Handling missing values (deletion, imputation strategies): Worked examples

This page is the worked-example companion to Handling Missing Values. The parent note gave you the formulas. Here we grind through every kind of situation a missing-value problem can be — different missingness mechanisms, tiny toy tables, degenerate inputs (all-missing columns, single neighbour), time-series gaps, and an exam-style trap.

Before we compute anything, one promise: no symbol appears before it is earned. If you have not yet read the parent, that is fine — every term is rebuilt below in plain words with a picture.

Vocabulary rebuilt from zero

The scenario matrix

Every missing-value problem falls into one of these cells. Our examples below cover all of them — each example is labelled with the cell it hits.

Cell	Data shape	Mechanism	Method that fits	Trap to expose
C1	Numeric, symmetric	MCAR	Mean	variance shrinks
C2	Numeric, skewed / outlier	MCAR	Median	mean gets pulled
C3	Two rows survive	MAR	Listwise deletion	bias by deletion
C4	Feature correlated with another	MAR	KNN / conditional	mean breaks correlation
C5	Time series, short gap	MCAR-ish	Forward fill	flat during a trend
C6	Categorical	MCAR	Mode	mean is meaningless
C7	Degenerate: whole column NaN	—	drop column	no value to borrow
C8	Degenerate: only 1 neighbour	MAR	KNN with K=1	collapses to copy
C9	Real-world word problem	MAR	choose + justify	reasoning, not formula
C10	Exam twist: leakage	MAR	fit on train only	leakage

Ten cells. Ten examples. Let's go.

C1 — Mean imputation, symmetric data (MCAR)

Worked example Fill the hole with the balance point

Data: [10, 15, NaN, 20, 25]. The sensor dropped one reading at random (MCAR).

Forecast: guess the fill value and whether the spread goes up, down, or stays after filling.

Step 1 — Compute the mean of the observed numbers. $\overset{x}{ˉ} = \frac{10 + 15 + 20 + 25}{4} = \frac{70}{4} = 17.5$ Why this step? MCAR means the hole is a typical value, so the balance point of what we did see is our best single guess.

Step 2 — Drop 17.5 into the hole. → [10, 15, 17.5, 20, 25]. Why this step? We assume the missing value came from the same distribution as the others.

Step 3 — Measure spread before and after. Observed variance (over the 4 seen values): $σ^{2} = 31.25$ . After filling, variance over 5 values: $σ^{2} = 25.0$ . Why this step? To expose the trap: a value placed exactly at the mean adds zero distance-from-mean, so it dilutes the average distance. See the figure — the amber bar sits on the balance line and contributes nothing to spread.

Verify: Mean unchanged? New mean $= \frac{70 + 17.5}{5} = 17.5$ ✓. Variance dropped from $31.25 \to 25.0$ ✓. Mean imputation always shrinks variance — models become over-confident.

C2 — Median beats mean when an outlier lurks (MCAR)

Worked example One giant value poisons the mean

Data: [12, 14, 15, NaN, 500]. The 500 is a genuine (rare) reading; the NaN is random.

Forecast: will mean or median give a fill closer to the "typical" 12–15 cluster?

Step 1 — Mean of observed: $\overset{x}{ˉ} = \frac{12 + 14 + 15 + 500}{4} = \frac{541}{4} = 135.25$ . Why this step? The mean uses every value with equal weight, so the lone 500 drags it far above the crowd.

Step 2 — Median of observed: sort → [12, 14, 15, 500], middle two are 14 and 15, median $= \frac{14 + 15}{2} = 14.5$ . Why this step? The median only cares about position, not magnitude, so the 500 sits at the end and barely moves the middle. This is exactly why the parent says median minimizes $\sum ∣ x_{i} - c ∣$ (least absolute deviation) — absolute distance doesn't reward chasing outliers.

Verify: 135.25 lies outside the 12–15 cluster; 14.5 sits inside it ✓. For skewed data or outliers, use the median.

C3 — Listwise deletion and its hidden bias (MAR)

Worked example Delete rows, count the damage

Data: the parent's patient table.

ID	Age	Income	BP
1	45	60k	120
2	52	NaN	130
3	38	NaN	NaN
4	61	80k	140

Forecast: how many rows survive listwise deletion, and what does the surviving average Age become?

Step 1 — Keep only rows with no NaN anywhere. Rows 1 and 4 survive; 2 and 3 die. That's $2/4 = 50%$ data loss. Why this step? Listwise deletion is all-or-nothing per row — one hole kills the whole row.

Step 2 — Compare average Age before vs after. All four ages: $\frac{45 + 52 + 38 + 61}{4} = 49.0$ . Survivors only: $\frac{45 + 61}{2} = 53.0$ . Why this step? If younger patients skip Income (MAR), deletion systematically removes the young — mean age drifts up from 49 to 53. Your dataset now over-represents older people, which is bias you injected yourself.

Verify: $49.0 \to 53.0$ , an upward shift of 4 years ✓. Under MAR, deletion is not neutral.

C4 — KNN imputation respects correlation (MAR)

Worked example Borrow from your neighbours, not the crowd

Data: Weight missing for patient A.

Patient	Height	Weight	Age
A	170	NaN	30
B	168	70	28
C	172	75	32
D	165	65	45

Forecast: which two patients look most like A, and will KNN give a weight above or below the global mean of 70?

