Covariance and correlation: Worked examples — AI-ML deep dive

This page is a case atlas. The parent note gave you the formulas; here we walk through every kind of situation those formulas can meet — every sign, the zero cases, the degenerate cases, the limiting values, a real-world word problem, and an exam twist. Nothing is left as "you'll figure out the rest".

If you have not yet met $E [\cdot]$ (the average of a random variable) or $Var$ (its spread), pause and read 1.3.05-Expectation-and-variance first — every step below leans on them. The idea of two variables having a joint table of probabilities comes from 1.3.08-Joint-and-marginal-distributions, and the notion of "moving together" ties back to 1.3.01-Random-variables.

Two worlds: population vs sample — pin down every symbol

Before any example we must be crystal-clear about which world we are in, because covariance and correlation each have two versions and the page will use both.

Definition The four symbols, defined once

Population world (we know the full probability distribution, so we use the true average $E [\cdot]$ ):

$Cov (X, Y) = E [(X - μ_{X}) (Y - μ_{Y})]$ — the population covariance. Here $μ_{X} = E [X]$ , $μ_{Y} = E [Y]$ are the true means.
$ρ_{X, Y} = \frac{Cov ( X , Y )}{σ _{X} σ _{Y}}$ — the population correlation (Greek $ρ$ , "rho"). $σ_{X} = Var (X)$ .

Sample world (we only have $n$ data points $(x_{1}, y_{1}), \dots, (x_{n}, y_{n})$ , so we replace $E [\cdot]$ with empirical sums):

$s_{X Y} = \frac{1}{n - 1} \sum_{i = 1}^{n} (x_{i} - \overset{x}{ˉ}) (y_{i} - \overset{y}{ˉ})$ — the sample covariance (Roman $s$ ). $\overset{x}{ˉ}, \overset{y}{ˉ}$ are the data means.
$s_{X} = \frac{1}{n - 1} \sum_{i = 1}^{n} (x_{i} - \overset{x}{ˉ})^{2}$ and $s_{Y} = \frac{1}{n - 1} \sum_{i = 1}^{n} (y_{i} - \overset{y}{ˉ})^{2}$ — the sample standard deviations. In plain words: $s_{X}$ is the typical distance of an $x$ -value from the data centre $\overset{x}{ˉ}$ (square each gap, average with the $n - 1$ divisor, un-square with the root). It measures how wide the $X$ -cloud is.
$r_{X Y} = \frac{s _{X Y}}{s _{X} s _{Y}}$ — the sample correlation (Roman $r$ , "Pearson's r"). It divides the raw teamwork $s_{X Y}$ by the two cloud-widths so units cancel.

Intuition The Cauchy–Schwarz bound, sketched (why

$∣ ρ ∣ \leq 1$ ) Correlation can never leave $[- 1, 1]$ , and Ex 3 hits both ends — so we need to see why that ceiling exists.

Picture each variable, after centring, as an arrow: list all the deviations $x_{i} - \overset{x}{ˉ}$ as the components of an arrow $u$ , and all the $y_{i} - \overset{y}{ˉ}$ as an arrow $v$ . Then the correlation numerator $\sum (x_{i} - \overset{x}{ˉ}) (y_{i} - \overset{y}{ˉ})$ is exactly the dot product $u \cdot v$ , and each denominator root is a length $∥ u ∥, ∥ v ∥$ . So $r = \frac{u \cdot v}{∥ u ∥ ∥ v ∥} = cos θ$ , the cosine of the angle between the two arrows. A cosine is always between $- 1$ and $+ 1$ — that is the Cauchy–Schwarz inequality $∣ u \cdot v ∣ \leq ∥ u ∥ ∥ v ∥$ in disguise. Equality (cosine $= \pm 1$ ) happens only when the arrows are parallel, i.e. one variable is an exact linear function of the other.

The scenario matrix

Now list every distinct case covariance/correlation can present. Each row is a "cell" the reader must handle; each worked example is tagged with the cell it fills and the world it uses.

