Report#
Answer the following questions based on the analyses performed in the notebook and the notes you made. Explain your reasoning clearly. No Python code is required.
Background and key results#
The farmer’s peat parcel is observed with two independent techniques: InSAR (average height change of the whole parcel, every 6 days, \(\sigma_{\text{InSAR}} = 3\) mm) and an extensometer (height change at one point, daily, \(\sigma_{\text{ext}} = 8\) mm). Two functional models were estimated with BLUE:
Model 1 (offset + velocity): \(d(t) = d_0 + v\,t\), with \(t\) in years since 1 January 2020.
Model 2 (weather model): \(d(t) = d_0 + x_I\,D(t) + k\,S_{10}(t)\), with \(D(t)\) the number of dry days and \(S_{10}(t)\) the water surplus with \(\tau = 10\) days.
You may use the following results obtained in the notebook (estimate \(\pm\) standard deviation):
InSAR (\(m = 145\)) |
Extensometer (\(m = 1461\)) |
|
|---|---|---|
Model 1: \(\hat{d}_0\) [mm] |
\(6.0 \pm 0.5\) |
\(2.4 \pm 0.4\) |
Model 1: \(\hat{v}\) [mm/yr] |
\(-5.37 \pm 0.21\) |
\(-7.95 \pm 0.18\) |
Model 1: std of residuals [mm] |
\(11.5\) |
\(16.9\) |
Model 2: \(\hat{d}_0\) [mm] |
\(7.7 \pm 0.7\) |
\(5.8 \pm 0.6\) |
Model 2: \(\hat{x}_I\) [mm/dry day] |
\(-0.0422 \pm 0.0016\) |
\(-0.0613 \pm 0.0014\) |
Model 2: \(\hat{k}\) [-] |
\(0.463 \pm 0.012\) |
\(0.684 \pm 0.011\) |
Model 2: std of residuals [mm] |
\(5.7\) |
\(9.9\) |
On average, the parcel has \(N_{\text{dry}} \approx 131\) dry days per year in the observation period (2020–2023).
Question 1: Functional and stochastic model#
a) For Model 1, write the observation equations in matrix form \(\mathbb{E}(Y) = A\,x\) (no numerical values needed, use \(t_1, t_2, \dots\)). You do not need to write out the whole matrix. State the units of the unknowns.
Solution:
\(d_0\) is in mm and \(v\) is in mm/yr (because \(t\) is in years).
b) Sketch the structure of the covariance matrix \(\Sigma_Y\) for the InSAR observations. State the dimension of the matrix. You do not need to write out the whole matrix.
How is the assumption that the observations are independent included?
Solution:
The dimension is \(m \times m\) (\(145 \times 145\) for InSAR). The diagonal contains the variances of the observations; the off-diagonal elements are the covariances. Independence means all covariances are zero, so \(\Sigma_Y\) is diagonal.
c) Write down the BLU estimator \(\hat{X}\) and its covariance matrix \(\Sigma_{\hat{X}}\). BLUE is a special case of weighted least-squares. Which weight matrix \(W\) does BLUE use, and why does a more precise observation receive more weight?
Solution:
BLUE uses the inverse of the covariance matrix of the observations as weight matrix: \(W = \Sigma_Y^{-1}\). A precise observation has a small variance, so a large weight (\(1/\sigma^2\)). It therefore contributes more to the estimate, which makes sense because it contains more reliable information. With this choice the estimator is the best: it has the smallest variance of all linear unbiased estimators.
Question 2: Results of Model 1 (offset + velocity)#
a) Precision of the observations and of the estimates:
Which sensor gives the more precise individual observations?
Which sensor gives the more precise estimate of the velocity \(v\)?
Does this make sense? Explain the relation between the precision of an estimated parameter and the number of observations. Use the redundancy of both datasets in your answer.
