Log and power transformations: Exercises — AI-ML deep dive

This is a self-test page for the parent topic (2.1.9). Every problem below has a fully worked solution hidden inside a collapsible [!recall]- callout — try first, then reveal. Problems climb from L1 Recognition (spot the right tool) to L5 Mastery (combine everything into a pipeline decision).

Before we start, a few pieces of notation used throughout (nothing here is assumed):

Definition Symbols you will meet

$x$ — a single raw data value (a positive or possibly negative number).
$x^{'}$ — the transformed value (what the model actually sees).
$lo g$ — natural logarithm (base $e \approx 2.718$ ) unless we write $lo g_{10}$ .
$λ$ (Greek "lambda") — the power knob of Box-Cox / Yeo-Johnson. It picks how strongly we bend the data. $λ = 1$ = no bending, $λ = 0$ = full log.
skewness — a single number saying "how lopsided is the histogram?" Positive = long tail to the right; $0$ = symmetric; negative = long tail to the left.
std dev (standard deviation, symbol $σ$ ) — the typical distance of values from their average; a measure of spread.
log1p — a function meaning "log of one plus": $log1p (x) = lo g (1 + x)$ . It exists so you can log data that contains zeros: $lo g (0)$ is undefined, but $lo g (1 + 0) = lo g (1) = 0$ is perfectly fine. So log1p shifts every value right by $1$ before taking the log, quietly rescuing the zero entries.
MLE (maximum likelihood estimation) — a recipe for picking a parameter (here $λ$ ): try every candidate value and keep the one under which the observed data is most probable. scipy's boxcox uses MLE to find the $λ$ that makes the transformed data look most normal.

The picture below shows what all these transforms have in common: they take a long right tail and squeeze it while stretching the small values apart.

Look at the red log curve: near $x = 0$ it climbs steeply (small values get spread apart), and for large $x$ it flattens (big values get squeezed together). That single shape is the engine behind every problem on this page.

Level 1 — Recognition

L1.1 — Which transform is even legal?

You have a feature temp_anomaly with values [-8.2, -3.1, 0.5, 2.3, 7.8]. A teammate writes np.log(temp_anomaly). What breaks, and which of {Log, Box-Cox, Yeo-Johnson} is the only valid choice?

Recall Solution

$lo g$ is only defined for strictly positive inputs. Here $- 8.2$ and $- 3.1$ are negative, and $lo g$ of a negative number is undefined (you get NaN). Box-Cox also requires $x > 0$ , so it is out too. The only tool that legally accepts negatives is Yeo-Johnson — it has a separate branch for $x < 0$ . See 2.1.08-Handling-Outliers for the sibling problem of extreme values.

Answer: Yeo-Johnson.

L1.2 — Name the $λ$

Match each Box-Cox $λ$ to the plain-English transform it equals: $λ \in {1, 0.5, 0, - 1}$ .

Recall Solution

Plug into $\frac{x ^{λ} - 1}{λ}$ (with $λ = 0$ meaning $lo g x$ ):

$λ = 1$ → $\frac{x ^{1} - 1}{1} = x - 1$ → essentially no transform (linear shift).
$λ = 0.5$ → square root shape.
$λ = 0$ → log.
$λ = - 1$ → $\frac{x ^{- 1} - 1}{- 1} = 1 - \frac{1}{x}$ → reciprocal shape.

L1.3 — Why does Box-Cox become log at $λ = 0$ ?

The formula $\frac{x ^{λ} - 1}{λ}$ has a $λ$ in the denominator — so at $λ = 0$ you divide by zero. Yet the definition says it "equals $lo g x$ " there. Show why by taking the limit $λ \to 0$ .

Recall Solution

At $λ = 0$ the top is $x^{0} - 1 = 1 - 1 = 0$ and the bottom is $0$ : a $\frac{0}{0}$ form. That's exactly the case L'Hôpital's rule was built for — when top and bottom both vanish, the ratio's limit equals the ratio of their derivatives (with respect to $λ$ ).