Cell	Situation	World	Filled by
A	Positive: $s_{X Y} > 0$	sample	Ex 1
B	Negative: $s_{X Y} < 0$	sample	Ex 2
C	Limit $r = + 1$	sample	Ex 3
D	Limit $r = - 1$	sample	Ex 3
E	Zero because independent, $Cov = 0$	population	Ex 4
F	Zero despite dependence, $ρ = 0$	population	Ex 5
G	Degenerate constant, $ρ$ undefined	sample	Ex 6
H	Scale change (units trap)	sample	Ex 7
I	Real-world word problem	sample	Ex 8
J	Exam twist: linear transform	population	Ex 9

Figure s01 is a 2×2 gallery, one panel per cell, with each panel titled by its cell letter and its target $ρ$ , and with the mean-lines drawn in so you can literally see which quadrants the dots fall in. Read it like this:

top-left, Cell A ( $ρ > 0$ ): dots hug a rising trend — most sit in the "both above mean / both below mean" quadrants (products positive).
top-right, Cell B ( $ρ < 0$ ): dots hug a falling trend — opposite quadrants (products negative).
bottom-left, Cell E ( $ρ = 0$ , independent): a shapeless round blob, no tilt.
bottom-right, Cell F ( $ρ = 0$ , but a parabola): a perfect ∪ curve — fully determined, yet symmetric left-right so the tilt cancels to zero. That last panel is the trap.

Alt text: four scatter panels arranged 2×2. Top-left, rising blue cloud labelled Cell A rho positive. Top-right, falling orange cloud labelled Cell B rho negative. Bottom-left, round green blob labelled Cell E rho zero independent. Bottom-right, a red U-shaped parabola labelled Cell F rho zero curve. Dashed grey mean-lines cross each panel dividing it into four quadrants.

Cell A — Positive sample covariance

Worked example Ex 1 · Study hours vs test score (Cell A, sample world)

A tiny class has four students. Hours studied $X$ and score $Y$ : $(2, 50), (4, 60), (6, 65), (8, 85)$ Compute the sample covariance $s_{X Y}$ and confirm its sign.

Forecast: More study → more score, so guess $s_{X Y} > 0$ .

Step 1 — Data means. $\overset{x}{ˉ} = \frac{2 + 4 + 6 + 8}{4} = 5, \overset{y}{ˉ} = \frac{50 + 60 + 65 + 85}{4} = 65$ Why this step? Sample covariance measures deviation from the data centre; we need $\overset{x}{ˉ}, \overset{y}{ˉ}$ first.

Step 2 — Deviation products $(x_{i} - \overset{x}{ˉ}) (y_{i} - \overset{y}{ˉ})$ . $(- 3) (- 15) = 45, (- 1) (- 5) = 5, (1) (0) = 0, (3) (20) = 60$ Why this step? Each product is positive when both coordinates sit on the same side of their mean — that "same-side" agreement is exactly what covariance rewards.

Step 3 — Sample covariance (divide by $n - 1 = 3$ , Bessel's correction): $s_{X Y} = \frac{45 + 5 + 0 + 60}{3} = \frac{110}{3} \approx 36.67$

Verify: Positive as forecast. Units: hours × points, positive — every non-negative product means no student bucked the trend, consistent with a clean upward cloud.

Cell B — Negative sample covariance

Worked example Ex 2 · Altitude vs temperature (Cell B, sample world)

Climb a mountain: altitude $X$ (km) vs temperature $Y$ (°C): $(0, 20), (1, 14), (2, 8), (3, 2)$ Find $s_{X Y}$ .

Forecast: Higher → colder, so guess $s_{X Y} < 0$ .

Step 1 — Data means. $\overset{x}{ˉ} = 1.5$ , $\overset{y}{ˉ} = 11$ . Why this step? Same reason as Ex 1 — locate the data centre.

Step 2 — Deviation products. $(- 1.5) (9) = - 13.5, (- 0.5) (3) = - 1.5, (0.5) (- 3) = - 1.5, (1.5) (- 9) = - 13.5$ Why this step? When $x$ is below its mean but $y$ is above (or vice-versa), the product is negative — opposite-side disagreement is what a negative covariance records.

Step 3 — Sample covariance. $s_{X Y} = \frac{- 13.5 - 1.5 - 1.5 - 13.5}{3} = \frac{- 30}{3} = - 10$

Verify: Negative as forecast. Units °C·km, negative. Every product was negative → a perfectly consistent downhill trend.