Solution: InSAR gives the more precise individual observations (\(\sigma = 3\) mm vs \(8\) mm). Still, the extensometer gives the more precise velocity (\(0.18\) vs \(0.21\) mm/yr). This makes sense: from \(\Sigma_{\hat{X}} = (A^T \Sigma_Y^{-1} A)^{-1}\), the precision of the estimates depends on
the covariance matrix \(\Sigma_Y\): more precise observations give more precise estimates;
the design matrix \(A\): the measurement set-up, i.e. how many observations there are and how they are spread in time.
The redundancy (\(m - n\)) is \(145 - 2 = 143\) for InSAR and \(1461 - 2 = 1459\) for the extensometer. More redundant observations average out the random errors. The extensometer has about 10 times more observations, which more than compensates for its worse precision. (Roughly, \(\sigma_{\hat{v}} \propto \sigma/\sqrt{m}\): \(3/\sqrt{145} \approx 0.25\) vs \(8/\sqrt{1461} \approx 0.21\).)
b) Compare the estimated velocities of InSAR and the extensometer. Do you think the difference is significant, given their standard deviations? Give a possible explanation.
Solution: The difference is about \(2.6\) mm/yr, while the standard deviations are only about \(0.2\) mm/yr each (std of the difference \(\approx \sqrt{0.21^2 + 0.18^2} \approx 0.28\) mm/yr). The difference is therefore many times larger than its standard deviation, so it is very unlikely to be caused by measurement noise alone. A plausible explanation is that the extensometer spot really subsides faster than the parcel average. InSAR measures the average height change of the whole parcel, whereas the extensometer measures at one point. One spot in a parcel (e.g. wetter, more peat, closer to a ditch) does not have to behave exactly like the parcel average. Neither is wrong: they observe different quantities.
c) The figure below shows the observations, the fitted Model 1 and the 98% confidence interval of \(\hat{Y}\) for both sensors.
The 98% confidence interval of \(\hat{Y}\) is very narrow, but many observations lie far outside it. Does this mean the confidence interval is computed wrongly? Explain what the confidence interval of \(\hat{Y}\) represents, and why this is not a contradiction.
Solution: No, the confidence interval is not computed wrongly. The confidence interval of \(\hat{Y}\) (here \(\pm 2.33\,\sigma_{\hat{Y}}\)) describes the uncertainty of the fitted line due to the uncertainty in the estimated parameters. Because there are many observations, the line is estimated very precisely. It does not describe the spread of individual observations around the line. Moreover, it is only meaningful if the functional model is correct; here the observations deviate from the line mainly because of the seasonal signal that the model does not contain.
d) Using the figure in 2c, sketch the residuals of Model 1 for the extensometer as a function of time. The standard deviation of the extensometer residuals is \(16.9\) mm, while \(\sigma_{\text{ext}} = 8\) mm. What does this tell you about Model 1?
Solution: The residuals show a clear seasonal (yearly) pattern instead of random noise around zero: positive in winter, negative in summer, with amplitudes of up to 3–4 cm. If the model were correct, the residuals should be random with a spread comparable to (slightly smaller than) \(\sigma\). A residual spread about twice as large as \(\sigma\) (and almost four times for InSAR: \(11.5\) vs \(3\) mm), together with the systematic pattern, shows that the functional model is incomplete: it misses the seasonal swelling and shrinking of the peat. (Alternatively, the precision of the observations could be too optimistic, but the clear pattern points to the functional model.)
Question 3: Model 2 (weather model)#
a) Explain the physical meaning and the sign of the estimated parameters \(\hat{x}_I\) and \(\hat{k}\).
Solution: \(\hat{x}_I < 0\) is the irreversible subsidence per dry day: each dry day the surface goes down by about \(0.04\) mm (InSAR) or \(0.06\) mm (extensometer) due to oxidation. \(\hat{k} > 0\) scales the reversible motion: a larger water surplus (wetter soil) gives an upward movement (swelling), a negative surplus gives a downward movement (shrinkage). The extensometer has larger values of both, i.e. that spot subsides faster and swells/shrinks more than the parcel average.
b) Convert \(\hat{x}_I\) for InSAR to a subsidence rate in mm/yr and compare it with the velocity \(\hat{v}\) of Model 1. Also convert the standard deviation. What is the difference in interpretation between the two models?
