothetical cross-embedded experiment E*. By (C) and the ancillarity of the indicator for h, Ev(E*,(h,x

^{0}))=Ev(E,x^{0}). By (C) and the ancillarity of the indicator for x^{0}, Ev(E*,(h,x^{0}))=Ev(B,h).By the same series of steps, replacing x’s with y’s and Es with E’s, one can show that Ev(E’,y

^{0})=Ev(B,h) for any (E’,y^{0}) such that y^{0}has the same likelihood function in E’ as x^{0}has in E. The likelihood principle follows immediately.
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