advanced39 min studyLesson 867 in Mathematics

Likelihood and innovation algorithms

Factor a finite Gaussian likelihood through sequential prediction errors, keeping its initial distribution and estimated mean explicit.

On this page

What you’ll understand

  • Derive the innovation factorization of a finite Gaussian likelihood.
  • Distinguish exact stationary initialization from a conditional likelihood.
  • Compute prediction-error variances and likelihood contributions for short records.

A Gaussian likelihood evaluates both the size of prediction errors and the uncertainty attached to them. Errors from an early observation can have a different variance from errors after a long history, even under a stationary model. Innovation algorithms organize that finite-history information without repeatedly inverting a large covariance matrix. They also reveal why a conditional sum of squared residuals, a stationary exact likelihood and a Gaussian approximation to non-Gaussian data are different statistical objects.

Sequential projections factor the covariance

Consider a jointly Gaussian observation vector with constant mean mu and a positive definite covariance matrix Gamma depending on model parameters. Sequentially predict each centered observation from those preceding it. The residuals are orthogonal, and joint Gaussianity makes them independent. Their variances are positive. Algebraically this is a lower triangular factorization Gamma=L D L-transpose, where L has unit diagonal and D contains the innovation variances; the residual vector is L-inverse times the centered observations.

The determinant of L is one, so the covariance determinant is the product of those variances. The inverse-covariance quadratic form becomes the sum of each squared residual divided by its variance. This factorization produces exactly the multivariate Gaussian density, using the mean and covariance of the entire finite vector. Natural logarithms appear in its log likelihood; the two pi normalization belongs to a normal density, not to the earlier choice of Fourier frequency units.

Γ=LDLT,e=L−1(y−μ1),D=diag⁡(v1,…,vn),ℓ=−n2ln⁡(2π)−12∑t=1n(ln⁡vt+et2vt)\Gamma=LDL^\mathsf T,\quad e=L^{-1}(y-\mu\mathbf1),\quad D=\operatorname{diag}(v_1,\ldots,v_n),\qquad\ell=-\frac n2\ln(2\pi)-\frac12\sum_{t=1}^n\left(\ln v_t+\frac{e_t^2}{v_t}\right)
The covariance must be positive definite. For jointly Gaussian observations the sequential projection residuals are independent normal variables, making this an exact finite-sample likelihood.↔ Scroll wide equations horizontally.

Initialization specifies a different probability model

For a stationary AR(1), the first observation has variance sigma-squared divided by one minus phi-squared. Later observations have conditional variance sigma-squared and conditional mean mu plus phi times the previous centered value. An exact stationary likelihood includes the first density, whose variance depends on the parameters. A likelihood conditioned on the first observed value drops that factor. Neither convention should be silently substituted for the other during a finite-sample comparison.

For moving averages, setting unobserved presample shocks to zero creates a convenient conditional residual recursion. Those zero values are not the random presample shocks integrated out by the stationary finite-vector density. Invertibility can make initialization effects decay and support asymptotic approximations under further regularity, but it does not make the two finite likelihoods identical. A declared mean must also remain consistent: an AR intercept equals phi of one times mu, rather than generally equaling mu itself.

Xt−μ=ϕ(Xt−1−μ)+εt,∣ϕ∣<1:v1=σ21−ϕ2,e1=y1−μ,vt=σ2,et=yt−μ−ϕ(yt−1−μ) (t≥2)X_t-\mu=\phi(X_{t-1}-\mu)+\varepsilon_t,\quad |\phi|<1:\qquad v_1=\frac{\sigma^2}{1-\phi^2},\quad e_1=y_1-\mu,\quad v_t=\sigma^2,\quad e_t=y_t-\mu-\phi(y_{t-1}-\mu)\ (t\ge2)
These exact innovation quantities use independent Gaussian innovations and stationary initialization. Conditioning on y-one removes its density rather than changing the later conditional errors.↔ Scroll wide equations horizontally.

Likelihood accuracy depends on more than an algorithm

If a series is not jointly Gaussian, orthogonal linear innovations need not be independent, and a product of their normal densities is generally not its exact distribution. It can instead define a Gaussian quasi-likelihood criterion. Consistency or standard errors for that criterion require assumptions beyond the second-moment identities. Similarly, an estimated innovation sequence looking uncorrelated does not prove independence, correct conditional variance or a correct tail model.

Finite innovation recursions also distinguish available observations from the infinite history in Wold's theorem. Their earliest variances incorporate the missing past. Positive definiteness guarantees mathematical divisions are possible, but a covariance near singularity can still cause numerical difficulty. Practical fitting should document mean handling, initialization, parameter constraints and convergence, then examine residual behavior and forecast performance. An optimization routine returning a number is not evidence that every statistical assumption used to interpret that number holds.

Retain the first stationary observation

For a zero-mean stationary Gaussian AR(1) with phi=0.5 and innovation variance one, evaluate the exact likelihood contributions for y=(2,1,0).

  1. The first prediction is zero with variance 4/3, so e-one=2 and its standardized squared contribution is 4/(4/3)=3.

  2. The next prediction is one with variance one, giving e-two=0. The third prediction is one half with variance one, giving e-three=−1/2 and squared contribution 1/4.

  3. The total quadratic term is 13/4 and the covariance log determinant is ln(4/3). Therefore the exact log likelihood is −(3/2)ln(2pi)−[ln(4/3)+13/4]/2.

The result

Conditioning on the first value instead gives −ln(2pi)−1/8. The omitted initial density is parameter dependent and cannot be treated as a universal constant.

Explore the model

Interactive lab

Interactive lab · matrix

Read a Gaussian conditional variance from a factor

Pause and reset, then set both controls to a = 1 and b = 0.5. For independent standard normal Z, the selected map gives Y-one=Z-one+0.5Z-two and Y-two=Z-two, so predict conditional mean 0.5Y-two and conditional variance one for Y-one given Y-two. For a fresh comparison, pause and reset, then set a = 0.5 and b = 0.5; that variance becomes 0.25. Covariance determinant is a², while the instrument reports determinant a for A. This illustrates Gaussian conditioning, not a full temporal innovation solver.

Read a Gaussian conditional variance from a factorThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. 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 forA triangular factor separates predictable dependence from a remaining conditional variance.

Check your understanding

Which statement correctly compares exact stationary and conditional ARMA likelihoods?

Try it yourself

A zero-mean Gaussian MA(1) has theta=0.5 and unit innovation variance. For observations (1,1), find the two innovation variances, the second prediction, covariance determinant and standardized squared-error sum.

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

Give me a hint

The two-by-two covariance has diagonal 5/4 and off-diagonal 1/2. Project the second observation on the first.

Compare with the solution

The first variance is 5/4. The second prediction coefficient is (1/2)/(5/4)=2/5, so the prediction is 2/5 and residual 3/5. Its variance is 5/4−(1/2)²/(5/4)=21/20. The determinant is (5/4)(21/20)=21/16. The quadratic sum is 1/(5/4)+(3/5)²/(21/20)=4/5+12/35=8/7. A zero-presample-shock recursion gives residuals one and one half, whose unweighted squares total 5/4; that is a different initialization criterion.

Take it with you

  • Gaussian innovations factor the exact finite covariance determinant and quadratic form.
  • Stationary and conditional initialization give different finite-sample likelihoods.
  • A Gaussian quasi-likelihood for other distributions requires separate inferential assumptions.

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