Two different density distributions inside the Earth produce exactly the same gravity field at every point on the surface. Which of Hadamard’s conditions for a well-posed problem does this violate?

Continuous dependence of the solution on the data

Linearity of the forward operator in the model parameters

A linear system G·m = d has condition number κ = 10⁶, and the data d are accurate to a relative error of 10⁻⁴. What is the largest relative error that the solution m can have?

As much as about 10², so the solution can be worthless

About 10⁻¹⁰, because the error is divided by the condition number

About 10⁻⁴, the same as the data error, because solving the system averages out the noise

As much as about 10², so the solution can be worthless

About 10⁻², because only the square root of κ matters for the error

A linear problem has 6 data and 6 unknowns, but the matrix G has rank 4. How is it classified?

Mixed-determined: two parameter combinations are unconstrained and two data combinations cannot be fitted

Even-determined, because the number of data equals the number of unknowns

Over-determined, because the 6 data exceed the rank of 4 and so no exact fit exists

Under-determined, because the rank is below the number of unknowns and there are infinitely many exact fits

In Tikhonov (damped least-squares) inversion, what is the effect of increasing the damping parameter λ?

The variance of the solution falls, but the bias grows and the resolution worsens, so the misfit increases

The data misfit falls to zero and the model resolution improves, because the solution is forced closer to the data

Only the null-space components of the model are changed, while the well-determined components stay as in least squares

The condition number of G itself decreases, because damping changes its singular values permanently

The variance of the solution falls, but the bias grows and the resolution worsens, so the misfit increases

In simulated annealing with the Metropolis rule, a trial model that raises the misfit by ΔE is accepted with probability exp(−ΔE/T). Why is the temperature T lowered slowly during the search?

Because early uphill moves let the search escape local minima, and cooling too fast would freeze it in one

Because a slow fall keeps every accepted model exactly on the contour of the lowest misfit found so far, which convergence requires

Because early uphill moves let the search escape local minima, and cooling too fast would freeze it in one

Because the derivative of the misfit is needed at each step and it is only defined at low temperature

Because a lower temperature makes uphill moves more likely, which steadily improves the final fit

Which pairing of a geophysical quantity with the probability distribution normally used to model it is most appropriate?

Gamma photons counted in a fixed time window at constant logging speed: a Poisson distribution

Rock permeability, which is positive and spans several orders of magnitude: a Gaussian distribution

Random picking errors of arrival times, which are symmetric about zero: a Poisson distribution

Gamma photons counted in a fixed time window at constant logging speed: a Poisson distribution

Ore grade, which is positive and right-skewed with a long tail: a Poisson distribution

With a Gaussian likelihood, which choice of prior makes the maximum a posteriori (MAP) estimate equal to the Tikhonov (L2) solution, and which prior leads to sparsity-promoting L1 regularisation?

A zero-mean Gaussian prior gives the L2 solution, with λ² = σ_d²/σₘ², and a Laplace prior gives the L1 solution

A Laplace prior gives the L2 solution and a zero-mean Gaussian prior gives the L1 solution

A uniform prior gives the L2 solution and a log-normal prior gives the L1 solution

A Poisson prior gives the L2 solution and a Gaussian prior with a very large variance gives the L1 solution, as in the limit of heavy damping

A zero-mean Gaussian prior gives the L2 solution, with λ² = σ_d²/σₘ², and a Laplace prior gives the L1 solution

A misfit J(m) = ½‖P·u − d‖² depends on a field u that solves a wave equation A(m)·u = s, and the model m has one million parameters. For a single source, how does the adjoint-state method obtain the gradient ∂J/∂m with respect to all the parameters?

From one forward solve and one adjoint solve, driven by the data residual at the receivers, whatever the number of parameters

From one million and one forward solves, perturbing the parameters one at a time by a small finite amount and differencing the misfit

From one million adjoint solves, one for each parameter, each driven by a unit source placed at that parameter’s location

From a single forward solve alone, because the adjoint field is equal to the forward field and needs no separate calculation

Which of the following statements about non-linear inversion methods are correct?

A genetic algorithm evolves a population of models by selection, crossover and mutation and needs no derivatives of the forward operator

In particle swarm optimisation each particle’s velocity combines its inertia, attraction to its own best position and attraction to the swarm’s best position

In simulated annealing an uphill move is accepted with probability exp(−ΔE/T), which falls as the temperature T is lowered

A genetic algorithm evolves a population of models by selection, crossover and mutation and needs no derivatives of the forward operator

In particle swarm optimisation each particle’s velocity combines its inertia, attraction to its own best position and attraction to the swarm’s best position

A damped Gauss–Newton (Levenberg–Marquardt) iteration is guaranteed to reach the global minimum from any starting model

In simulated annealing an uphill move is accepted with probability exp(−ΔE/T), which falls as the temperature T is lowered

Which of the following statements about distributions and misfit statistics are correct?