Cells C & D — the extremes $r = \pm 1$

Figure s02 draws both perfect lines used in this example. Look at the blue rising line (Cell C) and the orange falling line (Cell D): every data dot sits exactly on its line — no scatter at all. That perfect fit is what forces $r$ (the cosine of the angle between the two centred arrows from the box above) to $\pm 1$ : perfectly parallel arrows, angle $0°$ or $180°$ .

Alt text: an X-Y plane with two straight lines of four dots each. A blue line rises left-to-right through the points one-three, two-five, three-seven, four-nine, labelled Y equals two X plus one, r equals plus one, Cell C. An orange line falls through one-minus-one, two-minus-three, three-minus-five, four-minus-seven, labelled Y equals minus two X plus one, r equals minus one, Cell D. A grey horizontal axis marks y equals zero.

Worked example Ex 3 · Points exactly on a line (Cells C and D, sample world)

Part (a) rising line: $Y = 2 X + 1$ on data $x = 1, 2, 3, 4$ → $(1, 3), (2, 5), (3, 7), (4, 9)$ . Part (b) falling line: $Y = - 2 X + 1$ → $(1, - 1), (2, - 3), (3, - 5), (4, - 7)$ . Compute sample correlation $r$ for each.

Forecast: On a perfect line, $r = + 1$ (rising) and $r = - 1$ (falling) — the cosine hits its extremes.

Step 0 — Why the $n - 1$ vanishes. Sample correlation is $r = \frac{s _{X Y}}{s _{X} s _{Y}}$ . Written out with the divisors (using the definitions of $s_{X}, s_{Y}$ from the symbol box), $r = \frac{\frac{1}{n - 1} \sum ( x _{i} - x ˉ ) ( y _{i} - y ˉ )}{\frac{1}{n - 1} \sum ( x _{i} - x ˉ ) ^{2} \frac{1}{n - 1} \sum ( y _{i} - y ˉ ) ^{2}} = \frac{\frac{1}{n - 1} \sum ( \dots )}{\frac{1}{n - 1} \sum ( x _{i} - x ˉ ) ^{2} \sum ( y _{i} - y ˉ ) ^{2}} .$ Why this step? The denominator's two square roots each carry $1/ (n - 1)$ ; multiplied they give $1/ (n - 1)$ , the same factor as the numerator. It cancels top-and-bottom, so $r$ reduces to the divisor-free form $r = \frac{\sum ( x _{i} - x ˉ ) ( y _{i} - y ˉ )}{\sum ( x _{i} - x ˉ ) ^{2} \sum ( y _{i} - y ˉ ) ^{2}} .$ This is why $r$ is identical whether you use $n$ or $n - 1$ — we may work with raw sums below.

Step 1 — Deviations of $x$ : $\overset{x}{ˉ} = 2.5$ , so $x_{i} - \overset{x}{ˉ} = - 1.5, - 0.5, 0.5, 1.5$ ; $\sum (x_{i} - \overset{x}{ˉ})^{2} = 5$ . Why this step? $r$ 's denominator needs the spread of each variable separately.

Step 2 — Part (a). $\overset{y}{ˉ} = 6$ , deviations $- 3, - 1, 1, 3$ ; $\sum (y - \overset{y}{ˉ})^{2} = 20$ . Cross-sum $\sum (x - \overset{x}{ˉ}) (y - \overset{y}{ˉ}) = 4.5 + 0.5 + 0.5 + 4.5 = 10$ . $r = \frac{10}{5 20} = \frac{10}{100} = \frac{10}{10} = 1$ Why this step? When $y$ is an exact linear function of $x$ with positive slope, the two centred arrows are parallel — the dot product equals the product of lengths, so the cosine is exactly $+ 1$ . That is why the ceiling is hit here.

Step 3 — Part (b). $\overset{y}{ˉ} = - 4$ , deviations $3, 1, - 1, - 3$ ; $\sum (y - \overset{y}{ˉ})^{2} = 20$ . Cross-sum $= - 4.5 - 0.5 - 0.5 - 4.5 = - 10$ . $r = \frac{- 10}{5 20} = \frac{- 10}{10} = - 1$

Verify: Exactly $\pm 1$ , matching the Cauchy–Schwarz bound $∣ ρ ∣ \leq 1$ where equality occurs precisely when one variable is a linear function of the other. Slope sign sets the sign; slope magnitude is irrelevant (scale cancels).

