Worked examples — Covariance and correlation

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The scenario matrix

Every problem below is one cell of this table. If a cell is filled, you have seen how to handle it.

Cell	Relationship type	What is special / degenerate	Example
A	Positive, non-extreme	ordinary discrete joint pmf	Ex 1
B	Negative	$\rho <0$, deviations disagree	Ex 2
C	Exactly zero, linear boundary	$\rho =0$ from a genuine symmetry, no dependence issue	Ex 3
D	Zero cov but dependent	the famous trap $Y=X^2$	Ex 4
E	Perfect $\rho =+1$	equality case of Cauchy–Schwarz	Ex 5
F	Perfect $\rho =-1$	negative-slope exact line	Ex 5
G	Degenerate input	one variable is a constant ($\sigma =0$) — $\rho$ undefined!	Ex 6
H	Real-world, with units	covariance carries kg·cm; $\rho$ strips units	Ex 7
I	Exam twist (toolbox, not sum)	use bilinearity / $\mathrm{Var}(X+Y)$ instead of raw $E[XY]$	Ex 8

Before we compute anything, one picture to fix what the sign of covariance means geometrically — it will be our compass for the whole page.

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Example 1 — Cell A: ordinary positive relationship

1. Marginals. Sum the joint over the other variable. $P(X=1)=0.2+0.4=0.6$, so $E[X]=0.6$. $P(Y=1)=0.1+0.4=0.5$, so $E[Y]=0.5$. Why this step? Expectation of a 0/1 variable is just $P(\text{variable}=1)$; we need both means before any deviation talk.

2. $E[XY]$. The product $XY=1$ only when both are 1. $E[XY]=1\cdot P(1,1)=0.4$. Why this step? $E[XY]$ is the piece our working formula needs; every other $(x,y)$ pair kills the product.

3. Covariance. $\mathrm{Cov}=E[XY]-E[X]E[Y]=0.4-0.6\cdot 0.5=0.4-0.30=0.10$. Why this step? This is the shortcut formula $E[XY]-E[X]E[Y]$ from the parent — no need to build deviations by hand.

4. Standard deviations. $X$ is Bernoulli$(0.6)$: $\mathrm{Var}(X)=0.6\cdot 0.4=0.24$, ${\sigma }_X=\sqrt{0.24}$. $Y$ is Bernoulli$(0.5)$: ${\sigma }_Y=0.5$. Why this step? $\sigma$ rescales covariance into $\rho$.

5. Correlation. $\rho =\frac{0.10}{\sqrt{0.24}\cdot 0.5}=\frac{0.10}{0.2449}\approx 0.408$.

Verify: $\rho \approx 0.41\in [-1,1]$ ✓, positive as forecast ✓. Sanity: if we'd loaded all mass on $(0,0)$ and $(1,1)$ we'd expect $\rho \to 1$; here the "disagree" corners still carry $0.3$ of the mass, so a moderate $0.41$ is reasonable.

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Example 2 — Cell B: negative relationship

1. Marginals. $P(X=1)=0.4+0.1=0.5\Rightarrow E[X]=0.5$. By the symmetry of the numbers $E[Y]=0.5$. Why this step? Same as before — get the means first.

2. $E[XY]$. Only $(1,1)$ contributes: $E[XY]=1\cdot 0.1=0.1$.

3. Covariance. $\mathrm{Cov}=0.1-0.5\cdot 0.5=0.1-0.25=-0.15$. Why this step? Negative product-average confirms the compass: the disagree quadrants won.

4. Correlation. Both are Bernoulli$(0.5)$, so ${\sigma }_X={\sigma }_Y=0.5$. $\rho =\frac{-0.15}{0.5\cdot 0.5}=-0.6$.

Verify: $\rho =-0.6\in [-1,1]$ ✓, negative ✓. Note it is the mirror image of Example 1 in the parent note (which had $\rho =+0.6$): swapping the weight from agree-corners to disagree-corners flips the sign of the whole relationship.

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Example 3 — Cell C: exactly zero from genuine symmetry

1. Means. $E[X]=\frac{1}{2}(-1)+\frac{1}{2}(+1)=0$, likewise $E[Y]=0$. Why this step? Symmetric values around 0 make the mean 0 — this simplifies everything.

2. $E[XY]$. The four products $(-1)(-1),(-1)(1),(1)(-1),(1)(1)=+1,-1,-1,+1$, each weight $1/4$: $E[XY]=\frac{1}{4}(1-1-1+1)=0$.

3. Covariance. $\mathrm{Cov}=0-0\cdot 0=0$.

4. Correlation. ${\sigma }_X,{\sigma }_Y>0$ (each variance is $1$), so $\rho =0/(1\cdot 1)=0$.

Verify: $\rho =0$ ✓. Cross-check independence: $P(X=1,Y=1)=1/4$ equals $P(X=1)P(Y=1)=\frac{1}{2}\cdot \frac{1}{2}=1/4$, and similarly for all cells — so here independence genuinely holds. Contrast this with the next example.