For a Poisson variable with mean λ the variance is also λ, so the relative scatter of a count falls as 1/√λ

If ln X is Gaussian then X is log-normal, and its mean lies above its median

For a Poisson variable with mean λ the variance is also λ, so the relative scatter of a count falls as 1/√λ

If ln X is Gaussian then X is log-normal, and its mean lies above its median

Independent Gaussian errors make the L1 misfit the maximum-likelihood criterion

A reduced chi-square far above 1 shows that the data are fitted better than their stated errors allow

Which of the following statements about Bayesian inversion, sparsity and ambiguity are correct?

Ambiguity can persist even for perfect noise-free data, whereas uncertainty from noise shrinks as the data errors shrink

An L1 penalty λ‖m‖₁ promotes sparse models because the L1 ball has corners on the coordinate axes

Ambiguity can persist even for perfect noise-free data, whereas uncertainty from noise shrinks as the data errors shrink

An L1 penalty λ‖m‖₁ promotes sparse models because the L1 ball has corners on the coordinate axes

The posterior equals the likelihood alone, because the prior only rescales it by a constant

A model that fits the data to within the noise is thereby proved to be the true structure of the Earth

Geophysical Inversion: Ill-posedness, Norms and Regularisation, Global Search, Statistics, Bayesian and Sparse Methods, and the Adjoint-State Method

This chapter of Part B2 of the Geology and Geophysics (GG) paper answers the heading “Geophysical Inversion”: “Basic concepts of forward and inverse problems, Ill-posedness of inverse problems, condition number, non-uniqueness and stability of solutions; L1, L2, and Lp norms, overdetermined, underdetermined and mixed determined inverse problems, quasi-linear and non-linear methods including Tikhonov’s regularization method, Singular Value Decomposition, Backus-Gilbert method, simulated annealing, genetic algorithms, swarm intelligence, Statistics of misfit and likelihood, Random Variables and Processes, Probability distribution functions - Gaussian, Poissonian, log-normal; Spatial analysis, Hypothesis testing; Bayesian construction of posterior probabilities, sparsity promoting L1 optimization; Ambiguity and uncertainty in geophysical interpretation; PDE constrained optimization, Adjoint-state method.” The numericals are a small least-squares system, condition number, the Tikhonov solution, residual norms, chi-square, a z-test and a Bayesian posterior.

1. Forward and inverse problems, ill-posedness, condition number and non-uniqueness

The forward problem predicts data from a model: d = g(m), where m holds density, velocity or resistivity values and g is the physics (for a linear case d = G·m, with N data and M parameters). The inverse problem estimates m from the observed data. Forward problems are usually well-posed, but inverse problems are not. Hadamard called a problem well-posed if a solution exists (noise can put real data outside the range of G), is unique, and depends continuously on the data.

Non-uniqueness is physical: outside a body, gravity depends only on its total mass for a spherically symmetric distribution, so infinitely many density models fit equally, and a shallow small body and a deeper larger one can give the same anomaly. Instability comes from smoothing forward operators (diffusion, upward continuation): they damp fine detail, so the inverse amplifies noise at fine scales. The condition number κ = σ_max/σ_min (ratio of the largest to the smallest singular value of G) measures this: the relative error in m can be as large as κ times the relative error in d, and about log₁₀κ digits are lost. Worked example: singular values 10, 4 and 0.01 give κ = 1000, so data accurate to 0.1% can leave the model uncertain by 100%.

ℹ️ How the ill-posedness is treated

A problem is made usable by adding information: a prior model, smoothness or positivity constraints, regularisation, more data or joint inversion. Each is an assumption that selects one answer from the many that fit, so state it.

2. L1, L2 and Lp norms; over-, under- and mixed-determined problems

The Lp norm of a residual vector e is ‖e‖ₚ = (Σ|eᵢ|ᵖ)^(1/p). L1 sums the absolute values, L2 is the Euclidean length, and L∞ is the largest |eᵢ|. For e = (1, −2, 2, 0): L1 = 5, L2 = √9 = 3, L∞ = 2, and always L∞ ≤ L2 ≤ L1. L2 gives the closed-form least-squares solution and is the maximum-likelihood choice for Gaussian errors, but squaring lets outliers dominate. L1 is robust (the estimate of a constant is the median, not the mean: for 1, 2, 3, 100 the mean is 26.5, while any value from 2 to 3 minimises the L1 misfit) and suits long-tailed errors, but needs iterative solution. L∞ is the minimax fit.

