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A finite-horizon likelihood-ratio test supermartingale

Example

Fix an integer K≥1. Let Ω={0,1}K and let P0 be uniform on Ω, so the coordinate variables X1,…,XK are independent Bernoulli(1/2). Let Q be the product Bernoulli(3/4) law and let Ft=σ(X1,…,Xmin⁡(t,K)). The likelihood ratio of Q to P0 on Ft is Et=∏i=1min⁡(t,K)ℓ(Xi),ℓ(1)=32,ℓ(0)=12, with the empty product E0=1. It is a nonnegative test martingale, hence a nonnegative test supermartingale, for the simple null {P0}.

Facts & Assumptions

Given: K≥1, the finite sample space, laws, filtration, and process in the Example. Assume AC for the test-supermartingale convention.

[F1]

Nonnegative test supermartingales: the process must be real-valued, adapted, integrable, nonnegative, supermartingale under the null, and have initial expectation at most one.

[A1]

The Axiom of Choice: AC supplies the conditional-expectation convention used in F1.

Verification

1.1givenconstruct

For a prefix x1,…,xt with t≤K, the ratio of its Q-probability to its P0-probability is ∏i=1t(Q(Xi=xi)/P0(Xi=xi))=∏i=1tℓ(xi), because every P0 prefix has positive probability. Thus the displayed process is the finite-prefix likelihood ratio; after time K it is held constant.

2.1givenstep 1.1algebra

Each Et is a finite nonnegative function of the first min⁡(t,K) coordinates, hence is adapted and integrable. Conditional on any prefix before time K, the next bit is fair and EP0[ℓ(Xt+1)∣Ft]=(3/2+1/2)/2=1, so EP0[Et+1∣Ft]=Et. For t≥K the process is constant.

3.1F1A1step 1.1step 2.1∎

Since E0=1, steps 1.1 and 2.1 verify every condition in F1 under the null law P0; hence E is a test martingale and a test supermartingale.

Depends on

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