advanced32 min studyLesson 706 in Mathematics

Contiguity and Le Cam’s lemmas

Transfer local limits by checking likelihood mass and joint convergence.

Before you begin

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The question for today

What you’ll understand

Distinguish sequential contiguity from finite-sample absolute continuity.
Check a sufficient likelihood-ratio form of the first lemma.
Use a joint Gaussian limit and exponential tilting to transfer a statistic.

A statistic's limit under a reference model is not automatically its limit under a sequence of alternatives. Contiguity controls which negligible events remain negligible when the probability law changes with sample size. Le Cam’s lemmas connect that property to likelihood ratios and then transfer joint weak limits. These tools explain why local alternatives in a regular model have nontrivial power rather than behaving either identically to the null or as completely separated alternatives. Naming both sequences of probability measures is essential throughout the argument.

Negligible events under changing laws

Write $Q_n\triangleleft P_n$ when every sequence of measurable events whose probability tends to zero under the reference laws also has probability tending to zero under the alternative laws. This is a one-way implication. Mutual contiguity requires the reverse implication too. Absolute continuity at every finite sample size concerns exact zero probabilities separately at each size; it does not control a moving sequence of increasingly rare events. Thus even everywhere-positive normal densities can form sequences that are not contiguous.

A useful sufficient version of the first lemma starts with likelihood ratios $L_n=dQ_n/dP_n$ . If they converge in distribution under the reference laws to a nonnegative variable whose expectation is one, no likelihood mass is lost in the limit, and the alternatives are contiguous to the reference sequence. If that limit is also strictly positive almost surely, the sequences are mutually contiguous. These conditions are stated as a sufficient criterion here, with domination explicit; expectation one for each finite likelihood ratio alone is insufficient.

Q_n\triangleleft P_n\quad\Longleftrightarrow\quad [P_n(A_n)\to0\ \Longrightarrow\ Q_n(A_n)\to0]\quad\text{for every event sequence }A_n.

Contiguity compares asymptotically negligible events, rather than all event probabilities.↔ Scroll wide equations horizontally.

The third lemma needs a joint limit

Suppose a statistic and the log likelihood ratio converge jointly under the reference law. When the exponential of the limiting log ratio has expectation one, the statistic's alternative limit is obtained by weighting the joint limit by that exponential. Joint convergence is indispensable: the dependence between the statistic and likelihood ratio determines how the law changes. Knowing the two marginal limits separately does not determine the tilt. For bounded continuous test functions this gives an explicit formula for the transferred distribution.

If the statistic and log ratio are jointly Gaussian, and the log ratio has mean minus half its variance, exponential weighting shifts the statistic's mean by its covariance with the log ratio. Its covariance remains unchanged. This is the familiar Gaussian local-power calculation, derived from a change of measure rather than an assumption that the alternative has the same centered limit. A statistic can therefore have different limiting event probabilities under mutually contiguous laws without contradicting the definition.

E_Q f(T)=E_P\{f(T)e^V\},\qquad (T,V)\ \text{jointly Gaussian}\ \Longrightarrow\ T\text{ under the tilt}\sim N\!\bigl(E_PT+\operatorname{Cov}(T,V),\operatorname{Var}_P(T)\bigr).

The covariance with the limiting log likelihood ratio determines the local mean shift.↔ Scroll wide equations horizontally.

What contiguity transfers and what it does not

Contiguity carries convergence in probability to zero from the reference sequence to the alternative sequence. In particular, an asymptotic expansion with a negligible remainder under the reference law retains that negligible remainder under contiguous alternatives. It does not make arbitrary nonzero limiting probabilities equal, and it does not assert that total variation distance tends to zero. The exact normal example below has a persistent change in a simple rejection probability while all asymptotically negligible events still transfer.

A common diagnostic error is to inspect only the finite expectation of the likelihood ratio. Rare, enormous ratios can carry all its mean while its weak limit is zero. This lost mass signals the failure of the sufficient first-lemma criterion and often accompanies distinguishable alternatives. The practice example makes this concrete with a mean shift that grows without bound. The reference measure in every convergence statement matters; an expression that is small under one sequence need not be small under another without a transfer argument.