Classification uses N data, M unknowns and the rank p of G. Over-determined (N > M, p = M): no exact fit; minimise ‖d − G·m‖₂ to get the normal equations GᵀG·m = Gᵀd. Worked example: a line m₁ + m₂x through (0, 1), (1, 2), (2, 2) has G = [[1, 0], [1, 1], [1, 2]], GᵀG = [[3, 3], [3, 5]], Gᵀd = (5, 6), so m = (7/6, 1/2): intercept 1.17, slope 0.5. Under-determined (M > N, p = N): infinitely many exact fits, so a criterion is added, such as the minimum-norm solution m = Gᵀ(GGᵀ)⁻¹d. Mixed-determined (p < N and p < M): some parameter combinations are unconstrained and some data combinations cannot be fitted, and the solution needs damping or SVD.

⚠️ Counting equations is not enough

A system with as many data as unknowns can still be mixed-determined if its rank is lower, for instance when two measurements sample the same combination of parameters. Always ask for the rank, not the shape of G.

3. Singular value decomposition, Tikhonov regularisation and the Backus–Gilbert method

The SVD writes G = U·Σ·Vᵀ, with singular values σ₁ ≥ σ₂ ≥ … ≥ σₚ > 0, data-space vectors uᵢ and model-space vectors vᵢ. The generalised inverse gives m = Σ (uᵢᵀd/σᵢ)·vᵢ. Model directions with σ = 0 form the null space: the data are blind to them, which is non-uniqueness. Directions with small σᵢ amplify noise by 1/σᵢ. Truncated SVD keeps only the largest singular values. The model resolution matrix R = Vₚ·Vₚᵀ shows how blurred the estimate is (R = I means perfect resolution), and κ = σ₁/σₚ.

Tikhonov regularisation minimises ‖G·m − d‖² + λ²‖m‖², giving the damped solution m = (GᵀG + λ²I)⁻¹Gᵀd. In SVD form each component is multiplied by the filter factor f = σ²/(σ² + λ²), close to 1 where σ ≫ λ and close to 0 where σ ≪ λ, so poorly determined directions fade out smoothly. Replacing ‖m‖² by ‖L·m‖² (L a difference operator) penalises roughness. The parameter λ trades variance (noise amplification) against bias (loss of resolution) and is chosen by the L-curve, the discrepancy principle (misfit ≈ noise level) or cross-validation. Worked example: G = 0.1, d = 0.5 gives m = 5 by least squares, but m = σd/(σ² + λ²) = 2.5 with λ = 0.1; a data change of 0.01 moves the first by 0.1 and the damped one by only 0.05.

The Backus–Gilbert method, developed for Earth-structure problems, does not solve for every parameter. For a chosen point it builds the combination of data whose averaging (resolution) kernel A(x₀, x) is as narrow and delta-like as possible, so the estimate is a localised average of the true model. A trade-off parameter balances the kernel’s narrowness against the variance of the estimate.

🎯 Damping and truncation do the same job

Both discard directions the data cannot determine, whose 1/σ factors would multiply the noise: truncated SVD cuts them sharply, Tikhonov damping fades them smoothly. The smallest possible misfit is a wrong goal, since it fits the noise.

4. Quasi-linear and non-linear methods: simulated annealing, genetic algorithms and swarms

When d = g(m) is non-linear, the quasi-linear approach linearises about the current model mₖ with the Jacobian J = ∂g/∂m and updates mₖ₊₁ = mₖ + (JᵀJ + λ²I)⁻¹Jᵀ[d_obs − g(mₖ)], the damped Gauss–Newton or Levenberg–Marquardt step. It is fast, but it converges only to a minimum near the starting model, and a misfit surface with many local minima traps it.

Global search methods need no derivatives, only many forward calculations. Simulated annealing copies the slow cooling of a metal: a trial model m′ is proposed, ΔE = E(m′) − E(m) is computed, and the move is accepted if ΔE ≤ 0, or otherwise with the Metropolis probability P = exp(−ΔE/T). A high temperature T explores freely, a low one descends. The worked example ΔE = 1 gives P = e^(−0.5) = 0.61 at T = 2 but e^(−5) = 0.007 at T = 0.2. Cooling too fast freezes the search in a local minimum.

A genetic algorithm evolves a population of coded models: fitness (low misfit) decides which parents are selected, crossover swaps parts of their codes, mutation randomly changes a few entries to keep diversity, and the best may be copied unchanged (elitism). Fitness evaluations are independent, so they run in parallel.
Swarm intelligence (particle swarm optimisation) sends particles, each a model with a velocity, through the model space: v ← w·v + c₁r₁(p_best − x) + c₂r₂(g_best − x), x ← x + v, combining inertia, the particle’s own best position and the best found by the swarm (r₁, r₂ are random numbers between 0 and 1).

⚠️ Global does not mean cheap or exact

These methods need 10⁴ to 10⁶ forward runs and guarantee no global minimum. They suit tens or hundreds of parameters, not millions, where adjoint gradients are used. Their by-product is an ensemble of acceptable models that shows uncertainty.