Cell E — zero because truly independent

Worked example Ex 4 · A fair coin and a fair die (Cell E, population world)

$X$ = coin ( $0$ or $1$ , each prob $\frac{1}{2}$ ), $Y$ = die ( $1..6$ , each $\frac{1}{6}$ ), rolled independently. Find the population covariance $Cov (X, Y)$ .

Forecast: Independent → covariance $0$ .

Step 1 — Marginal means. $E [X] = \frac{1}{2}$ , $E [Y] = \frac{1 + 2 + \dots + 6}{6} = 3.5$ . Why this step? We will use the computational shortcut $Cov = E [X Y] - E [X] E [Y]$ (derived in the formula box above), which needs both marginals. We are in the population world because the full distribution is given.

Step 2 — $E [X Y]$ under independence. Because the events don't influence each other, the joint probability factorises, so $E [X Y] = E [X] E [Y] = \frac{1}{2} \cdot 3.5 = 1.75$ Why this step? This factorisation is the definition consequence of 1.2.04-Independence: knowing $X$ tells you nothing about $Y$ .

Step 3 — Covariance. $Cov (X, Y) = 1.75 - 1.75 = 0$ Why this step? Plug the two numbers into the shortcut $E [X Y] - E [X] E [Y]$ ; because $E [X Y]$ equalled $E [X] E [Y]$ (Step 2), the subtraction lands exactly on $0$ .

Verify: Zero — and here zero really does mean unrelated. This is the "safe" zero.

Cell F — zero despite perfect dependence

Worked example Ex 5 ·

$Y = X^{2}$ on symmetric points (Cell F, population world) Let $X$ take values $- 2, - 1, 1, 2$ each with probability $\frac{1}{4}$ , and $Y = X^{2}$ . Find $Cov (X, Y)$ and $ρ_{X, Y}$ .

Forecast: $Y$ is completely determined by $X$ , yet the link is nonlinear and symmetric — guess $Cov = 0$ .

Step 1 — Means. $E [X] = \frac{- 2 - 1 + 1 + 2}{4} = 0$ . $Y$ values are $4, 1, 1, 4$ , so $E [Y] = \frac{10}{4} = 2.5$ . Why this step? Symmetry of $X$ around $0$ is the engine that will kill the covariance.

Step 2 — $E [X Y] = E [X^{3}]$ . Values $X^{3} = - 8, - 1, 1, 8$ , mean $= \frac{- 8 - 1 + 1 + 8}{4} = 0$ . Why this step? $X \cdot X^{2} = X^{3}$ is an odd function; a symmetric distribution makes its average vanish. This is the exact mechanism behind "nonlinear ⇒ can dodge correlation".

Step 3 — Covariance & correlation. $Cov (X, Y) = 0 - (0) (2.5) = 0 \Rightarrow ρ_{X, Y} = 0$ Why this step? Same shortcut $E [X Y] - E [X] E [Y]$ ; with $E [X] = 0$ the whole second term dies, and $E [X Y] = 0$ too, so covariance (and hence $ρ$ ) is exactly $0$ .

Verify: Zero correlation, yet if you know $X$ you know $Y$ exactly. This is the danger case: zero correlation does not mean independence. (Contrast Ex 4, where the zero was honest.) Relevant when choosing features — see 4.1.03-Feature-selection.

Cell G — degenerate: a constant variable

Worked example Ex 6 · One variable never changes (Cell G, sample world)

$X$ takes values $3, 5, 9$ ; $Y = 7$ always. Compute $s_{X Y}$ and then attempt $r$ .

Forecast: A flat $Y$ can't co-vary with anything → covariance $0$ ; but correlation will misbehave.

Step 1 — Deviations of $Y$ . Since $Y$ is constant, $y_{i} - \overset{y}{ˉ} = 0$ for every $i$ . Why this step? Sample covariance sums $(x_{i} - \overset{x}{ˉ}) (y_{i} - \overset{y}{ˉ})$ ; multiplying by zero every time forces the total to $0$ .

Step 2 — Covariance. $s_{X Y} = \frac{1}{n - 1} \sum (x_{i} - \overset{x}{ˉ}) \cdot 0 = 0$

Step 3 — Correlation. $r = \frac{0}{s _{X} \cdot s _{Y}} = \frac{0}{s _{X} \cdot 0} = \frac{0}{0} — undefined.$ Why this step? $s_{Y} = \frac{1}{n - 1} \sum (y_{i} - \overset{y}{ˉ})^{2} = 0$ for a constant (its arrow has zero length), so we divide by zero. Correlation does not exist, it is not "zero".