Step 1 — Distance to each patient, ignoring the missing feature (Weight). Distance = straight-line gap in the Height–Age plane: $d (A, i) = (Δ Height)^{2} + (Δ Age)^{2}$ . $d (A, B) = (170 - 168)^{2} + (30 - 28)^{2} = 4 + 4 = 8 \approx 2.83$ $d (A, C) = (170 - 172)^{2} + (30 - 32)^{2} = 4 + 4 = 8 \approx 2.83$ $d (A, D) = (170 - 165)^{2} + (30 - 45)^{2} = 25 + 225 = 250 \approx 15.81$ Why this step? "Similar people have similar weight." We measure similarity as closeness in feature space — the figure shows A hugging B and C while D floats far off in the Age direction.

Step 2 — Pick $K = 2$ nearest → B and C (both $\approx 2.83$ ; D at 15.8 is far).

Step 3 — Average their weights (uniform weights). $Weight_{A} = \frac{70 + 75}{2} = 72.5 kg$ Why this step? We exclude D entirely, so its low weight (65) can't drag our guess down. The global mean $\frac{70 + 75 + 65}{3} \approx 70$ would.

Verify: KNN gives 72.5 vs global mean $\approx 70$ ✓; KNN is pulled toward the two similar patients ✓. KNN keeps the Height–Weight relationship that mean imputation destroys.

Worked example Carry the last price — and see it fail

Data: daily stock price [100, 105, NaN, NaN, 112].

Forecast: what do the two holes get filled with, and by how much does forward fill undershoot the last real value 112?

Step 1 — Forward fill: each hole copies the last known value. Last known before the gap is 105, so both holes get 105 → [100, 105, 105, 105, 112]. Why this step? Time series have autocorrelation — today ≈ yesterday. Carrying 105 forward beats a global mean.

Step 2 — Measure the error against the true trend. A straight line from 105 (day 2) to 112 (day 5) would place day 3 at 105 + (112−105)/3 ≈ 107.33 and day 4 at ≈ 109.67. Forward fill kept both at 105 — flat during a clear climb. Undershoot on day 4: $109.67 - 105 = 4.67$ . Why this step? To expose forward fill's blind spot: it freezes the series, so a rising trend is systematically under-filled. See the amber flat segment sitting below the cyan trend line.

Verify: both holes = 105 ✓; flat fill sits below the linear guess (undershoot ≈ 4.67) ✓. Forward fill fails on long or trending gaps.

C6 — Mode imputation for a categorical column (MCAR)

C7 — Degenerate case: the whole column is missing

C8 — Degenerate case: only one neighbour ( $K = 1$ )

C9 — Real-world word problem (MAR): the survey trap

Worked example Income missing for the young

A bank surveys customers. 40% of under-25s left Income blank; over-25s almost always answered. You want a model for loan approval. Which strategy?

Forecast: deletion, mean, or conditional — and what's the risk of the wrong choice?

Step 1 — Identify the mechanism. Missingness depends on Age, an observed variable → this is MAR, not MCAR. Why this step? MAR tells us the observed variables (Age) carry the information to reconstruct the hole honestly.

Step 2 — Reject deletion. Deleting blank-Income rows removes ~40% of the young → the model barely sees young customers (exactly the C3 bias). Reject.

Step 3 — Reject plain mean. One global income ignores that young people earn less on average → over-estimates their income → wrongly approves loans. Reject.

Step 4 — Choose conditional imputation (KNN or MICE) that predicts Income from Age and other features. It restores the age–income relationship. Why this step? Under MAR, conditioning on the observed cause removes the bias. This is the parent's core lesson: the mechanism picks the method.

Verify: mechanism = MAR (depends on observed Age) ⇒ conditional imputation is the unbiased choice ✓.

C10 — Exam twist: imputation must not leak (MAR)

Worked example Where do you compute the mean?

You will split data into train and test (split note), then mean-impute a feature. Question: do you compute the imputation mean from (a) the whole dataset or (b) the training set only?

Forecast: which is correct, and what breaks if you pick wrong?

Step 1 — Recall the rule: the test set must simulate unseen future data. Any statistic touching test rows before evaluation is data leakage.

Step 2 — Compute the mean on train only, then apply that same number to fill holes in both train and test. Why this step? If you averaged over all data, the test rows influenced the fill value — the model peeks at the test set, and your reported accuracy is optimistically inflated.

Tiny numbers: train Income [40, 60, NaN] → train mean of observed $= \frac{40 + 60}{2} = 50$ . Test row with NaN gets filled with 50, not with any test-derived number.

Verify: train mean $= \frac{40 + 60}{2} = 50$ , applied to the test hole ✓. Fit imputers on train, transform everywhere.

Recall

Recall Why does mean imputation shrink variance?

Because the fill value sits exactly at the mean, adding zero distance-from-mean, which dilutes the average squared distance. ::: A value at the balance point contributes 0 to the spread, so average spread drops.

Recall Under MAR, why is deletion biased but conditional imputation not?

Deletion removes rows non-randomly (e.g. the young), skewing the sample. ::: Conditioning on the observed cause (Age) reconstructs the missing values honestly.

Recall With

$K = 1$ , KNN imputation equals what? A direct copy of the single nearest neighbour's value (weights cancel). ::: $\frac{w \cdot v}{w} = v$ .

Recall On which set do you compute an imputation statistic to avoid leakage?

The training set only, then apply it to both train and test. ::: Fit on train, transform everywhere.

Related: Outlier handling · Feature scaling · Parent: Handling Missing Values.