5. Random variables, distributions, misfit and likelihood

A random variable X has a probability density f(x), mean μ and variance σ². The Gaussian N(μ, σ²) has f = [1/(σ√(2π))]·exp[−(x − μ)²/(2σ²)], with 68.3%, 95.4% and 99.7% of values within 1σ, 2σ and 3σ. The central limit theorem (sums of many small independent errors tend to a Gaussian) justifies Gaussian noise. The Poisson law P(k) = λᵏe^(−λ)/k! governs counts in a fixed interval, such as radioactive decays or gamma-ray counts; its mean and variance both equal λ, so the relative scatter is 1/√λ. A variable is log-normal if ln X is Gaussian: positive, right-skewed, with median e^μ and mean e^(μ + σ²/2) above it, as for permeability, grain size or ore grade, which are products of many factors.

A random process is a family of random variables indexed by time or position. It is stationary if its statistics do not change under a shift, and ergodic if averages along one long record equal ensemble averages. The autocovariance and power spectrum are a Fourier pair (Wiener–Khinchin); white noise is uncorrelated.

For residuals rᵢ = dᵢ − gᵢ(m) with independent Gaussian errors σᵢ, the likelihood is L(m) ∝ exp(−χ²/2), where the chi-square misfit is χ² = Σ(rᵢ/σᵢ)². Maximum likelihood is therefore weighted least squares (Laplace errors would give L1). For a good model with correct error estimates, χ² ≈ ν = N − M, so the reduced χ²/ν ≈ 1; a value far above 1 means a poor model or underestimated errors, and far below 1 means overestimated errors or fitting of the noise. Worked example: N = 10, M = 2, σ = 0.5 and a residual sum of squares 3.0 give χ² = 3.0/0.25 = 12, ν = 8 and χ²/ν = 1.5. For independent noise σ, the covariance of the least-squares estimate is σ²(GᵀG)⁻¹.

6. Spatial analysis and hypothesis testing

Geophysical values at nearby places are correlated. The semivariogram γ(h) = ½E[(Z(x + h) − Z(x))²] rises with separation h from the nugget (the jump at h → 0 from measurement error and fine-scale variability) to the sill (about the variance) at the range, beyond which values are uncorrelated; for a stationary field the covariance is C(h) = sill − γ(h). Kriging is the best linear unbiased estimator built from the variogram: its weights sum to 1 in ordinary kriging, it accounts for clustering of stations, and it returns an estimation variance as well as the estimate.

In hypothesis testing a null hypothesis H₀ is tested against H₁ using a test statistic. The p-value is the probability, if H₀ were true, of a result at least as extreme as that observed, and H₀ is rejected if p < α, the significance level. A type I error (rejecting a true H₀, a false alarm) has probability α, a type II error (missing a real effect) has probability β, and the power is 1 − β. For a mean with known σ, z = (x̄ − μ₀)/(σ/√n). Worked example: x̄ = 12.6, μ₀ = 12, σ = 3, n = 25 give z = 0.6/0.6 = 1, below the two-sided 5% limit 1.96, so H₀ is not rejected. The χ² test judges whether a misfit fits the noise, the F-test whether a more complex model is significantly better.

⚠️ Not rejecting is not proving

A large p-value says only that the data do not contradict H₀; with few data a real effect is easily missed. A small p-value says chance is unlikely, not that the effect is large or important.

7. Bayesian posteriors, sparsity-promoting L1 optimisation, ambiguity and uncertainty

Bayes’ theorem gives the posterior p(m|d) = p(d|m)·p(m)/p(d): likelihood times prior, normalised by the evidence. The answer to the inverse problem is the whole posterior; it is summarised by the most probable model (MAP), the mean and credible intervals, and sampled by Markov-chain Monte Carlo when it is not Gaussian. For one parameter with prior N(m₀, σₘ²) and a datum d = m + noise of standard deviation σ_d, the precisions (1/variance) add: 1/σ_post² = 1/σₘ² + 1/σ_d², and the mean is the precision-weighted average of m₀ and d. Worked example: m₀ = 5, σₘ = 2, d = 9, σ_d = 1 give precision 0.25 + 1 = 1.25, σ_post² = 0.8 and mean (5 × 0.25 + 9 × 1)/1.25 = 8.2, pulled towards the more certain datum.

This links regularisation to prior information. With a Gaussian likelihood, a zero-mean Gaussian prior makes the MAP estimate the Tikhonov solution with λ² = σ_d²/σₘ², and a Laplace (double-exponential) prior gives L1 regularisation. Sparsity-promoting L1 optimisation minimises ½‖G·m − d‖² + λ‖m‖₁. The L1 ball has corners on the coordinate axes, so the optimum tends to sit exactly on an axis, giving true zeros, while L2 shrinks every component but rarely zeroes one. With G = I the solution is the soft threshold mᵢ = sign(dᵢ)·max(|dᵢ| − λ, 0). Uses: sparse-spike deconvolution, compressive interpolation of seismic data, total-variation imaging of blocky models; the model must really be sparse in some basis.