Derivative of the top $x^{λ} - 1$ w.r.t. $λ$ : write $x^{λ} = e^{λ l o g x}$ , whose derivative is $lo g x \cdot e^{λ l o g x} = x^{λ} lo g x$ .
Derivative of the bottom $λ$ w.r.t. $λ$ : just $1$ .

So $lim_{λ \to 0} \frac{x ^{λ} - 1}{λ} = lim_{λ \to 0} \frac{x ^{λ} l o g x}{1} = x^{0} lo g x = lo g x .$ That is why the $- 1$ and the $\div λ$ are in the formula: they glue the power family smoothly onto the log at $λ = 0$ , leaving no gap. Without them the curve would jump.

Level 2 — Application

L2.1 — Log10 of housing prices

Compute $x^{'} = lo g_{10} (x)$ for $x = 250000$ . Round to 2 decimals.

Recall Solution

$lo g_{10} (250000) = lo g_{10} (2.5 \times 1 0^{5}) = lo g_{10} (2.5) + 5 = 0.39794 + 5 = 5.40$ . Answer: $5.40$ . (This is why log compresses: a quarter-million becomes a number near 5.)

L2.2 — Box-Cox by hand

Apply Box-Cox with $λ = - 0.1$ to $x = 450$ . Formula: $x^{'} = \frac{x ^{λ} - 1}{λ}$ .

Recall Solution

Rewrite $\frac{x ^{- 0.1} - 1}{- 0.1} = 10 (1 - x^{- 0.1})$ . Now $45 0^{- 0.1} = e^{- 0.1 l n 450} = e^{- 0.1 \times 6.1092} = e^{- 0.61092} = 0.5428$ . So $x^{'} = 10 (1 - 0.5428) = 10 \times 0.4572 = 4.57$ . Answer: $x^{'} \approx 4.57$ . (The parent note's $4.52$ used a slightly rounded intermediate; the exact value is $4.57$ .)

L2.3 — Yeo-Johnson on a negative value

Apply Yeo-Johnson with $λ = 0.8$ to $x = - 8.2$ . For $x < 0, λ \neq = 2$ : $x^{'} = - \frac{( - x + 1 ) ^{2 - λ} - 1}{2 - λ} .$

Recall Solution

Here $- x + 1 = 8.2 + 1 = 9.2$ and $2 - λ = 1.2$ . $9. 2^{1.2} = e^{1.2 l n 9.2} = e^{1.2 \times 2.2192} = e^{2.6631} = 14.34$ . $x^{'} = - \frac{14.34 - 1}{1.2} = - \frac{13.34}{1.2} = - 11.12$ . Answer: $x^{'} \approx - 11.12$ .

L2.4 — The forgotten edge case: Yeo-Johnson at $λ = 2$ , $x < 0$

The formula in L2.3 divides by $2 - λ$ . What happens at $λ = 2$ , and what does the transform become for a negative $x$ there? Evaluate it for $x = - 8.2$ .

Recall Solution

At $λ = 2$ the denominator $2 - λ = 0$ — division by zero, another $\frac{0}{0}$ gap, exactly like Box-Cox at $λ = 0$ . Taking the limit $λ \to 2$ (same L'Hôpital move) collapses the negative branch to $x^{'} = - lo g (- x + 1) .$ This is the fourth line of the full Yeo-Johnson definition, the mirror-image partner of the $x \geq 0, λ = 0$ log case. For $x = - 8.2$ : $- lo g (8.2 + 1) = - lo g (9.2) = - 2.2192$ . Answer: at $λ = 2$ the negative branch is $- lo g (- x + 1)$ ; for $x = - 8.2$ this gives $\approx - 2.219$ .

Level 3 — Analysis

L3.1 — Log vs. sqrt: which is gentler?

Data [1, 4, 100] is right-skewed. Compute the ratio (max − min) after $x$ vs after $ln (x)$ . Which compresses the tail more, and why might "more" be too much?

