Question 3 of 6: Missing-Data Strategy and Log-Normal Flood-Frequency Analysis

Question 3: Missing-Data Strategy and Log-Normal Flood-Frequency Analysis (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Eight years of peak-discharge instrumentation with three years missing (sensor/battery failure — not just peak flows but the whole record for those years):

Find. (a) A justified strategy for the three missing years. (b) $Q_2$ and $Q_{10}$ assuming a log-normal distribution. (c) $P(Q_{2016}\gt 40)$. (d) $P(Q_{2016}\gt 40\ \text{and}\ Q_{2017}\gt 40)$.

a. Missing-data strategy. Because the failures are attributed to the sensor's power supply — a mechanism unrelated to the size of the flow that happened to be occurring — the three gaps can reasonably be treated as missing completely at random rather than as a bias toward missing high or low flows; that is the hydrologically important distinction, because it means simply working with the five complete years (as done in part b) is unbiased, if statistically weaker than a full eight-year record. The stronger option, where available, is record extension: identify a nearby, hydrologically similar gauge with a long, continuous record and build a regression (or the MOVE.1 technique) between the two stations' concurrent years of overlap, then use that relationship to estimate the three missing peaks from the neighbouring station's record for those same years. This is preferable to simple deletion because annual peak flows over a region are driven by the same synoptic storm systems and correlate strongly between nearby, similarly sized watersheds, so a well-correlated neighbour genuinely recovers information rather than fabricating it. A second, complementary option is a regional flood-frequency (index-flood) approach: rather than estimate the distribution's shape (mean, variance, skew) purely from this station's short record, borrow a regional growth curve or regional skew coefficient developed from many long-record stations in hydrologically similar catchments, and scale it by this station's own index flood (e.g. its mean annual peak). This directly addresses the fact that higher-order moments such as skew are notoriously unstable when estimated from fewer than about 20–30 years of record. In the absence of either a correlated neighbour or a regional study, the fallback used below — proceed with the five available years and explicitly flag the resulting wide confidence interval — is defensible but should never be presented as equivalent in reliability to a full eight-year analysis.

Approach. Fit a log-normal distribution to the five available annual peaks by working with $y=\ln Q$, then use the standard-normal frequency factor $z_T$ for each return period $T$ to estimate quantiles and exceedance probabilities, and combine independent-year probabilities by multiplication for part (d).

1. Log-transform the five available years and fit the normal distribution of $y=\ln Q$. The available data are $Q=12,\,15,\,42,\,22,\,33\ \text{m}^3/\text{s}$ ($n=5$), giving $$\bar y=\frac{1}{5}\sum\ln Q_i=3.104\qquad s_y=\sqrt{\frac{1}{n-1}\sum(\ln Q_i-\bar y)^2}=\boxed{s_y=0.523}$$
2. $Q_2$ and $Q_{10}$ from $Q_T=\exp(\bar y+z_T\,s_y)$. For $T=2$ yr, $z_2=0$ (the median of a normal distribution coincides with its mean), so $Q_2$ is simply the geometric mean of the data: $$Q_2=\exp(3.104+0)=\boxed{22.3\ \text{m}^3/\text{s}}$$ For $T=10$ yr, the standard-normal value at the 90th percentile is $z_{10}=1.282$: $$Q_{10}=\exp(3.104+1.282\times0.523)=\boxed{43.6\ \text{m}^3/\text{s}}$$
3. Probability of exceeding 40 m³/s in 2016. Standardize $Q=40$ and read the upper-tail area of the standard normal distribution: $$z=\frac{\ln 40-\bar y}{s_y}=\frac{3.689-3.104}{0.523}=1.118\qquad P(Q_{2016}\gt 40)=1-\Phi(1.118)=\boxed{0.132\ (13.2\%)}$$ which corresponds to an equivalent return period of about $1/0.132\approx7.6$ years.
4. Probability of exceeding 40 m³/s in both 2016 and 2017. Successive annual peak flows are treated as independent (each year's flood-generating storms are essentially unrelated events once one year has passed), so the joint probability is the product of the two identical single-year probabilities: $$P(Q_{2016}\gt 40\ \cap\ Q_{2017}\gt 40)=0.132\times0.132=\boxed{0.0174\ (1.7\%)}$$