Ambiguity is non-uniqueness that survives perfect, noise-free data: the depth–density trade-off in gravity, or equivalence in resistivity soundings, where a thin conductive layer is fixed only by its conductance S = h/ρ and a thin resistive layer only by its transverse resistance T = h·ρ, so h and ρ cannot be separated. Uncertainty is the spread of acceptable models caused by noise and limited sampling, and it shrinks as the data improve. Both are appraised through the resolution matrix, the covariance matrix or posterior samples; ambiguity is reduced by independent constraints (boreholes, geology) and joint inversion. A good fit is necessary for a model to be right, never sufficient.

8. PDE-constrained optimisation and the adjoint-state method

Many forward problems are solutions of a partial differential equation A(m)·u = s: the wave, diffusion, Laplace or Maxwell equation, with field u (the state), parameters m (velocity, conductivity), and source s. The data are samples of the field at receivers, d = P·u. The misfit J(m) = ½‖P·u − d‖² (plus regularisation) is to be minimised subject to the PDE as a constraint: PDE-constrained optimisation. A gradient method needs ∂J/∂m for perhaps 10⁶ parameters. Finite differences would need M + 1 forward solves per gradient, which is impossible.

The adjoint-state method avoids this. Form the Lagrangian ℒ = ½‖P·u − d‖² + ψᵀ(A(m)·u − s). Setting its derivative with respect to u to zero gives the adjoint equation Aᵀψ = −Pᵀ(P·u − d): the same kind of PDE with the transposed operator, driven by the data residual injected at the receivers. Then the gradient for all parameters at once is ∂J/∂m = ψᵀ(∂A/∂m)·u. The cost is one forward and one adjoint solve per source, independent of M, so S sources need 2S solves.

In seismic full-waveform inversion the adjoint field is the residual injected at the receivers and propagated backwards in time, and the gradient is the zero-lag correlation, summed over time, of the forward and adjoint wavefields. The forward field must be stored or recomputed (checkpointing) because the two run in opposite time directions. If the starting model predicts arrivals more than half a period off, the misfit has local minima (cycle skipping), so inversion starts from low frequencies and moves up.