Recall Solution

$x$ : $[1, 2, 10]$ → spread $= 10 - 1 = 9$ .
$ln (x)$ : $[0, 1.386, 4.605]$ → spread $= 4.605 - 0 = 4.605$ . Log compresses more (spread $4.6$ vs $9$ ). But "more" isn't always better: over-compressing can flatten real structure the model needs. When skew is only moderate, $x$ preserves more relative differences — hence the parent's rule "moderately skewed → $x$ ". Answer: log compresses more; use $x$ when you don't want that much bending.

L3.2 — Reading a skewness table

Given the candidate scan below, pick $λ$ and justify. $λ = 1.0 \to 2.1, 0.5 \to 1.3, 0.0 \to 0.4, - 0.5 \to - 0.2$

Recall Solution

We want skewness closest to $0$ (most symmetric). $∣2.1∣, ∣1.3∣, ∣0.4∣, ∣0.2∣$ → smallest absolute value is $0.2$ at $λ = - 0.5$ . But note $λ = 0.0$ gives $0.4$ (still small) and is far more interpretable (it's plain log). In practice you'd interpolate: the optimum sits between $- 0.5$ and $0$ , near $λ \approx - 0.1$ where skewness crosses zero. Answer: optimum near $λ \approx - 0.1$ ; among the listed values, $λ = - 0.5$ is most symmetric.

Level 4 — Synthesis

L4.1 — Round-trip a log prediction

You trained on $lo g$ -price. The model predicts $\overset{y}{^}_{l o g} = 12.4$ . What is the raw-dollar point prediction before any correction?

Recall Solution

Inverse of natural log is $exp$ : $\overset{y}{^} = e^{12.4}$ . $e^{12.4} = e^{12} \cdot e^{0.4} = 162754.79 \times 1.49182 = 242801$ . Answer: $\hat y \approx \$ 242{,}801$.

L4.2 — Jensen bias correction

Continuing L4.1: residual variance in log-space is $σ^{2} = 0.16$ . Apply the bias correction $\overset{y}{^}_{mean} = exp (\overset{y}{^}_{l o g} + \frac{σ ^{2}}{2})$ . By what percentage does the corrected mean exceed the naive point estimate? And why is the factor $e^{σ^{2} /2}$ ?

Recall Solution

Where the factor comes from. In log-space the residual is a symmetric bell (a normal) $Z$ with mean $μ = \overset{y}{^}_{l o g}$ and variance $σ^{2}$ . The dollar prediction is $y = e^{Z}$ , so $y$ is log-normal. We want its average, $E [e^{Z}]$ . Look at the figure: $e^{Z}$ is a curved-up (convex) function. Jensen's inequality says that for a convex curve, the average of the curve sits above the curve of the average — pushing bumps up more than dips pull down. So $E [e^{Z}] > e^{E [Z]} = e^{μ}$ ; the naive $e^{μ}$ under-shoots. The exact gap for a normal $Z$ is the standard result $E [e^{Z}] = e^{μ + \frac{σ ^{2}}{2}},$ so the correction factor over the naive estimate is $e^{σ^{2} /2}$ — bigger variance, bigger upward push, exactly as the convex picture predicts.

The number. Correction factor $= e^{σ^{2} /2} = e^{0.08} = 1.08329$ . So the corrected mean is $8.33%$ larger than the naive point estimate. Numerically: $242801 \times 1.08329 = \$ 263{,}028 $. * * A n s w er :$ +8.33% $, g i v in g$ \approx $263{,}028$.**

L4.3 — Invert Box-Cox

A Box-Cox model ( $λ = 0.5$ ) outputs $x^{'} = 6$ . Recover $x$ using $x = (λ x^{'} + 1)^{1/ λ}$ .

