Covariance and Correlation

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Covariance and Correlation #

Probability & StatisticsDifficulty: ★★★☆☆Depth: 7Unlocks: 6

Measures of linear relationship between random variables.

Interactive Visualization #

⏮◀◀▶▶STEP0.25x1xZOOM

t=0s

Core Concepts #

Key Symbols & Notation #

Cov(X,Y)rho_{X,Y} (correlation coefficient)

Essential Relationships #

Prerequisites (2) #

Variance5 atomsJoint Distributions6 atoms

Unlocks (2) #

Principal Component Analysislvl 4Time Series Foundationslvl 3

Referenced by (9) #

Where this concept shows up in the operating-finance and personal-finance graphs.

From Business (8) #

[personal financeBusiness

Portfolio diversification is the canonical real-world application of covariance reduction - Markowitz's entire insight was that portfolio risk depends on asset covariance, making this the exact math behind diversification](/business/personal-finance/)[Execution RiskBusiness

Correlated execution risk is literally positive covariance between project failure modes; understanding Cov(X,Y) > 0 when X and Y share an underlying factor (the team) is the mathematical foundation](/business/execution-risk/)[Markowitz Portfolio TheoryBusiness

The covariance matrix of asset returns is the central mathematical object in Markowitz. Portfolio variance equals w^T Sigma w, so understanding covariance structure is prerequisite to constructing the efficient frontier.](/business/markowitz-portfolio-theory/)[Operating InvestmentsBusiness

The entire Markowitz insight is that portfolio variance = w^T Σ w, so risk depends on the covariance matrix between investments, not just individual variances - this is the mathematical core](/business/operating-investments/)[Risk-Adjusted ReturnBusiness

The covariance matrix between asset returns is the core mathematical object for handling correlations and computing portfolio variance (w'Σw) - without this, risk-adjusted return is single-instrument only](/business/risk-adjusted-return/)[PortfolioBusiness

The covariance matrix of asset returns is THE central mathematical object in Markowitz - portfolio variance is w^T Sigma w, and the entire efficient frontier emerges from the structure of Sigma](/business/portfolio/)[Portfolio ConstructionBusiness

Portfolio construction fundamentally depends on the covariance matrix between assets - 'handling correlations' from the definition is precisely what covariance and correlation measure, and the math of portfolio variance reduction through diversification is built on this](/business/portfolio-construction/)[CFABusiness

Markowitz mean-variance portfolio optimization - core CFA Level 1-2 material - requires computing covariance matrices between asset returns](/business/cfa/)

From Money (1) #

[DiversificationMoney

Diversification benefit comes from low or negative correlation between assets](/money/diversification/)

Advanced Learning Details

Graph Position #

91

Depth Cost

6

Fan-Out (ROI)

2

Bottleneck Score

7

Chain Length

Cognitive Load #

6

Atomic Elements

30

Total Elements

L1

Percentile Level

L4

Atomic Level

All Concepts (14) #

Teaching Strategy #

Multi-session curriculum - substantial prior knowledge and complex material. Use mastery gates and deliberate practice.

A city analyst plots two monthly time series: ice cream sales and drownings. The scatterplot slopes upward—strongly. A headline writes itself: “Ice cream causes drownings.”

But the analyst pauses. Covariance and correlation can tell you how tightly two variables move together in a straight-line way. They cannot, by themselves, tell you why (causation), nor whether a relationship is real vs driven by a third variable (season/temperature), nor whether a relationship is nonlinear (e.g., U-shaped).

In this lesson you’ll learn:

TL;DR:

Covariance is the expected product of two variables’ deviations from their means: Cov(X,Y) = E[(X−E[X])(Y−E[Y])]. Its sign indicates direction of linear co-movement, and its magnitude depends on units. Correlation ρ₍X,Y₎ standardizes covariance by dividing by σXσY, producing a unitless number in [−1,1] that measures linear association. Zero covariance means “no linear relationship” but does not generally mean independence (except in special cases like jointly normal variables).

What Is Covariance and Correlation? #

The motivation (why we need them) #

Variance tells you how a single random variable spreads around its mean. But many real questions are about pairs of variables:

We want a number that captures co-movement.

Covariance: “do deviations move together?” #

Covariance looks at whether XXX and YYY are simultaneously above their means or simultaneously below their means.