Key takeaways

Inverse problems violate Hadamard’s existence, uniqueness or stability; κ = σ_max/σ_min bounds noise amplification, and non-uniqueness (null space, gravity depth–density trade-off) is physical, so added information is unavoidable.
L2 is least squares (Gaussian errors, outlier-sensitive), L1 is robust (median-like), and L∞ ≤ L2 ≤ L1. Classify by rank: over-determined (least squares), under-determined (minimum norm), mixed-determined (damping or SVD).
SVD gives m = Σ(uᵢᵀd/σᵢ)vᵢ; Tikhonov m = (GᵀG + λ²I)⁻¹Gᵀd multiplies each term by σ²/(σ² + λ²), trading variance against resolution; Backus–Gilbert builds a localised average with the narrowest averaging kernel.
Non-linear problems use damped Gauss–Newton (local) or global search: simulated annealing accepts uphill moves with exp(−ΔE/T), genetic algorithms use selection, crossover and mutation, swarms combine inertia, personal best and global best.
Gaussian errors give likelihood exp(−χ²/2) and reduced χ² ≈ 1 for a good fit; Poisson counts have mean = variance; log-normal variables have mean above median. Kriging uses the variogram; tests control type I error α against power.
Posterior ∝ likelihood × prior (Gaussian case: precisions add); a Gaussian prior is L2 regularisation, a Laplace prior is sparse L1. The adjoint-state method gives the full gradient from one forward and one adjoint solve per source; ambiguity survives perfect data, uncertainty does not.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Two different density distributions inside the Earth produce exactly the same gravity field at every point on the surface. Which of Hadamard’s conditions for a well-posed problem does this violate?
1. Existence of a solution
2. Uniqueness of the solution
3. Continuous dependence of the solution on the data
4. Linearity of the forward operator in the model parameters
Show answer
Answer: B — Uniqueness of the solution
If two models give identical data, the data cannot choose between them, so the inverse solution is not unique. Existence is not at issue, since both models explain the data, and stability concerns how the solution changes when the data are perturbed. Linearity is not one of Hadamard’s conditions, and even a linear problem can be non-unique.
A linear system G·m = d has condition number κ = 10⁶, and the data d are accurate to a relative error of 10⁻⁴. What is the largest relative error that the solution m can have?
1. About 10⁻¹⁰, because the error is divided by the condition number
2. About 10⁻⁴, the same as the data error, because solving the system averages out the noise
3. As much as about 10², so the solution can be worthless
4. About 10⁻², because only the square root of κ matters for the error
Show answer
Answer: C — As much as about 10², so the solution can be worthless
The bound is (relative error in m) ≤ κ × (relative error in d) = 10⁶ × 10⁻⁴ = 10², which is a hundred times the size of the solution itself, so no digit of m can be trusted in the worst case. The condition number multiplies the error, it does not divide it, and averaging does not help when the noise falls on directions that have small singular values.
A linear problem has 6 data and 6 unknowns, but the matrix G has rank 4. How is it classified?
1. Mixed-determined: two parameter combinations are unconstrained and two data combinations cannot be fitted
2. Even-determined, because the number of data equals the number of unknowns
3. Over-determined, because the 6 data exceed the rank of 4 and so no exact fit exists
4. Under-determined, because the rank is below the number of unknowns and there are infinitely many exact fits
Show answer
Answer: A — Mixed-determined: two parameter combinations are unconstrained and two data combinations cannot be fitted
With rank p = 4 < N = 6 some combinations of the data lie outside the range of G and cannot be fitted, and with p = 4 < M = 6 a two-dimensional null space leaves two parameter combinations unconstrained: the problem is mixed-determined. Equal counts do not make it even-determined, which needs p = N = M. The over- and under-determined labels each describe only one of the two defects, and the under-determined one also wrongly promises exact fits.
In Tikhonov (damped least-squares) inversion, what is the effect of increasing the damping parameter λ?
1. The data misfit falls to zero and the model resolution improves, because the solution is forced closer to the data
2. Only the null-space components of the model are changed, while the well-determined components stay as in least squares
3. The condition number of G itself decreases, because damping changes its singular values permanently
4. The variance of the solution falls, but the bias grows and the resolution worsens, so the misfit increases
Show answer
Answer: D — The variance of the solution falls, but the bias grows and the resolution worsens, so the misfit increases
Larger λ shrinks the filter factors σ²/(σ² + λ²), suppressing noisy directions with small σ, so the variance falls; but components of the true model are also shrunk, which adds bias, blurs the resolution and raises the data misfit. Zero misfit and best resolution belong to λ → 0, not to heavy damping. Damping acts on all components (most strongly on small σ), not only the null space, and G itself is unchanged: only the matrix GᵀG + λ²I is better conditioned.
In simulated annealing with the Metropolis rule, a trial model that raises the misfit by ΔE is accepted with probability exp(−ΔE/T). Why is the temperature T lowered slowly during the search?
1. Because a slow fall keeps every accepted model exactly on the contour of the lowest misfit found so far, which convergence requires
2. Because early uphill moves let the search escape local minima, and cooling too fast would freeze it in one