Verify: Covariance $= 0$ is legitimate; correlation is genuinely undefined. In code, a constant column returns NaN correlation — always drop zero-variance features before correlating (a real preprocessing rule in 3.2.06-Multicolinearity).

Cell H — the units trap

Worked example Ex 7 · Minutes vs hours (Cell H, sample world)

Take Ex 1's data but express hours in minutes instead: multiply every $x$ by $60$ . New $X^{'} = 60 X$ , same $Y$ . What happens to $s_{X^{'} Y}$ and to $r_{X^{'} Y}$ ?

Forecast: Covariance should scale by $60$ ; correlation should not move at all.

Step 1 — Covariance scaling. Every deviation $x_{i} - \overset{x}{ˉ}$ becomes $60 (x_{i} - \overset{x}{ˉ})$ , so each product scales by $60$ : $s_{X^{'} Y} = 60 \cdot s_{X Y} = 60 \cdot \frac{110}{3} = 2200$ Why this step? Sample covariance is linear in each variable's scale — this is why its magnitude is meaningless across unit choices.

Step 2 — Correlation scaling. Numerator scales by $60$ ; but $s_{X^{'}} = 60 s_{X}$ (the cloud-width stretches by the same factor), so the denominator also scales by $60$ : $r_{X^{'} Y} = \frac{60 s _{X Y}}{60 s _{X} s _{Y}} = \frac{s _{X Y}}{s _{X} s _{Y}} = r_{X Y}$ Why this step? The $60$ cancels top and bottom — that cancellation is the whole point of normalising. (In the arrow picture, stretching one arrow doesn't change its angle, so the cosine $r$ is untouched.)

Verify: $s$ jumped from $36.67$ to $2200$ ; $r$ is identical before and after. Use correlation to compare strengths — this scale-invariance is why 2.4.01-Principal-component-analysis often works on the correlation matrix.

Cell I — real-world word problem

Figure s03 plots the four data points below with their mean-lines, so you can see every dot land in the top-right or bottom-left quadrant — the visual signature of a strong positive $r$ before any arithmetic.

Alt text: a scatter of four green dots rising steeply from lower-left to upper-right, at ice-cream sales two-one, four-two, six-four, eight-five. Dashed grey vertical line at mean x equals five and horizontal line at mean y equals three split the plane; every dot sits in the lower-left or upper-right quadrant, signalling positive correlation.

Worked example Ex 8 · Ice-cream sales vs drownings (Cell I, sample world)

Over four summer weeks: ice-cream sales $X$ (thousands) and drownings $Y$ : $(2, 1), (4, 2), (6, 4), (8, 5)$ Find $r$ and discuss.

Forecast: Both rise with heat → strong positive $r$ , near $+ 0.99$ (look at the figure — every dot in the same-side quadrants).

Step 1 — Data means. $\overset{x}{ˉ} = \frac{2 + 4 + 6 + 8}{4} = 5$ , $\overset{y}{ˉ} = \frac{1 + 2 + 4 + 5}{4} = 3$ . Why this step? Standard first move for the sample correlation formula — locate the centre so we can measure deviations, exactly the mean-lines drawn in the figure.

Step 2 — The three sums (using the divisor-free $r$ from Ex 3, Step 0). Deviations $x$ : $- 3, - 1, 1, 3$ ; deviations $y$ : $- 2, - 1, 1, 2$ . Numerator (agreement), product term-by-term: $(- 3) (- 2) = 6, (- 1) (- 1) = 1, (1) (1) = 1, (3) (2) = 6 \Rightarrow \sum (x - \overset{x}{ˉ}) (y - \overset{y}{ˉ}) = 14$ Spreads: $\sum (x - \overset{x}{ˉ})^{2} = 9 + 1 + 1 + 9 = 20, \sum (y - \overset{y}{ˉ})^{2} = 4 + 1 + 1 + 4 = 10$ Why this step? $r$ is numerator (do the dots agree?) divided by the two cloud-widths; we build each piece separately, and every product being positive already tells us the sign is $+$ .

Step 3 — Correlation. $r = \frac{14}{20 10} = \frac{14}{200} = \frac{14}{14.142} \approx 0.9899$ Why this step? Combine the pieces in the divisor-free formula; the near- $1$ value confirms the almost-perfect line the figure already shows.