Define the mean-centered variables:

Then covariance is the expected product:

Cov⁡(X,Y)=E[(X−E[X])(Y−E[Y])].\operatorname{Cov}(X,Y) = E\big[(X - E[X])(Y - E[Y])\big].Cov(X,Y)=E[(X−E[X])(Y−E[Y])].

Interpretation by sign:

A key subtlety: covariance is unitful. If XXX is in dollars and YYY is in seconds, covariance is in dollar·seconds. Changing units (e.g., dollars → cents) scales covariance.

Correlation: “covariance without units” #

To compare relationships across different scales, we standardize by each variable’s standard deviation.

ρX,Y=Corr⁡(X,Y)=Cov⁡(X,Y)σX σY,\rho_{X,Y} = \operatorname{Corr}(X,Y) = \frac{\operatorname{Cov}(X,Y)}{\sigma_X,\sigma_Y},ρX,Y​=Corr(X,Y)=σX​σY​Cov(X,Y)​,

where σX=Var⁡(X)\sigma_X = \sqrt{\operatorname{Var}(X)}σX​=Var(X)​ and σY=Var⁡(Y)\sigma_Y = \sqrt{\operatorname{Var}(Y)}σY​=Var(Y)​.

Correlation is:

Interpretation:

A concrete picture: centered scatter and “cosine similarity” #

If you take paired samples (xi,yi)(xᵢ, yᵢ)(xi​,yi​) and center them, each data point becomes a 2D vector from the mean: zi=[xi−xˉ,  yi−yˉ]\mathbf{z}_i = [xᵢ-\bar{x},; yᵢ-\bar{y}]zi​=[xi​−xˉ,yi​−yˉ​].

A helpful geometric intuition is that correlation behaves like a normalized “alignment” between the centered coordinates of XXX and YYY.

Here’s a simple ASCII scatter showing a positive association. The “+” is the mean; arrows suggest centered deviations.

Y
^               *
|            *
|         *
|      *
|   *
| +-----------------> X
|   *
|      *

When points trend from bottom-left to top-right, products (xi−xˉ)(yi−yˉ)(xᵢ-\bar{x})(yᵢ-\bar{y})(xi​−xˉ)(yi​−yˉ​) are often positive, boosting covariance and correlation.

What they can and cannot conclude (preview) #

Covariance/correlation can:

They cannot by themselves:

Core Mechanic 1: Computing Covariance (and its key identities) #

Why the definition looks the way it does #

We want a measure that’s:

  1. 1)Positive when both variables tend to be on the same side of their means.
  2. 2)Negative when they tend to be on opposite sides.
  3. 3)Larger when these deviations are larger.

The product (X−E[X])(Y−E[Y])(X-E[X])(Y-E[Y])(X−E[X])(Y−E[Y]) does exactly that:

Averaging (expectation) makes it a stable summary.

Expanding covariance: a very useful formula #

Start from the definition:

Cov⁡(X,Y)=E[(X−E[X])(Y−E[Y])].\operatorname{Cov}(X,Y) = E\big[(X - E[X])(Y - E[Y])\big].Cov(X,Y)=E[(X−E[X])(Y−E[Y])].

Expand step-by-step:

Cov⁡(X,Y)=E[XY−X E[Y]−E[X] Y+E[X]E[Y]]=E[XY]−E[Y]E[X]−E[X]E[Y]+E[X]E[Y]=E[XY]−E[X]E[Y].\begin{aligned}
\operatorname{Cov}(X,Y)
&= E\big[XY - X,E[Y] - E[X],Y + E[X]E[Y]\big] \
&= E[XY] - E[Y]E[X] - E[X]E[Y] + E[X]E[Y] \
&= E[XY] - E[X]E[Y].
\end{aligned}Cov(X,Y)​=E[XY−XE[Y]−E[X]Y+E[X]E[Y]]=E[XY]−E[Y]E[X]−E[X]E[Y]+E[X]E[Y]=E[XY]−E[X]E[Y].​

So:

Cov⁡(X,Y)=E[XY]−E[X]E[Y].\boxed{\operatorname{Cov}(X,Y) = E[XY] - E[X]E[Y].}Cov(X,Y)=E[XY]−E[X]E[Y].​

This identity is especially handy in calculations.

Covariance with constants and scaling #

These properties matter constantly in modeling:

  1. Adding constants doesn’t change covariance:

Cov⁡(X+a, Y+b)=Cov⁡(X,Y).\operatorname{Cov}(X+a,, Y+b) = \operatorname{Cov}(X,Y).Cov(X+a,Y+b)=Cov(X,Y).