3. Because the derivative of the misfit is needed at each step and it is only defined at low temperature
4. Because a lower temperature makes uphill moves more likely, which steadily improves the final fit
Show answer
Answer: B — Because early uphill moves let the search escape local minima, and cooling too fast would freeze it in one
At high T uphill moves are accepted often, so the search roams widely and can leave local minima; as T falls, exp(−ΔE/T) shrinks and the method turns into a descent that settles into a minimum, so cooling too fast traps it in whichever basin it is in. The method uses only misfit values, no derivatives. Lower T makes uphill moves less likely, not more.
Which pairing of a geophysical quantity with the probability distribution normally used to model it is most appropriate?
1. Rock permeability, which is positive and spans several orders of magnitude: a Gaussian distribution
2. Random picking errors of arrival times, which are symmetric about zero: a Poisson distribution
3. Gamma photons counted in a fixed time window at constant logging speed: a Poisson distribution
4. Ore grade, which is positive and right-skewed with a long tail: a Poisson distribution
Show answer
Answer: C — Gamma photons counted in a fixed time window at constant logging speed: a Poisson distribution
Counts of independent events in a fixed interval follow the Poisson law, with mean equal to variance. Permeability and ore grade, being positive, right-skewed and products of many factors, are modelled as log-normal, not Gaussian or Poisson, and symmetric picking errors are naturally Gaussian. A Poisson variable takes only non-negative integer values, so it cannot describe continuous grades or timing errors.
With a Gaussian likelihood, which choice of prior makes the maximum a posteriori (MAP) estimate equal to the Tikhonov (L2) solution, and which prior leads to sparsity-promoting L1 regularisation?
1. A Laplace prior gives the L2 solution and a zero-mean Gaussian prior gives the L1 solution
2. A uniform prior gives the L2 solution and a log-normal prior gives the L1 solution
3. A Poisson prior gives the L2 solution and a Gaussian prior with a very large variance gives the L1 solution, as in the limit of heavy damping
4. A zero-mean Gaussian prior gives the L2 solution, with λ² = σ_d²/σₘ², and a Laplace prior gives the L1 solution
Show answer
Answer: D — A zero-mean Gaussian prior gives the L2 solution, with λ² = σ_d²/σₘ², and a Laplace prior gives the L1 solution
The negative log of the posterior is the data misfit plus the negative log of the prior. A Gaussian prior contributes ‖m‖²/(2σₘ²), which is an L2 penalty (λ² = σ_d²/σₘ² after scaling by 2σ_d²), while a Laplace prior contributes a term proportional to ‖m‖₁, the L1 penalty, whose peaked density at zero favours exact zeros. The option that gives L2 to the Laplace prior has the two assignments swapped, and a uniform prior adds no penalty, so it gives plain least squares.
A misfit J(m) = ½‖P·u − d‖² depends on a field u that solves a wave equation A(m)·u = s, and the model m has one million parameters. For a single source, how does the adjoint-state method obtain the gradient ∂J/∂m with respect to all the parameters?
1. From one forward solve and one adjoint solve, driven by the data residual at the receivers, whatever the number of parameters
2. From one million and one forward solves, perturbing the parameters one at a time by a small finite amount and differencing the misfit
3. From one million adjoint solves, one for each parameter, each driven by a unit source placed at that parameter’s location
4. From a single forward solve alone, because the adjoint field is equal to the forward field and needs no separate calculation
Show answer
Answer: A — From one forward solve and one adjoint solve, driven by the data residual at the receivers, whatever the number of parameters
The adjoint equation Aᵀψ = −Pᵀ(P·u − d) is a single PDE solve with the data residual as its source, and the gradient ∂J/∂m = ψᵀ(∂A/∂m)·u then follows for every parameter at once. Perturbing parameters one at a time is the finite-difference approach that needs about M + 1 forward solves, and an adjoint per parameter is not required. The adjoint field is a different field from u, driven by the residual and run backwards in time, so it cannot be dropped.
Which of the following statements about non-linear inversion methods are correct?
1. A genetic algorithm evolves a population of models by selection, crossover and mutation and needs no derivatives of the forward operator
2. In particle swarm optimisation each particle’s velocity combines its inertia, attraction to its own best position and attraction to the swarm’s best position
3. A damped Gauss–Newton (Levenberg–Marquardt) iteration is guaranteed to reach the global minimum from any starting model
4. In simulated annealing an uphill move is accepted with probability exp(−ΔE/T), which falls as the temperature T is lowered
Show answer
Answer: A — A genetic algorithm evolves a population of models by selection, crossover and mutation and needs no derivatives of the forward operator; B — In particle swarm optimisation each particle’s velocity combines its inertia, attraction to its own best position and attraction to the swarm’s best position; D — In simulated annealing an uphill move is accepted with probability exp(−ΔE/T), which falls as the temperature T is lowered
Genetic algorithms, particle swarms and simulated annealing are derivative-free global searches built exactly as these statements describe. Gauss–Newton and Levenberg–Marquardt follow local gradient information and so converge to a minimum near the starting model, which may be a local one; no guarantee of the global minimum exists.
Which of the following statements about distributions and misfit statistics are correct?