Work it through

An exact Gaussian exponential tilt

Let $P_n=N(0,4)^{\otimes n}$ and $Q_n=N(2/\sqrt n,4)^{\otimes n}$ . Set $T_n=\sqrt n\,\overline X/2$ . Determine its law under each sequence, establish mutual contiguity, and compare the probabilities of $T_n>0$ .

Under $P_n$ , $T_n$ is exactly standard normal. Direct normal likelihood algebra gives $\log L_n=T_n-1/2$ , so its joint limit with the statistic has covariance $1$ .
The limiting likelihood ratio is $\exp(Z-1/2)$ for standard normal $Z$ . It is strictly positive and has expectation $1$ , establishing mutual contiguity by the stated criterion.
Tilting shifts the statistic's mean by $1$ , giving $N(1,1)$ under $Q_n$ , which is also its exact law. Thus $P_n(T_n>0)=1/2$ while $Q_n(T_n>0)=\Phi(1)\approx0.841345$ .

The result

Different nonzero event probabilities coexist with mutual contiguity. Only probabilities tending to zero are constrained by its definition.

The living notebook

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Interactive lab

Interactive lab · sampling

A local shift changes probability without separation

Pause/Reset and set BOTH controls to $p=0.50$ , $n=100$ : predict variance $0.0025$ and standard deviation $0.05$ . For a fresh comparison Pause/Reset and set BOTH to $p=0.55$ , $n=100$ : variance becomes $0.002475$ while the mean shifts by one reference standard deviation. Controls retain the paused clock until Reset.

Exact binomial massNormal interval massHighlighted outcome / mean

Mean sample proportion: 0.5
Sample-mean variance: 0.0125
Uniform CDF error bound: ≤ 0.1062

Success probability p0.5Independent samples n20

t = 0 s

Exact binomial masses for the mean of n independent Bernoulli trials, compared with normal mass between adjacent half-count boundaries; normal tails outside the finite support remain outside the bars. Playback highlights outcomes, not random samples. The Berry–Esseen bound applies to standardized CDFs, not individual bars or continuity-corrected errors. At p = 0 or 1, normal comparison and that bound are disabled. Normal-CDF evaluation uses a numerical approximation with absolute error below 7.5×10⁻⁸ (at most 1.5×10⁻⁷ per interval, before rounding). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThese exact binomial distributions illustrate a finite local shift. A comparison at one sample size cannot establish contiguity of two sequences.

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Check your understanding

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Read the checkpoint explanation

Apply the negligible-event definition to the events where the remainder exceeds a fixed positive threshold. Contiguity does not equate ordinary limiting probabilities, imply vanishing total variation, or follow merely from finite-sample equivalence.

Your turn

Try it yourself

Take $P_n=N(0,1)$ and $Q_n=N(n,1)$ on the real line. Use $A_n=\{X>n/2\}$ to test contiguity. What happens to the likelihood ratio under the reference sequence despite its expectation being one?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compare the event with the center of each normal law, and then write the normal density ratio.

Compare with the solution

Under $P_n$ , $P_n(A_n)\to0$ ; under $Q_n$ , $Q_n(A_n)\to1$ . Therefore $Q_n$ is not contiguous to $P_n$ . The likelihood ratio is $L_n=\exp(nX-n^2/2)$ and converges to zero in $P_n$ -probability, although $E_{P_n}L_n=1$ for every sample size. Its weak limit has expectation zero: mean mass escapes into rare extreme ratios. This is why a finite-sample expectation calculation cannot replace the first lemma's limiting expectation condition.

Keep the idea

Take it with you

Contiguity transfers negligible events between named sequences of laws.
The first lemma checks whether likelihood mass survives the limit.
The third lemma uses a joint limit, with covariance determining Gaussian local shifts.

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