Reason: centering removes constants.

  1. Scaling scales covariance:

Cov⁡(cX, dY)=cd Cov⁡(X,Y).\operatorname{Cov}(cX,, dY) = cd,\operatorname{Cov}(X,Y).Cov(cX,dY)=cdCov(X,Y).

So if you convert meters to centimeters (×100), covariance changes by ×100 on that variable’s side.

Relationship to variance #

Variance is just covariance with itself:

Var⁡(X)=Cov⁡(X,X).\operatorname{Var}(X) = \operatorname{Cov}(X,X).Var(X)=Cov(X,X).

That makes covariance a true generalization of variance.

Independence implies zero covariance (but not conversely) #

If XXX and YYY are independent, then E[XY]=E[X]E[Y]E[XY] = E[X]E[Y]E[XY]=E[X]E[Y], so:

Cov⁡(X,Y)=0.\operatorname{Cov}(X,Y) = 0.Cov(X,Y)=0.

But the reverse is not generally true: Cov⁡(X,Y)=0\operatorname{Cov}(X,Y)=0Cov(X,Y)=0 does not guarantee independence. (We’ll build a concrete counterexample later.)

Sample covariance: estimating from data #

In practice, you usually have samples (xi,yi)(xᵢ, yᵢ)(xi​,yi​) for i=1..ni=1..ni=1..n.

Define sample means:

A common estimator (unbiased for i.i.d. samples) is:

sxy=1n−1∑i=1n(xi−xˉ)(yi−yˉ).s_{xy} = \frac{1}{n-1}\sum_{i=1}^n (xᵢ-\bar{x})(yᵢ-\bar{y}).sxy​=n−11​i=1∑n​(xi​−xˉ)(yi​−yˉ​).

You may also see 1n\frac{1}{n}n1​ instead of 1n−1\frac{1}{n-1}n−11​; that version is the maximum-likelihood estimator under a normal model but is biased in finite samples.

Interpreting magnitude: why covariance is hard to compare #

Suppose:

The relationship is identical, but scaling height by 100 scales covariance by 100. That’s why correlation is so widely used: it removes units.

Core Mechanic 2: Correlation as Standardized Covariance (and the geometry behind ±1) #

Why standardize? #

Covariance answers “do they move together?”, but not “how strong is that co-movement relative to each variable’s natural scale?”

Correlation fixes this by dividing by σXσY\sigma_X\sigma_YσX​σY​.

ρX,Y=Cov⁡(X,Y)σXσY.\rho_{X,Y} = \frac{\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y}.ρX,Y​=σX​σY​Cov(X,Y)​.

Because σX\sigma_XσX​ has units of XXX and σY\sigma_YσY​ has units of YYY, the units cancel.

Correlation is bounded between −1 and 1 #

This isn’t just a convention—it’s a theorem. One route uses Cauchy–Schwarz.

Let Xc=X−E[X]X_c = X - E[X]Xc​=X−E[X] and Yc=Y−E[Y]Y_c = Y - E[Y]Yc​=Y−E[Y].

Apply Cauchy–Schwarz to random variables:

∣E[XcYc]∣≤E[Xc2] E[Yc2].|E[X_c Y_c]| \le \sqrt{E[X_c^2]},\sqrt{E[Y_c^2]}.∣E[Xc​Yc​]∣≤E[Xc2​]​E[Yc2​]​.

But E[Xc2]=Var⁡(X)=σX2E[X_c^2]=\operatorname{Var}(X)=\sigma_X^2E[Xc2​]=Var(X)=σX2​ and similarly for YYY.

So:

∣Cov⁡(X,Y)∣≤σXσY.|\operatorname{Cov}(X,Y)| \le \sigma_X\sigma_Y.∣Cov(X,Y)∣≤σX​σY​.

Divide both sides by σXσY\sigma_X\sigma_YσX​σY​ (assuming nonzero):

∣ρX,Y∣≤1.|\rho_{X,Y}| \le 1.∣ρX,Y​∣≤1.

When do we get ±1? #

Equality in Cauchy–Schwarz occurs when one variable is an exact scalar multiple of the other (almost surely) after centering.

That means:

Equivalently, Y=aX+bY = aX + bY=aX+b almost surely for some constants a,ba, ba,b.

A visual: correlation as “angle” between centered coordinates (sample view) #

For a dataset, define centered vectors:

Then the sample correlation is:

r=x⋅y∥x∥ ∥y∥.r = \frac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}|,|\mathbf{y}|}.r=∥x∥∥y∥x⋅y​.

