The Multivariate Normal Density at Multiple Looks

Why the Joint Density Matters

At several interim looks, the test statistics are not independent: each later statistic includes information already used at earlier looks. A joint density describes the probabilities of their possible values together. It is therefore the mathematical object needed to reason about a complete vector of statistics, rather than treating each look as an unrelated test.

This tutorial uses the canonical joint normal model and the nested-look covariance established in the earlier material on the canonical joint normal model and covariance from shared data. The aim here is to write the density explicitly, understand its parts, and evaluate it without confusing a density value with a probability.

Model: Let \(k\geq 3\) be the number of planned looks, and let \(0<t_1<t_2<\cdots<t_k=1\) be their fixed, normalized information times. Under the canonical joint normal model, the vector \(\mathbf Z=(Z_1,\ldots,Z_k)^\mathsf{T}\) has distribution \(N_k(\boldsymbol{\eta},\Sigma)\), where \(\eta_i=\delta\sqrt{t_i}\) and \(\Sigma_{ij}=\sqrt{t_i/t_j}\) if \(i\leq j\), with \(\Sigma_{ij}=\Sigma_{ji}\) if \(i>j\). Here \(\delta\) is the standardized effect on the canonical scale; under \(H_0\), \(\delta=0\).

The model assumes that the looks use nested data, that the information times are fixed and strictly increasing, and that the standardized statistics have the stated joint normal distribution. For normal observations with known variance and the corresponding cumulative statistics, this model can be exact. In settings where the statistics are only approximately jointly normal, conclusions based on this density are approximate as well. The density is not justified merely by having several reported Z statistics: the shared-data covariance structure must also be appropriate.

Writing the Density and Reading Its Components

For a possible observed vector \(\mathbf z=(z_1,\ldots,z_k)^\mathsf{T}\), the multivariate normal density is

f(\mathbf z)=\frac{1}{(2\pi)^{k/2}|\Sigma|^{1/2}} \exp\left\{-\frac{1}{2}(\mathbf z-\boldsymbol{\eta})^\mathsf{T} \Sigma^{-1}(\mathbf z-\boldsymbol{\eta})\right\}.

The determinant \(|\Sigma|\) and inverse \(\Sigma^{-1}\) both depend on the look times. The factor \((2\pi)^{-k/2}|\Sigma|^{-1/2}\) normalizes the density over all possible vectors. The exponent contains the squared Mahalanobis distance from \(\mathbf z\) to the mean vector: it measures discrepancy while accounting for both the scales and the correlations. Thus, the exponent is not generally the sum of the squared residuals divided by separate variances.

The matrix entries have a direct interpretation. Each diagonal entry is \(\Sigma_{ii}=1\), because each \(Z_i\) is standardized. For \(i<j\), the correlation between the statistics at looks \(i\) and \(j\) is \(\sqrt{t_i/t_j}\). It is positive and less than one: the statistics share earlier data, but the later look also contains additional information.

Worked Example: Assembling a Three-Look Canonical Model

Suppose a hypothetical design has information times \(t_1=0.25\), \(t_2=0.60\), and \(t_3=1\). Write its canonical mean and covariance matrices, and identify the pieces that enter the density.

For a fixed standardized effect \(\delta\), the mean vector is \(\boldsymbol{\eta}=(0.5\delta,\sqrt{0.60}\delta,\delta)^\mathsf{T}\). The pairwise correlations are \(\Sigma_{12}=\sqrt{0.25/0.60}=0.6455\), \(\Sigma_{13}=\sqrt{0.25/1}=0.5000\), and \(\Sigma_{23}=\sqrt{0.60/1}=0.7746\). Therefore,

\Sigma= \begin{pmatrix} 1&0.6455&0.5000\\ 0.6455&1&0.7746\\ 0.5000&0.7746&1 \end{pmatrix}.

For \(k=3\), the density has the normalizing factor \((2\pi)^{-3/2}|\Sigma|^{-1/2}\) and the exponential term \(\exp\{-\tfrac12(\mathbf z-\boldsymbol{\eta})^\mathsf{T}\Sigma^{-1} (\mathbf z-\boldsymbol{\eta})\}\). Under \(H_0\), set \(\delta=0\), so the mean vector is zero; the covariance matrix remains the same. The matrix is not the identity, so replacing the joint density by a product of three independent standard normal densities would be incorrect.

The Determinant from the Information Times

The determinant is a key part of the density's normalization. For the canonical covariance matrix it has a convenient expression. Put \(t_0=0\). Then

|\Sigma|=\prod_{j=1}^{k}\left(1-\frac{t_{j-1}}{t_j}\right).

A short justification follows from the shared-data representation. Define \(W_j=\sqrt{t_j}Z_j\). The canonical covariance gives \(\operatorname{Cov}(W_i,W_j)=\min(t_i,t_j)\). Thus the increments \(W_j-W_{j-1}\), with \(W_0=0\), have variances \(t_j-t_{j-1}\) and are independent under the canonical Brownian representation. The linear transformation from \((W_1,\ldots,W_k)\) to these increments is triangular with determinant one, so the determinant of the covariance matrix of \(\mathbf W\) is \(\prod_j(t_j-t_{j-1})\). Since \(W_j=\sqrt{t_j}Z_j\), its covariance determinant is also \((\prod_j t_j)|\Sigma|\). Dividing gives the stated formula.

Every factor is positive when the information times strictly increase, so \(|\Sigma|>0\). If a look is repeated, one factor becomes zero and the covariance matrix is singular: the repeated statistic contributes no new random dimension, and the ordinary \(k\)-dimensional density formula does not apply.

