The Likelihood Ratio in Canonical Coordinates

Why Express the Likelihood Ratio in Canonical Coordinates?

At an interim analysis, the accumulated data can be summarized by a cumulative score and its information. The canonical statistic rescales that score so its null variance is one. Expressing a likelihood ratio in these coordinates makes its dependence on the observed evidence, the assumed fixed effect, and the information level explicit. It also shows why the full sequence of interim statistics is not needed to calculate the likelihood ratio for a fixed effect using data through the current look.

The result depends on a specified model. In the canonical joint normal model, the cumulative score increments are independent normal random variables, with means proportional to their information increments. The derivation below uses this structure; it does not claim that every test statistic in every model has an exact likelihood-ratio representation of this form.

Definition: Let $U_k$ be the cumulative score through look $k$, and let $I_k$ be the cumulative information through that look, with $0<I_1<\cdots<I_K$. Write $\Delta U_j=U_j-U_{j-1}$ and $\Delta I_j=I_j-I_{j-1}$, where $U_0=I_0=0$. In the canonical fixed-effect model, under effect $\theta$, the increments are independent and $$ \Delta U_j\sim N(\theta\Delta I_j,\Delta I_j). $$ The canonical statistic at look $k$ is $Z_k=U_k/\sqrt{I_k}$. Thus $Z_k\sim N(\theta\sqrt{I_k},1)$ marginally. The fixed-effect likelihood ratio through look $k$, comparing $\theta$ with the null effect $0$, is $L_k(\theta)=f_\theta/f_0$, the ratio of the joint densities of the observed data through that look under those two specified effects.

Here $\theta$ is a fixed, specified effect, not an estimate chosen after seeing the data. The information levels are treated as fixed and known for the likelihood calculation. The increment variances $\Delta I_j$ must be positive, and the same score and information convention must be used at every look. The model may be exact, as in a normal location model with known variance, or an approximation justified by a suitable large-sample model.

Derivation From the Score Increments

Under effect $\theta$, the density of increment $\Delta U_j$ is normal with mean $\theta\Delta I_j$ and variance $\Delta I_j$. Under the null, its mean is zero and its variance is the same. The ratio of these two normal densities can be simplified directly.

\frac{f_{\theta,j}(\Delta U_j)}{f_{0,j}(\Delta U_j)} = \exp\left\{ -\frac{(\Delta U_j-\theta\Delta I_j)^2}{2\Delta I_j} +\frac{(\Delta U_j)^2}{2\Delta I_j} \right\} = \exp\left\{\theta\Delta U_j-\frac{\theta^2\Delta I_j}{2}\right\}.

Independence of the increments means their joint densities are products. Consequently, the joint likelihood ratio through look $k$ is the product of the increment ratios. Multiplying exponentials adds their exponents, and the increments telescope: $\sum_{j=1}^k\Delta U_j=U_k$ and $\sum_{j=1}^k\Delta I_j=I_k$.

L_k(\theta) = \prod_{j=1}^k \exp\left\{\theta\Delta U_j-\frac{\theta^2\Delta I_j}{2}\right\} = \exp\left\{\theta U_k-\frac{\theta^2 I_k}{2}\right\}.

Since $U_k=\sqrt{I_k}Z_k$, the same ratio in canonical coordinates is

L_k(\theta) = \exp\left\{\theta\sqrt{I_k}\,Z_k-\frac{\theta^2I_k}{2}\right\}, \qquad \log L_k(\theta) = \theta\sqrt{I_k}\,Z_k-\frac{\theta^2I_k}{2}.

Key result: For a fixed effect $\theta$, the likelihood ratio through look $k$ depends on the observed sequence through its cumulative score $U_k$, or equivalently through $Z_k$ and $I_k$. The earlier look-specific values do not appear separately in the final expression, although they may matter for a group-sequential stopping decision.

The expression also makes the direction of evidence clear. If $\theta>0$, larger positive $Z_k$ increases $L_k(\theta)$; if $\theta<0$, more negative $Z_k$ increases it. The term $-\theta^2I_k/2$ is always nonpositive. It prevents the ratio from increasing just because the proposed effect is nonzero: the observed score must provide evidence in a direction compatible with that effect.

Worked Calculations

Worked Example: Combining Two Score Increments

Suppose a hypothetical two-look experiment follows the canonical model. The fixed effect of interest is $\theta=0.20$. At the first look, the observed score increment is $\Delta U_1=4.0$ with $\Delta I_1=15$; between the first and second looks, $\Delta U_2=5.6$ with $\Delta I_2=25$. Calculate the likelihood ratio through look 2 against the null effect.

The cumulative values are $U_2=4.0+5.6=9.6$ and $I_2=15+25=40$. The increment formula gives log-likelihood-ratio contributions

\ell_1=(0.20)(4.0)-\frac{(0.20)^2(15)}{2}=0.50, \qquad \ell_2=(0.20)(5.6)-\frac{(0.20)^2(25)}{2}=0.62.

Their sum is $1.12$, so $L_2(0.20)=\exp(1.12)\approx 3.065$. Equivalently, $Z_2=9.6/\sqrt{40}\approx1.518$, and

\log L_2(0.20) =(0.20)\sqrt{40}(1.518)-\frac{(0.20)^2(40)}{2} \approx 1.12.

Under the specified canonical model, the observed data through look 2 are about 3.065 times as likely under the fixed effect $0.20$ as under effect 0. This is a likelihood comparison between two specified effects, not a probability that either effect is true.

Worked Example: A Fixed Negative Effect

In another hypothetical analysis, suppose $I_k=100$, $Z_k=-1.40$, and the specified effect is $\theta=-0.10$. Calculate the likelihood ratio against the null.