That is exactly the cosine of the angle between x and y in Rn\mathbb{R}^nRn.

Inline “plot” of the vector-angle intuition:

           y (centered)
           ^
          /|
         / |   cos(θ) = r
        /θ |
       +---+----------> x (centered)

This is a powerful mental model: correlation is alignment between patterns of deviation-from-mean.

Sample correlation #

Using the sample covariance sxys_{xy}sxy​ and sample standard deviations sx,sys_x, s_ysx​,sy​:

r=sxysxsy.r = \frac{s_{xy}}{s_x s_y}.r=sx​sy​sxy​​.

Correlation is not robust to outliers #

Because it depends on products of deviations, a single extreme point can dominate.

Practical note:

Application/Connection: What Covariance and Correlation Enable (and what they don’t) #

1) The “ice cream and drownings” lesson: correlation ≠ causation #

A strong correlation can arise from:

In the ice-cream example, temperature (a confounder) increases both ice cream consumption and swimming, which increases drowning risk.

Covariance/correlation quantify association; causal claims require additional design/assumptions.

2) Zero covariance: what it means, and the classic trap #

Cov⁡(X,Y)=0\operatorname{Cov}(X,Y)=0Cov(X,Y)=0 means the linear term in their relationship is absent in a precise sense:

But XXX can still strongly determine YYY nonlinearly.

We’ll show an explicit example in the worked examples: X∼Uniform(−1,1)X \sim \text{Uniform}(-1,1)X∼Uniform(−1,1) and Y=X2Y=X^2Y=X2. Then covariance is 0, yet YYY is completely determined by XXX.

Special case worth knowing: if (X,Y)(X,Y)(X,Y) are jointly normal, then zero covariance does imply independence. (Not true in general.)

3) Covariance matrix: scaling up to many variables #

With ddd random variables, stack them into a random vector X ∈ Rd\mathbb{R}^dRd.

Define the covariance matrix:

Σ=Cov⁡(X)=E[(X−E[X])(X−E[X])T].\Sigma = \operatorname{Cov}(\mathbf{X}) = E\big[(\mathbf{X}-E[\mathbf{X}])(\mathbf{X}-E[\mathbf{X}])^T\big].Σ=Cov(X)=E[(X−E[X])(X−E[X])T].

Key facts:

This matrix is the central object in multivariate statistics.

4) Connection to PCA (what this node unlocks) #

PCA looks for directions in feature space with maximum variance.

If you have centered data vectors x ∈ Rd\mathbb{R}^dRd, PCA finds unit vectors v maximizing:

Var⁡(vTX)=vTΣv.\operatorname{Var}(\mathbf{v}^T\mathbf{X}) = \mathbf{v}^T \Sigma \mathbf{v}.Var(vTX)=vTΣv.

The solutions are eigenvectors of the covariance matrix Σ\SigmaΣ.

So learning covariance is not just about pairwise relationships—it’s about understanding the geometry of data clouds and the linear structure PCA extracts.

5) A quick “what to use when” table #

GoalUse covariance?Use correlation?Notes
Keep physical units (e.g., risk in \·days)✅❌Covariance preserves scale
Compare relationships across different units❌✅Correlation is unitless
Build PCA on raw feature scales✅❌/✅Often you choose covariance PCA or correlation PCA (standardized)
Detect any dependence (including nonlinear)❌❌Need other tools (MI, plots, kernels, etc.)

Worked Examples (3) #

Compute covariance and correlation from a small dataset (by hand) #

You observe n=5 paired measurements:

X: [1, 2, 3, 4, 5]

Y: [2, 1, 4, 3, 6]

Compute the sample covariance s_xy (with 1/(n−1)) and sample correlation r.

  1. Compute means:

    \bar{x} = (1+2+3+4+5)/5 = 15/5 = 3

    \bar{y} = (2+1+4+3+6)/5 = 16/5 = 3.2

  2. Compute centered values and products:

    For each i, compute (xᵢ−\bar{x}), (yᵢ−\bar{y}), and product.

    i=1: x=1 → −2; y=2 → −1.2; product = (−2)(−1.2)= 2.4

    i=2: x=2 → −1; y=1 → −2.2; product = (−1)(−2.2)= 2.2

    i=3: x=3 → 0; y=4 → 0.8; product = 0·0.8= 0

    i=4: x=4 → 1; y=3 → −0.2; product = 1·(−0.2)= −0.2

    i=5: x=5 → 2; y=6 → 2.8; product = 2·2.8= 5.6

  3. Sum of products:

    ∑(xᵢ−\bar{x})(yᵢ−\bar{y}) = 2.4+2.2+0−0.2+5.6 = 10.0

  4. Sample covariance:

    sxy=1n−1∑(xi−xˉ)(yi−yˉ)=104=2.5.s_{xy} = \frac{1}{n-1}\sum (xᵢ-\bar{x})(yᵢ-\bar{y}) = \frac{10}{4} = 2.5.sxy​=n−11​∑(xi​−xˉ)(yi​−yˉ​)=410​=2.5.