Recall Solution

$x = (0.5 \times 6 + 1)^{1/0.5} = (3 + 1)^{2} = 4^{2} = 16$ . Answer: $x = 16$ . (Check forward: $\frac{1 6 ^{0.5} - 1}{0.5} = \frac{4 - 1}{0.5} = 6$ . ✓)

Level 5 — Mastery

L5.1 — Design a leak-free pipeline

You have income (all positive, right-skewed) as target and a train/test split. Order these into a correct, leak-free sequence, and state where $λ$ is learned: {transform test, split, learn λ on train, transform train, fit model, inverse-transform predictions}.

Recall Solution

Correct order:

Split into train / test.
Learn $λ$ on train only — via MLE (maximum likelihood estimation: pick the $λ$ under which the training data looks most normal; scipy.stats.boxcox does this). This is the one place a parameter is fit.
Transform train with that $λ$ .
Fit model on transformed train.
Transform test using the same $λ$ (never refit).
Inverse-transform predictions back to dollars (with Jensen correction if reporting means).

The leak everyone fears is computing $λ$ on all data before splitting — that lets the test distribution whisper into training. Learning $λ$ on train only is correct, not a mistake. Mirrors the fit-on-train rule from 2.1.05-Feature-Scaling.

L5.2 — Transform or switch models?

Your only model option becomes a decision tree. Your teammate still wants to Box-Cox the features "for normality." Should you? Explain in one line what trees care about.

Recall Solution

No — skip it for pure tree models. Trees split on thresholds ("is $x > t$ ?"), which depend only on the rank order of values. Any monotonic transform (log, Box-Cox with the ranks preserved) leaves every possible split point equivalent, so the tree's decisions are unchanged. Transforms matter for distance/linearity-based models, not rank-based ones. (Contrast with 2.2.03-Polynomial-Features, which do change what trees and linear models can express.) Answer: don't transform for a lone decision tree — no benefit.

L5.3 — Choose the transform, end to end

A feature is a proportion $p \in (0, 1)$ , e.g. click-through rate, and you're feeding a linear model. Log? Box-Cox? Something else? Transform $p = 0.2$ .

Recall Solution

Neither log nor Box-Cox is ideal: they don't respect the ceiling at $1$ . For proportions use the logit: $logit (p) = lo g \frac{p}{1 - p}$ , which maps $(0, 1)$ onto the whole real line $(- \infty, \infty)$ — exactly what a linear model wants. For $p = 0.2$ : $lo g \frac{0.2}{0.8} = lo g (0.25) = - 1.3863$ . Answer: use logit; $logit (0.2) \approx - 1.386$ .

Recall Self-test checklist (reveal to grade yourself)

Legal transform for negatives ::: Yeo-Johnson (log & Box-Cox need $x > 0$ ) Why Box-Cox = log at $λ = 0$ ::: L'Hôpital limit of $(x^{λ} - 1) / λ$ gives $lo g x$ $lo g_{10} (250000)$ ::: $5.40$ Box-Cox $λ = - 0.1$ on $x = 450$ ::: $\approx 4.57$ Yeo-Johnson $λ = 0.8$ on $x = - 8.2$ ::: $\approx - 11.12$ Yeo-Johnson $λ = 2$ , $x = - 8.2$ ::: $- lo g (9.2) \approx - 2.219$ Inverse-log of $12.4$ ::: $\approx \$ 242{,}801 $J e n se n f a c t or$ e^{\sigma^2/2} $,$ \sigma^2=0.16 $:::$ +8.33% $($ \approx $263{,}028 $) W h y$ e^{\sigma^2/2} $:::$ e^Z $i sco n v e x, J e n se n l i f t s t h e m e anab o v e$ e^{\mu} $I n v er tB o x - C o x$ \lambda=0.5 $,$ x'=6 $:::$ x=16 $‘ l o g 1 p ‘ m e anin g :::$ \log(1+x) $, s a f e a t$ x=0 $M L E m e anin g ::: ma x im u m l ik e l ih oo d es t ima t i o n — p i c k$ \lambda $mak in g d a t am os tp r o bab l e l o g i t$ (0.2) $:::$ \approx -1.386$