Solution: \(\hat{x}_I \cdot N_{\text{dry}} = -0.0422 \times 131 \approx -5.5\) mm/yr, with standard deviation \(0.0016 \times 131 \approx 0.2\) mm/yr. This is very close to \(\hat{v} = -5.37\) mm/yr. (Extensometer: \(-0.0613 \times 131 \approx -8.0\) mm/yr vs \(-7.95\) mm/yr.) Both models give a similar long-term subsidence rate, but Model 1 assumes subsidence happens at a constant rate, whereas Model 2 says it only happens on dry days, and in addition explains the seasonal up-and-down motion.
Question 4: Model comparison and advice to the farmer#
The figure below shows the residuals of Model 2 for the extensometer as a function of time (left) and the histogram of the normalized residuals \(\hat{\epsilon}/\sigma\) together with the standard normal distribution \(N(0,1)\) (right).
a) Compare the residuals of Model 2 in the figure with your sketch of the Model 1 residuals from Question 2d. Look at the size of the residuals (see also the std of the residuals in the table) and the pattern in time. Explain why Model 2 describes the data better.
Solution: Model 1 residuals show a clear yearly up-and-down pattern with amplitudes of several centimetres, because Model 1 cannot describe the seasonal swelling and shrinking. Model 2 includes this reversible motion through \(k\,S_{10}(t)\) and lets subsidence happen only on dry days through \(x_I\,D(t)\). As a result, the residuals are much smaller and the seasonal pattern has largely disappeared: they now scatter around zero without a clear yearly wave. The std of the residuals drops from \(16.9\) to \(9.9\) mm (extensometer) and from \(11.5\) to \(5.7\) mm (InSAR).
b) Model 2 is better, but is it “correct”? Use the figure (residuals and histogram) and the table (compare the std of the residuals with \(\sigma\) for both sensors) to argue whether the model is complete, and suggest one way to improve it.
Solution: It is not yet complete: the residual spread is still larger than \(\sigma\) (\(9.9\) vs \(8\) mm for the extensometer, \(5.7\) vs \(3\) mm for InSAR), and more residuals fall outside the 98% confidence interval than the 2% that would be expected (about 6%). Some pattern remains in the residuals (e.g. mostly negative residuals in the second half of 2022, followed by mostly positive residuals in the winter of 2022/2023), and the histogram of the normalized residuals is wider than \(N(0,1)\), with heavier tails. Possible improvements: estimate \(\tau\) from the data instead of fixing it at 10 days (the soil memory need not be 10 days, nor the same for the parcel and the extensometer spot), which requires non-linear estimation; other options are to add processes that are not modelled (e.g. groundwater level, temperature) or to reconsider the stochastic model.
c) Write a short advice (3–4 sentences) to the farmer: is his parcel sinking, how fast, and what is driving it? Which dataset would you use to describe the parcel as a whole?
Solution: Yes, the parcel is sinking: the parcel as a whole subsides by about 5–6 mm/yr (InSAR), and the spot with the extensometer by about 8 mm/yr. Superimposed on this trend, the surface moves up and down by several centimetres per year due to swelling and shrinking of the peat. The irreversible subsidence happens during dry periods, so keeping the soil wetter (e.g. a higher ditch/groundwater level in summer) would likely slow down the subsidence. InSAR is the most suitable to describe the parcel as a whole, since it measures the parcel average; the extensometer only represents one spot.
How did things go? (Optional)#
Use this space to let us know if you ran into any challenges while working on this GA, and if you have any feedback to report.
By Lina Hagenah and Isabel Slingerland, Delft University of Technology. CC BY 4.0, more info on the Credits page of Workbook.