1. For a Poisson variable with mean λ the variance is also λ, so the relative scatter of a count falls as 1/√λ
2. If ln X is Gaussian then X is log-normal, and its mean lies above its median
3. Independent Gaussian errors make the L1 misfit the maximum-likelihood criterion
4. A reduced chi-square far above 1 shows that the data are fitted better than their stated errors allow
Show answer
Answer: A — For a Poisson variable with mean λ the variance is also λ, so the relative scatter of a count falls as 1/√λ; B — If ln X is Gaussian then X is log-normal, and its mean lies above its median
The Poisson variance equals the mean, so a count of N events has scatter √N and relative scatter 1/√N, and a log-normal distribution has mean e^(μ + σ²/2) above its median e^μ. Gaussian errors give the L2 (least-squares) criterion, with L1 belonging to Laplace errors, and a reduced chi-square far above 1 means the fit is poor or the errors were underestimated, not that the fit is too good.
Which of the following statements about Bayesian inversion, sparsity and ambiguity are correct?
1. Ambiguity can persist even for perfect noise-free data, whereas uncertainty from noise shrinks as the data errors shrink
2. An L1 penalty λ‖m‖₁ promotes sparse models because the L1 ball has corners on the coordinate axes
3. The posterior equals the likelihood alone, because the prior only rescales it by a constant
4. A model that fits the data to within the noise is thereby proved to be the true structure of the Earth
Show answer
Answer: A — Ambiguity can persist even for perfect noise-free data, whereas uncertainty from noise shrinks as the data errors shrink; B — An L1 penalty λ‖m‖₁ promotes sparse models because the L1 ball has corners on the coordinate axes
Ambiguity is non-uniqueness of the physics (for example the depth–density trade-off), which better data cannot remove, while noise-driven uncertainty does shrink with better data. The corners of the L1 ball make exact zeros likely at the optimum. The prior shapes the posterior, and a Bayesian posterior is proportional to likelihood × prior, so it is not the likelihood alone. A good fit is necessary but never sufficient, since many other models fit equally well.
Solve G·m = d in the least-squares sense for G = [[1, 0], [0, 1], [1, 1]] (rows separated by commas) and d = (2, 1, 6). The minimum value of the sum of squared residuals ‖d − G·m‖² is ____.
Numerical answer — type the value.
Show answer
Answer: 3
GᵀG = [[2, 1], [1, 2]] and Gᵀd = (2 + 6, 1 + 6) = (8, 7). Solving 2m₁ + m₂ = 8 and m₁ + 2m₂ = 7 gives m = (3, 2). Then G·m = (3, 2, 5), the residual is d − G·m = (−1, −1, 1), and ‖d − G·m‖² = 1 + 1 + 1 = 3. Check: any other m, such as (3, 3) with residual (−1, −2, 0), gives a larger sum 5.
A matrix G has singular values 5, 2 and 0.02. Its condition number κ = σ_max/σ_min is ____.
Numerical answer — type the value.
Show answer
Answer: 250
The condition number is the ratio of the largest to the smallest singular value: κ = 5/0.02 = 250. The middle singular value 2 does not enter. The tempting error 5/2 = 2.5 divides by the wrong value.
For G = diag(2, 0.2) and d = (4, 0.4), the damped (Tikhonov) solution is m = (GᵀG + λ²I)⁻¹Gᵀd with λ = 0.2. The second component m₂ is ____ (a whole number).
Numerical answer — type the value.
Show answer
Answer: 1
G is diagonal, so each component separates: mᵢ = σᵢdᵢ/(σᵢ² + λ²). For the second component m₂ = 0.2 × 0.4/(0.2² + 0.2²) = 0.08/0.08 = 1. The undamped least-squares value is 0.4/0.2 = 2, so damping halves m₂. The first component is 2 × 4/(4 + 0.04) = 1.98, almost unchanged because σ₁ is much larger than λ.
A residual vector is e = (−2, 3, 6). Its L2 norm ‖e‖₂ = (Σ eᵢ²)^(1/2) is ____.
Numerical answer — type the value.
Show answer
Answer: 7
Squaring gives 4 + 9 + 36 = 49, and the square root is 7. For comparison the L1 norm is 2 + 3 + 6 = 11 and the L∞ norm is 6, consistent with L∞ ≤ L2 ≤ L1. Forgetting to square the negative entry, or summing the absolute values, are the usual slips.
A model with 2 fitted parameters is compared with 4 data whose independent Gaussian errors all have σ = 0.2. The residuals are 0.4, −0.6, 0.2 and 1.0. The reduced chi-square χ²/ν, with ν = N − M, is ____ to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 19.5
χ² = Σ(rᵢ/σ)² = (0.4/0.2)² + (0.6/0.2)² + (0.2/0.2)² + (1.0/0.2)² = 4 + 9 + 1 + 25 = 39. The degrees of freedom are ν = 4 − 2 = 2, so χ²/ν = 39/2 = 19.5. A value this far above 1 says the model is a poor fit or the errors were badly underestimated; dividing by N = 4 instead of ν would give 9.75.
A parameter m has a Gaussian prior with mean 10 and standard deviation 4. One measurement d = 14 of m is made, with Gaussian noise of standard deviation 2. The mean of the Gaussian posterior for m is ____ to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 13.2
The precisions are 1/4² = 0.0625 for the prior and 1/2² = 0.25 for the datum, which add to 0.3125 (posterior variance 3.2, standard deviation 1.79). The posterior mean is the precision-weighted average (10 × 0.0625 + 14 × 0.25)/0.3125 = (0.625 + 3.5)/0.3125 = 13.2, pulled towards the more precise datum. The simple average 12 ignores the different uncertainties.
A sample of n = 36 independent measurements has mean 2.8, and the population standard deviation is known to be σ = 0.6. For the null hypothesis μ₀ = 2.5, the test statistic z = (x̄ − μ₀)/(σ/√n) equals ____.
Numerical answer — type the value.
Show answer
Answer: 3
The standard error is σ/√n = 0.6/6 = 0.1, and the difference of means is 2.8 − 2.5 = 0.3, so z = 0.3/0.1 = 3. This lies beyond the two-sided 5% limit 1.96 (and the 1% limit 2.58), so H₀ would be rejected. Using σ instead of σ/√n as the denominator would give 0.5 and wrongly suggest no effect.