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Example 4 — Cell D: zero covariance, but fully dependent (the trap)

1. Mean of $X$. $E[X]=\frac{1}{4}(-2-1+1+2)=0$. Why this step? The symmetric spread makes $E[X]=0$, which will kill the correction term.

2. $E[XY]=E[X\cdot X^2]=E[X^3]$. Cubes: $(-8),(-1),(1),(8)$. $E[X^3]=\frac{1}{4}(-8-1+1+8)=0$. Why this step? Substituting $Y=X^2$ turns $E[XY]$ into a moment of $X$ alone.

3. Covariance. $\mathrm{Cov}(X,Y)=E[XY]-E[X]E[Y]=0-0\cdot E[Y]=0$.

4. Independence check — it FAILS. Take $X=2$: then $Y=4$ with certainty, so $P(Y=4\mid X=2)=1\ne P(Y=4)=\frac{1}{2}$. Knowing $X$ changes the distribution of $Y$ ⇒ not independent.

Verify: $\mathrm{Cov}=0$ ✓ yet dependence is total. This is the one-way street from the parent's [!mistake] callout: independence $\Rightarrow \mathrm{Cov}=0$, but $\mathrm{Cov}=0\nRightarrow$ independence. Covariance is blind to the symmetric (even) part of the relationship.

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Example 5 — Cells E & F: the perfect-line boundary $\rho =\pm 1$

1. Up-slope means. $\bar{x}=\frac{1+2+3+4}{4}=2.5$; $\bar{y}=\frac{5+7+9+11}{4}=8$. Why this step? Sample correlation needs the sample means to form deviations.

2. Deviation sums. Deviations of $x$: $-1.5,-0.5,0.5,1.5$; of $y$: $-3,-1,1,3$ (exactly $2\times$ the $x$-deviations, because slope $2$). $S_{xy}=\sum (x_i-\bar{x})(y_i-\bar{y})=(-1.5)(-3)+(-0.5)(-1)+(0.5)(1)+(1.5)(3)=4.5+0.5+0.5+4.5=10$. $S_{xx}=2.25+0.25+0.25+2.25=5$; $S_{yy}=9+1+1+9=20$. Why this step? $r=S_{xy}/\sqrt{S_{xx}{S}_{yy}}$ — the sample version of $\mathrm{Cov}/({\sigma }_X{\sigma }_Y)$.

3. Up-slope $r$. $r=\frac{10}{\sqrt{5\cdot 20}}=\frac{10}{\sqrt{100}}=\frac{10}{10}=1$.

4. Down-slope. Every $y$-deviation flips sign, so $S_{xy}\to -10$ while $S_{xx},S_{yy}$ are unchanged (squares). $r=\frac{-10}{10}=-1$. Why this step? This is property 5 of the toolbox: multiplying $Y$ by a negative scale flips $\mathrm{sign}(\rho )$, magnitude preserved.

Verify: $r=+1$ and $r=-1$ ✓, both on the boundary of $[-1,1]$. This is exactly where $U-tV=0$ in the Cauchy–Schwarz proof — the deviations are perfectly proportional. See Linear Regression: on a perfect line the fitted slope reproduces the data with zero error, so $r^2=1$.

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Example 6 — Cell G: degenerate input, $\rho$ undefined

1. Deviation of $Y$. $E[Y]=7$, so $Y-{\mu }_Y=7-7=0$ for every outcome. Why this step? A constant has zero variance — this is the degenerate feature of the cell.

2. Covariance. $\mathrm{Cov}(X,Y)=E[(X-{\mu }_X)\cdot 0]=0$. Why this step? Any factor of $0$ inside the expectation zeroes the whole thing.

3. Correlation attempt. ${\sigma }_Y=\sqrt{\mathrm{Var}(Y)}=\sqrt{0}=0$, so $\rho =\frac{0}{{\sigma }_X\cdot 0}=\frac{0}{0}$ — undefined (0/0, not "0").

Verify: $\mathrm{Cov}=0$ ✓. But do not report $\rho =0$: with ${\sigma }_Y=0$ the ratio is $0/0$, and correlation is simply not defined when either variable is constant. This is the boundary case people silently mishandle — a constant column in data has no correlation, not "zero correlation".

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Example 7 — Cell H: real-world word problem with units

1. Means. $\bar{h}=\frac{150+160+170+180}{4}=165$ cm; $\bar{w}=\frac{50+60+65+75}{4}=62.5$ kg. Why this step? Deviations need the sample means.