First, $\sqrt{I_k}=10$ and $U_k=Z_k\sqrt{I_k}=(-1.40)(10)=-14.0$. Substitution into the cumulative-score formula gives

\log L_k(-0.10) =(-0.10)(-14.0)-\frac{(-0.10)^2(100)}{2} =1.40-0.50=0.90, \qquad L_k(-0.10)=\exp(0.90)\approx2.460.

The likelihood is about 2.460 times as large under the specified negative effect as under the null. The negative observed statistic is aligned with the negative effect. Applying a positive-effect interpretation to this statistic would reverse the direction of the evidence.

Worked Example: Normal Observations With Known Variance

Suppose, hypothetically, that $n=16$ independent observations follow $X_i\sim N(\theta,100)$, so the known standard deviation is 10. The observed sample mean is $\bar{x}=2.5$, and the fixed effect being compared with the null mean 0 is $\theta=1.0$. Check the model quantities and calculate the likelihood ratio.

For this normal model, the log likelihood ratio for the observations is $\theta\sum_{i=1}^n X_i/100-n\theta^2/(2\cdot100)$. Thus the cumulative score and information are $U_n=\sum X_i/100=n\bar{x}/100=16(2.5)/100=0.40$ and $I_n=n/100=0.16$. The canonical statistic is $Z_n=U_n/\sqrt{I_n}=0.40/0.40=1.00$. These expressions satisfy the canonical model exactly because observations are independent normal with the same known variance and the effect is a mean shift.

\log L_n(1.0) =(1.0)(0.40)-\frac{(1.0)^2(0.16)}{2} =0.32, \qquad L_n(1.0)=\exp(0.32)\approx1.377.

The observed sample is about 1.377 times as likely under mean 1.0 as under mean 0 in this model. The canonical-coordinate calculation agrees: $(1.0)\sqrt{0.16}(1.00)-(1.0)^2(0.16)/2=0.32$.

What the Result Does—and Does Not—Say

A likelihood ratio compares the likelihoods of two specified models for the observed data. It is not itself a p-value, a posterior probability, or a stopping boundary. For example, $L_k(\theta)=3$ means the data through look $k$ have three times the likelihood under the specified effect than under the null, given the model. It does not mean there is a three-in-four probability that $\theta$ is correct.

The canonical statistic and the likelihood ratio also answer different questions. $Z_k$ reports the cumulative score in standard-error units, whereas $L_k(\theta)$ compares two fixed effects and depends on the proposed effect and information as well as the observed statistic. At a fixed $I_k$ and fixed $\theta$, the log likelihood ratio is a linear function of $Z_k$. Across different information levels, the same $Z_k$ need not imply the same likelihood ratio.

In a group-sequential design, the likelihood ratio through the current look is computed from data accumulated so far. Whether to stop, continue, or reject a hypothesis is a separate design decision governed by the prespecified sequential procedure. Repeated looks do not change the algebra above, but they do affect how a test’s overall error rate is controlled.

Assumptions, Diagnostics, and Common Pitfalls

The calculation is reliable only when its model ingredients describe the analysis being performed. Check that the score increments correspond to nonoverlapping information increments, that the stated variances equal $\Delta I_j$, and that the fixed-effect means equal $\theta\Delta I_j$. Also check that information is cumulative, positive, and on the same scale at all looks. In applications where the canonical model is approximate, substantial departures from the assumed score means, variances, or increment independence can make the computed likelihood ratio only an approximation.

If successive increments share data in a way not accounted for by the model, their joint density does not factor into the product used in the derivation. If information is random or estimated from the observed outcomes, treating it as fixed may also require additional justification. In such cases, use a likelihood based on the actual joint model or explicitly justify the approximation; do not simply insert the observed quantities into the canonical formula and assume it remains exact.

Common mistake: One might see a positive $Z_k$ and conclude that every positive fixed effect has a likelihood ratio greater than one. That is not necessarily true: the penalty $\theta^2I_k/2$ can outweigh the score term. The correction is to calculate $\log L_k(\theta)=\theta\sqrt{I_k}Z_k-\theta^2I_k/2$ for the particular effect and information level being compared.

A second mistake is to replace the fixed $\theta$ in the formula with an estimate and then describe the result as the likelihood ratio for that prespecified effect. Those are different calculations: the formula here compares $H_0:\theta=0$ with one specified $H_a:\theta$, not with an effect selected from the observed data. Likewise, a large likelihood ratio alone does not establish that a sequential test has crossed its calibrated rejection boundary.

Specify the comparison.
Choose the fixed effect $\theta$ and compare it with the null effect 0; keep the direction and units of the effect clear.

Check the canonical model.
Confirm the independent normal increment assumptions, the information scale, and whether the representation is exact or approximate.

Calculate and interpret.
Use the cumulative score and information, or $Z_k$ and $I_k$, then describe the result as a likelihood comparison between the specified effects.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

Under the canonical model, write the likelihood ratio through look $k$ in terms of $U_k$ and $I_k$.
Suppose $U_k=6$, $I_k=25$, and the fixed effect is $\theta=0.20$. Calculate the log likelihood ratio and the likelihood ratio.
For a negative fixed effect, which direction of $Z_k$ increases the likelihood ratio relative to the null?
Why is the likelihood ratio not itself a p-value or a probability that the fixed effect is true?

Solutions. (1) $L_k(\theta)=\exp\{\theta U_k-\theta^2I_k/2\}$. (2) The log ratio is $(0.20)(6)-(0.20)^2(25)/2=1.20-0.50=0.70$, so the ratio is $\exp(0.70)\approx2.014$. (3) More negative $Z_k$ increases the ratio when $\theta<0$, because $\theta\sqrt{I_k}Z_k$ then increases. (4) It is a ratio of likelihoods under two specified effects; it is neither a tail probability under a null distribution nor a probability distribution over parameter values.