Verify: $r \approx 0.99$ (and $0 \leq ∣ r ∣ \leq 1$ as required), strong positive — matching the tight upward cloud in Figure s03. But ice-cream does not cause drowning — a hidden variable (summer heat) drives both. Correlation ≠ causation; this confounding is the classic trap flagged in the parent note and central to 3.2.05-Linear-regression.

Cell J — exam twist: linear transform

Worked example Ex 9 ·

$Cov (a X + b, c Y + d)$ (Cell J, population world) Given $Cov (X, Y) = 0.2$ (from the parent note's Example 1). Let $U = 3 X + 5$ and $V = - 2 Y + 7$ . Find $Cov (U, V)$ and the sign of $ρ_{U, V}$ versus $ρ_{X, Y}$ .

Forecast: The $+ 5$ and $+ 7$ shifts shouldn't matter; multipliers $3$ and $- 2$ scale it; the $- 2$ flips the sign.

Step 1 — Constants add nothing. Adding $b$ or $d$ shifts the mean equally, so deviations $U - μ_{U} = 3 (X - μ_{X})$ and $V - μ_{V} = - 2 (Y - μ_{Y})$ — the constants $5, 7$ vanish. Why this step? Covariance depends only on deviations from the mean; a rigid shift moves point and mean together, so the gap between them is unchanged.

Step 2 — Pull out the multipliers. $Cov (U, V) = E [3 (X - μ_{X}) \cdot (- 2) (Y - μ_{Y})] = 3 \cdot (- 2) Cov (X, Y)$ $= - 6 \times 0.2 = - 1.2$ Why this step? Constants pull out of the expectation (linearity), giving the general law $Cov (a X + b, c Y + d) = a c Cov (X, Y)$ .

Step 3 — Correlation sign. Standard deviations scale by $∣3∣$ and $∣ - 2∣$ (always positive), so $ρ_{U, V} = \frac{a c Cov ( X , Y )}{∣ a ∣ ∣ c ∣ σ _{X} σ _{Y}} = sign (a c) ρ_{X, Y}$ Here $sign (3 \cdot (- 2)) = - 1$ , so $ρ_{U, V} = - ρ_{X, Y}$ : the negative multiplier flips the correlation. Why this step? The magnitude $∣ a ∣∣ c ∣$ cancels between numerator and denominator (scale-invariance again); only the sign of $a c$ survives.

Verify: $Cov (U, V) = - 1.2$ ; correlation magnitude unchanged, sign flipped by the single negative factor. The additive constants never appear — exactly as forecast.

Recall

Recall Population vs sample symbols

Which symbols belong to which world? ::: Population: $E [\cdot], Cov, ρ, σ$ . Sample: $\overset{x}{ˉ}, s_{X Y}, r, s_{X}$ .

Recall Definition of

$s_{X}$ How is the sample standard deviation $s_{X}$ built? ::: $s_{X} = \frac{1}{n - 1} \sum (x_{i} - \overset{x}{ˉ})^{2}$ — the typical distance of $x$ -values from $\overset{x}{ˉ}$ .

Recall Does the

$n - 1$ affect $r$ ? Why does sample correlation not care whether you divide by $n$ or $n - 1$ ? ::: The $\frac{1}{n - 1}$ appears in numerator and (via the two square roots) in the denominator, so it cancels.

Recall Correlation as a cosine

What geometric object is $r$ ? ::: The cosine of the angle between the two centred deviation-arrows; Cauchy–Schwarz forces $∣ cos θ ∣ \leq 1$ .

Recall The two flavours of "zero"

Zero correlation from independence vs from nonlinearity — same? ::: No. Independence ⇒ zero, but zero ⇏ independence (Ex 5's $Y = X^{2}$ ).

Recall Constant variable

What is $r$ when one variable never changes? ::: Undefined ( $s_{Y} = 0$ → divide by zero), even though sample covariance is $0$ .

Recall Linear transform law

$Cov (a X + b, c Y + d) = ?$ ::: $a c Cov (X, Y)$ ; additive constants drop out, and $ρ$ picks up $sign (a c)$ .

Recall Units

Change units of $X$ — what moves, covariance or correlation? ::: Covariance scales; correlation stays fixed (scale-invariant).