2. Deviation products. $h$-devs: $-15,-5,5,15$; $w$-devs: $-12.5,-2.5,2.5,12.5$. $S_{hw}=(-15)(-12.5)+(-5)(-2.5)+(5)(2.5)+(15)(12.5)=187.5+12.5+12.5+187.5=400$. Sample covariance $=\frac{S_{hw}}{n-1}=\frac{400}{3}\approx 133.3$ cm·kg. Why this step? Sample covariance uses $n-1$ in the denominator; the product carries mixed units.

3. $r$. $S_{hh}=225+25+25+225=500$; $S_{ww}=156.25+6.25+6.25+156.25=325$. $r=\frac{400}{\sqrt{500\cdot 325}}=\frac{400}{\sqrt{162500}}\approx \frac{400}{403.11}\approx 0.9923$.

4. Switch to metres. Heights divide by $100$: by bilinearity $\mathrm{Cov}\to \frac{1}{100}\mathrm{Cov}$, so sample covariance becomes $\approx 1.333$ m·kg. But ${\sigma }_h$ also divides by $100$, so in $r$ the factors cancel: $r$ stays $\approx 0.9923$. Why this step? Property 5 — $\rho$ is scale-invariant; covariance is not.

Verify: covariance $\approx 133.3$ cm·kg (unit-bearing) ✓; $r\approx 0.99\in [-1,1]$ ✓. This is exactly the parent's second [!mistake]: the number covariance changed by 100× under a harmless unit change, while the strength $r$ did not budge. Always judge strength with $r$.

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Example 8 — Cell I: exam twist (use the toolbox, not the raw sum)

1. $\mathrm{Cov}(U,V)$ by bilinearity. $\mathrm{Cov}(2X-1,-Y+5)=(2)(-1)\mathrm{Cov}(X,Y)=-2\cdot 3=-6$. Why this step? Toolbox property 2: shifts $b,d$ vanish, scales $a,c$ multiply out. Faster and cleaner than any $E[XY]$ computation.

2. ${\rho }_{XY}$ first. ${\rho }_{XY}=\frac{\mathrm{Cov}(X,Y)}{{\sigma }_X{\sigma }_Y}=\frac{3}{\sqrt{4}\sqrt{9}}=\frac{3}{2\cdot 3}=0.5$. Why this step? We need it to apply the sign rule.

3. ${\rho }_{UV}$ by scale-invariance. $\mathrm{sign}(ac)=\mathrm{sign}(2\cdot (-1))=-1$, so ${\rho }_{UV}=-{\rho }_{XY}=-0.5$. Why this step? Property 5 — a negative scale flips the sign, magnitude preserved.

4. $\mathrm{Var}(U+V)$. First $\mathrm{Var}(U)=2^2\mathrm{Var}(X)=4\cdot 4=16$; $\mathrm{Var}(V)=(-1{)}^2\mathrm{Var}(Y)=9$. $\mathrm{Var}(U+V)=\mathrm{Var}(U)+\mathrm{Var}(V)+2\mathrm{Cov}(U,V)=16+9+2(-6)=13$. Why this step? Property 3, the variance-of-a-sum identity; the negative covariance reduces the spread of the sum.

Verify: $\mathrm{Cov}(U,V)=-6$; ${\rho }_{UV}=-0.5\in [-1,1]$ ✓; $\mathrm{Var}(U+V)=13>0$ ✓ (a variance must be non-negative). Direct check of ${\rho }_{UV}$: $\frac{-6}{\sqrt{16}\sqrt{9}}=\frac{-6}{4\cdot 3}=-0.5$ ✓ — matches step 3.

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Recall Which cell was which?

Positive ordinary ::: Ex 1 (cell A), $\rho \approx 0.41$. Negative ::: Ex 2 (cell B), $\rho =-0.6$. Honest zero (independent) ::: Ex 3 (cell C), $\rho =0$ AND independent. Zero-cov trap (dependent) ::: Ex 4 (cell D), $\mathrm{Cov}=0$ but $Y=X^2$. Perfect line ::: Ex 5 (cells E,F), $r=+1$ and $r=-1$. Constant input ::: Ex 6 (cell G), $\rho$ UNDEFINED (0/0). Units word problem ::: Ex 7 (cell H), cov changes with units, $r$ does not. Toolbox twist ::: Ex 8 (cell I), bilinearity + variance-of-sum.

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Connections

• Covariance and Correlation — this page is its worked-example companion.
• Expectation of Random Variables — every mean and $E[XY]$ here.
• Variance and Standard Deviation — the $\sigma$'s that rescale, and the degenerate $\sigma =0$ case.
• Independence of Random Variables — Ex 3 (genuine) vs Ex 4 (fails).
• Cauchy–Schwarz Inequality — the $\rho =\pm 1$ boundary in Ex 5.
• Linear Regression — perfect line ⇒ $r^2=1$.
• Covariance Matrix — Ex 8's toolbox scales to whole random vectors.