  5. Compute sample standard deviations.

    First compute sums of squares.

    For X:

    (xᵢ−\bar{x})²: 4, 1, 0, 1, 4 → sum = 10

    So s_x² = 10/(n−1)=10/4=2.5 → s_x = √2.5

    For Y:

    (yᵢ−\bar{y})²: (−1.2)²=1.44, (−2.2)²=4.84, 0.8²=0.64, (−0.2)²=0.04, 2.8²=7.84

    Sum = 1.44+4.84+0.64+0.04+7.84 = 14.8

    So s_y² = 14.8/4 = 3.7 → s_y = √3.7

  6. Compute correlation:

    r=sxysxsy=2.52.5 3.7=2.59.25.r = \frac{s_{xy}}{s_x s_y} = \frac{2.5}{\sqrt{2.5},\sqrt{3.7}} = \frac{2.5}{\sqrt{9.25}}.r=sx​sy​sxy​​=2.5​3.7​2.5​=9.25​2.5​.

    Numerically, √9.25 ≈ 3.041, so

    r ≈ 2.5 / 3.041 ≈ 0.822.

Insight: The covariance (2.5) says X and Y tend to be above/below their means together, but its magnitude depends on X and Y units. The correlation (~0.82) shows a fairly strong positive linear association on a standardized scale.

Zero covariance but strong dependence: Y = X² on a symmetric interval #

Let X ~ Uniform(−1, 1) and define Y = X². Compute Cov(X,Y). Are X and Y independent?

  1. Compute expectations.

    By symmetry of Uniform(−1,1), the distribution is symmetric around 0, so

    E[X]=0.E[X] = 0.E[X]=0.

  2. Compute E[Y]. Since Y=X²:

    E[Y]=E[X2].E[Y] = E[X^2].E[Y]=E[X2].

    For X ~ Uniform(−1,1) with density f(x)=1/2 on [−1,1]:

    E[X2]=∫−11x2⋅12 dx=12[x33]−11=12(13−(−13))=12⋅23=13.\begin{aligned}
    E[X^2]
    &= \int_{-1}^{1} x^2 \cdot \frac{1}{2},dx \
    &= \frac{1}{2}\left[\frac{x^3}{3}\right]_{-1}^{1} \
    &= \frac{1}{2}\left(\frac{1}{3} - \left(-\frac{1}{3}\right)\right) \
    &= \frac{1}{2}\cdot \frac{2}{3} = \frac{1}{3}.
    \end{aligned}E[X2]​=∫−11​x2⋅21​dx=21​[3x3​]−11​=21​(31​−(−31​))=21​⋅32​=31​.​

  3. Compute E[XY]. Since Y=X², we have XY = X·X² = X³. Then:

    E[XY]=E[X3]=∫−11x3⋅12 dx.E[XY] = E[X^3] = \int_{-1}^{1} x^3 \cdot \frac{1}{2},dx.E[XY]=E[X3]=∫−11​x3⋅21​dx.

    But x³ is an odd function and the interval is symmetric, so the integral is 0:

    E[X3]=0.E[X^3]=0.E[X3]=0.

  4. Compute covariance using E[XY] − E[X]E[Y]:

    Cov⁡(X,Y)=E[XY]−E[X]E[Y]=0−(0)(13)=0.\operatorname{Cov}(X,Y) = E[XY] - E[X]E[Y] = 0 - (0)\left(\frac{1}{3}\right) = 0.Cov(X,Y)=E[XY]−E[X]E[Y]=0−(0)(31​)=0.

  5. Check independence intuition.

    If X and Y were independent, knowing X would not give information about Y.

    But here Y is determined exactly by X via Y=X². For example:

    • •If X=0, then Y=0 with certainty.
    • •If X=±1, then Y=1 with certainty.

    So Y is not independent of X.

