Report#
Submit one set of handwritten answers per group at 12:30.
Computers, phones, online resources and AI tools are not permitted. You may use only the handwritten notes prepared earlier in the session. Show the formulas and reasoning that support your answers; numerical answers without an explanation will not receive full credit.
Given model and numerical reference values#
The simplified Stefan model for ice thickness after a cold spell is
where \(H_{\mathrm{ice},0}\) is the initial ice thickness, \(\Delta t\) is the duration in seconds, and \(\Delta T\) is the temperature deficit relative to the freezing point, in Kelvin.
Use \(\Delta t=5\times24\times3600\ \mathrm{s}\), \(\mu_{H_0}=0.20\ \mathrm{m}\), \(\sigma_{H_0}=0.03\ \mathrm{m}\), \(\mu_{\Delta T}=10\ \mathrm{K}\), and \(\sigma_{\Delta T}=4\ \mathrm{K}\). Unless otherwise specified, the two inputs are independent. In the Monte Carlo simulation, negative draws of \(\Delta T\) are set to zero (no further ice growth; melting is not modelled). The Taylor approximation is evaluated at the positive input means.
The following are reference results from the notebook (\(N=100000\), specified seed). Use them where relevant; answers should focus on the calculations and interpretation.
Quantity |
Independent inputs (\(\rho=0\)) |
Correlated inputs (\(\rho=0.5\)) |
|---|---|---|
\(q(\mu_{H_0},\mu_{\Delta T})\) [m] |
0.319700 |
0.319700 |
Second-order Taylor mean \(\mu_H\) [m] |
0.318188 |
0.317046 |
First-order Taylor standard deviation \(\sigma_H\) [m] |
0.043206 |
0.050962 |
Monte Carlo mean [m] |
0.318273 |
0.316976 |
Monte Carlo sample standard deviation [m] |
0.044155 |
0.052515 |
For the independent-input simulation, the event \(H_\mathrm{ice}<0.24\ \mathrm{m}\) has probability 0.04783 according to Monte Carlo and 0.03517 according to the Gaussian CDF with Taylor-propagated moments. Around 0.615% of temperature-deficit draws were negative and set to zero.
Part 1 — Mean and variance propagation#
1.1 Derive the first-order partial derivatives \(\partial q/\partial H_{\mathrm{ice},0}\) and \(\partial q/\partial(\Delta T)\), and state the sign of each derivative at the given means.
1.2 Derive the second-order partial derivatives \(\partial^2q/\partial H_{\mathrm{ice},0}^2\) and \(\partial^2q/\partial(\Delta T)^2\). Write the second-order Taylor approximation of the mean for independent inputs. Use the reference values to quantify the difference between this mean and \(q\) evaluated at the mean inputs.
1.3 Write the first-order Taylor variance propagation formula for independent inputs. Give the resulting standard deviation (in metres) using the provided numerical reference.
1.4 Explain why evaluating the non-linear model at the input means need not equal the output expectation, and why the Taylor result is an approximation rather than an exact value.
Solutions — Part 1#
Define \(S=\frac{2k_\mathrm{ice}}{\rho_\mathrm{ice}l}\Delta T\Delta t+H_{\mathrm{ice},0}^2\) as a temporary expression in the derivation, so \(q=\sqrt S\); no additional quantity is required in student answers.
1.1:
Both positive at the given means. Numerically about \(0.625586\) and \(0.009730\) m/K.
1.2:
Numerical correction \(0.318188-0.319700\approx-0.001512\ \mathrm{m}\) (about -1.5 mm).
1.3:
1.4: Curvature causes \(\mathbb E[q(Y)]\ne q(\mathbb E[Y])\) in general; terms of order 3 or higher are discarded, and the clipping rule is not represented by a smooth global Taylor expansion.
Part 2 — Dependence between input variables#
Now assume the initial ice thickness and temperature deficit have a Pearson correlation coefficient \(\rho=0.5\).
2.1 State \(\mathrm{Cov}(H_{\mathrm{ice},0},\Delta T)\) in terms of \(\rho\), \(\sigma_{H_0}\) and \(\sigma_{\Delta T}\) and write the extra term in the first-order output variance compared with independent inputs. Is this term positive or negative? Justify using the derivatives from Part 1.
2.2 Compare the Taylor standard deviations for \(\rho=0\) and \(\rho=0.5\). Explain physically why positive correlation has this effect on the propagated uncertainty. Does specifying the marginal distributions alone determine the output uncertainty?
Solutions — Part 2#
2.1: $\(\mathrm{Cov}(H_{\mathrm{ice},0},\Delta T)=\rho\sigma_{H_0}\sigma_{\Delta T}=0.5(0.03)(4)=0.06\ \mathrm{m\,K},\)$
2.2: \(\sigma_H\) increases from \(0.043206\) to \(0.050962\) m (increase of \(\approx0.007756\) m, approximately 18%). When initial thickness and the deficit increase together, their positive sensitivities reinforce each other, increasing variance. Marginals do not specify covariance, so they do not determine the variance of this multivariate transformation.
Part 3 — Lower-tail risk and Gaussian approximation#
3.1 For the event \(H_\mathrm{ice}<0.24\ \mathrm{m}\), compare the Monte Carlo and Gaussian non-exceedance probabilities from the reference values. State which estimate is higher and calculate their absolute difference in percentage points.
3.2 Give two specific reasons why a Gaussian PDF with Taylor-propagated moments need not describe the simulated output distribution exactly, even if its mean and standard deviation are similar.
3.3 Is agreement of the first two moments enough to validate a model for assessing exceptionally thin ice? Explain briefly.
Solutions — Part 3#
3.1: Monte Carlo \(0.04783=4.783\%\); Gaussian \(0.03517=3.517\%\). The MC estimate is higher by \(0.01266\) in probability, or 1.266 percentage points. Relative to the Gaussian estimate, it is approximately 36.0% larger; this relative comparison is optional and must not be confused with percentage points.
3.2: The Stefan square-root transformation is non-linear and can change the output distribution’s skewness/tails; moreover, clipping the few negative Gaussian draws of \(\Delta T\) changes the input distribution (including a point mass at zero), which Taylor propagation about a positive mean does not describe globally.
3.3: No: matching mean and variance does not ensure matching extreme quantiles or tail probabilities. The 0.24 m example demonstrates a materially different predicted risk.
Part 4 — Monte Carlo accuracy and model limitations#
4.1 Compare the Monte Carlo and first-order Taylor standard deviations for independent inputs. Calculate their relative difference (using Monte Carlo as reference). Would increasing the number of Monte Carlo realizations \(N\) necessarily eliminate this difference? Explain why or why not.
4.2 Identify one limitation of the input distributions and one limitation of the physical growth model. For each, propose a practical improvement.
Solutions — Part 4#
4.1: The Monte Carlo standard deviation (\(0.044155\) m) is larger than the Taylor standard deviation (\(0.043206\) m), with a relative difference of approximately \(2.15%\). Increasing \(N\) reduces Monte Carlo sampling variability, but does not eliminate the Taylor approximation error, since the first-order propagation neglects higher-order contributions from the non-linear transformation. A
4.2: Input example: a normal temperature deficit includes negative values, while clipping changes the distribution; improve with calibrated non-Gaussian or weather-data-based distributions. Model example: the constant-temperature Stefan model neglects thawing and daily temperature changes; improve with a time-resolved thermal model including melting.
How did things go? (Optional)#
After the session, your group may share feedback with us (i.e. the teaching team) about the notebook and the written questions.
By Lotfi Massarweh and Sandra Verhagen, Delft University of Technology. CC BY 4.0, more info on the Credits page of Workbook.