Worked Example: Evaluating a Density at Three Looks

Continue with the hypothetical times \(0.25,0.60,1\). Suppose \(\delta=-0.2\), giving \(\boldsymbol{\eta}=(-0.1000,-0.1549,-0.2000)^\mathsf{T}\). Consider the hypothetical statistic vector \(\mathbf z=(0.2000,-0.2549,0.3000)^\mathsf{T}\), constructed so that its residual vector is \(\mathbf r=\mathbf z-\boldsymbol{\eta}=(0.3000,-0.1000,0.5000)^\mathsf{T}\).

First, the determinant is \(|\Sigma|=(1-0.25/0.60)(1-0.60/1) =(0.5833)(0.4000)=0.2333\). For this three-look matrix, the inverse, rounded for display, is

\Sigma^{-1}\approx \begin{pmatrix} 1.7143&-1.1066&0\\ -1.1066&3.2143&-1.9365\\ 0&-1.9365&2.5000 \end{pmatrix}.

Multiplying the residuals through this matrix gives the quadratic form \[ \mathbf r^\mathsf{T}\Sigma^{-1}\mathbf r \approx (1.7143)(0.3000)^2+(3.2143)(-0.1000)^2 +(2.5000)(0.5000)^2 +2(-1.1066)(0.3000)(-0.1000) +2(-1.9365)(-0.1000)(0.5000) =1.0715. \] The density is consequently

f(\mathbf z)\approx (2\pi)^{-3/2}(0.2333)^{-1/2}\exp(-1.0715/2) \approx 0.07693.

This is a density value at the specified vector, not the probability of observing that exact vector. A probability is obtained by integrating the density over a region of possible statistic vectors.

Extending the Calculation to Four Looks

The same definitions apply for any \(k\); there is no new form of density when a fourth or later look is added. The covariance matrix has one row and column per look, its entries come from the information-time ratios, and the determinant product includes one factor per look. The quadratic form uses the full inverse covariance matrix.

Worked Example: Four Looks and the Density at the Origin

Consider hypothetical information times \(t=(0.20,0.40,0.70,1)\) under \(H_0\), so \(\boldsymbol{\eta}=\mathbf 0\). The correlations above the diagonal are \(\sqrt{0.20/0.40}=0.7071\), \(\sqrt{0.20/0.70}=0.5345\), \(\sqrt{0.20}=0.4472\), \(\sqrt{0.40/0.70}=0.7559\), \(\sqrt{0.40}=0.6325\), and \(\sqrt{0.70}=0.8367\). Together with diagonal entries one, these values specify the full \(4\times4\) canonical covariance matrix.

At the origin, \(\mathbf z=\mathbf 0\), the quadratic form is zero. The determinant is \[ |\Sigma|=(1-0)(1-0.20/0.40)(1-0.40/0.70)(1-0.70/1) =(1)(0.5000)(0.4286)(0.3000)=0.06429. \] Therefore, \[ f(\mathbf 0)=(2\pi)^{-2}(0.06429)^{-1/2} \approx (0.02533)(3.944)=0.09990. \] This value is the joint density per unit in four-dimensional statistic space at the zero vector. It is not a probability that all four statistics equal zero.

Checks, Alternatives, and Limitations

A reliable calculation separates the model check from the numerical evaluation. First verify that the times are positive and ordered, then check that the diagonal entries are one and that each off-diagonal entry is the square root of the earlier-to-later information ratio. The matrix must be symmetric and positive definite when the times are strictly increasing. Finally, verify that the determinant is positive and that the mean vector uses the same time ordering as the covariance matrix.

A direct multivariate normal calculation using a matrix inverse is the general approach. For hand calculations, the determinant product is more convenient than expanding a large determinant. Another possible representation uses independent increments of the underlying cumulative process; it can make later calculations more efficient, but it must preserve the same joint distribution and is not permission to treat the \(Z_i\) themselves as independent. Separate one-look normal densities are appropriate only when the statistics are independent, which nested canonical statistics are not.

Common warning signs include a determinant of zero for distinct, increasing times, correlations exceeding one, or a negative quadratic form. Such results usually signal a misordered time ratio, a transcription error, or an invalid covariance matrix; the quadratic form for a positive-definite \(\Sigma\) cannot be negative. The canonical density also has limits: if information times are random or the statistics are not adequately approximated by the canonical joint normal model, this density may not describe their joint behavior accurately. A numerical density calculation cannot repair a misspecified model.

Common mistake: Using \(\sqrt{t_j/t_i}\) for the correlation when \(i<j\), or multiplying marginal standard normal densities as if the looks were independent. The correct correlation is \(\sqrt{t_i/t_j}\), which is less than one, and the joint density uses the full canonical covariance matrix. Also, a density value is not the probability of an exact observed vector.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

For three looks at \(t=(0.25,0.50,1)\), what are \(\Sigma_{12}\), \(\Sigma_{23}\), and \(\Sigma_{13}\)?
For four strictly increasing information times, why is the canonical covariance determinant positive?
Under \(H_0\), what is the mean vector, and which parts of the density still depend on the look times?
What does a density value at a single vector represent, and why is it not the probability of observing exactly that vector?

Solutions. (1) The correlations are \(\sqrt{0.25/0.50}=0.7071\), \(\sqrt{0.50/1}=0.7071\), and \(\sqrt{0.25/1}=0.5000\). (2) The determinant is \(\prod_{j=1}^4(1-t_{j-1}/t_j)\), with \(t_0=0\); each factor is positive under strict increase, so their product is positive. (3) The mean is the zero vector. The covariance matrix, its determinant and inverse, and hence both the normalizing factor and the quadratic form generally depend on the look times. (4) It is density per unit volume at that point in the vector space. A point has zero volume, so its probability is zero for a continuous distribution; probabilities require integration over a region.