Insight: Covariance zero means “no linear association,” not “no relationship.” The relationship here is perfectly nonlinear: the scatter is a U-shape. Covariance and correlation miss that shape even though dependence is complete.

Correlation is invariant to shifting and positive scaling (but flips sign under negative scaling) #

Let X and Y be random variables with Corr(X,Y)=ρ. Define X' = 3X + 10 and Y' = −2Y + 5. Find Corr(X',Y').

  1. Use covariance and standard deviation scaling rules.

    Constants don’t affect covariance:

    Cov(X+a, Y+b)=Cov(X,Y).

    Scaling: Cov(cX, dY)=cd Cov(X,Y).

  2. Compute Cov(X',Y'):

    X' = 3X+10, Y' = −2Y+5

    Cov⁡(X′,Y′)=Cov⁡(3X+10, −2Y+5)=(3)(−2)Cov⁡(X,Y)=−6Cov⁡(X,Y).\operatorname{Cov}(X',Y') = \operatorname{Cov}(3X+10,,-2Y+5) = (3)(-2)\operatorname{Cov}(X,Y) = -6\operatorname{Cov}(X,Y).Cov(X′,Y′)=Cov(3X+10,−2Y+5)=(3)(−2)Cov(X,Y)=−6Cov(X,Y).

  3. Compute σ_{X'} and σ_{Y'}.

    Standard deviation scales by absolute value:

    σ_{3X+10} = |3|σ_X = 3σ_X

    σ_{−2Y+5} = |−2|σ_Y = 2σ_Y

  4. Compute correlation:

    Corr⁡(X′,Y′)=Cov⁡(X′,Y′)σX′σY′=−6Cov⁡(X,Y)(3σX)(2σY)=−Cov⁡(X,Y)σXσY=−ρ.\begin{aligned}
    \operatorname{Corr}(X',Y')
    &= \frac{\operatorname{Cov}(X',Y')}{\sigma_{X'}\sigma_{Y'}} \
    &= \frac{-6\operatorname{Cov}(X,Y)}{(3\sigma_X)(2\sigma_Y)} \
    &= -\frac{\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y} \
    &= -\rho.
    \end{aligned}Corr(X′,Y′)​=σX′​σY′​Cov(X′,Y′)​=(3σX​)(2σY​)−6Cov(X,Y)​=−σX​σY​Cov(X,Y)​=−ρ.​

Insight: Correlation ignores shifts and positive rescalings (unit changes), but a negative scaling flips the sign because it reverses one axis.

Key Takeaways #

Common Mistakes #

Practice #

easy

Let X have E[X]=2, Var(X)=9. Let Y = 5 − 2X. Compute Cov(X,Y) and Corr(X,Y).

Hint: Use Cov(X, a+bX) = b Var(X). For correlation, note Y is exactly linear in X.

Show solution

Compute covariance:

Y = 5 − 2X ⇒ Cov(X,Y) = Cov(X, 5 − 2X) = −2 Cov(X,X) = −2 Var(X) = −2·9 = −18.

For correlation, since Y is an exact decreasing linear function of X, Corr(X,Y)=−1.

(You can also compute σ_Y = |−2|σ_X = 2·3=6, so Corr = (−18)/(3·6)=−1.)

easy

Suppose E[X]=1, E[Y]=3, and E[XY]=10. Compute Cov(X,Y).

Hint: Use Cov(X,Y)=E[XY]−E[X]E[Y].

Show solution

Cov(X,Y) = 10 − (1)(3) = 7.

medium

Let X be Uniform(0,1) and Y = X. Compute Corr(X,Y). Then let Z = X² and reason (without heavy computation) whether Corr(X,Z) is closer to 1, closer to 0, or could be negative.

Hint: For Y=X, it’s perfect linear. For Z=X² on (0,1), it’s increasing but nonlinear; think about whether larger X tends to mean larger Z.

Show solution

For Y=X, correlation is 1 because Y is an exact positive linear function of X (Y=1·X+0).

For Z=X² with X∈(0,1), Z is strictly increasing in X, so larger X tends to correspond to larger Z. That suggests a positive covariance and positive correlation. However, because the relationship is nonlinear, Corr(X,Z) will be less than 1 (not a perfect straight line). It cannot be negative here because X and X² move in the same direction on (0,1). So Corr(X,Z) is closer to 1 than to 0, but still < 1.

Connections #

Next steps and related nodes:

Helpful background refreshers:

Quality: A (4.4/5)

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