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Week 5.3: Bivariate Normal Distribution

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# Week 5.3: Bivariate Normal Distribution > **Prerequisites:** Joint Continuous ([Week 5.1: Joint Continuous Distributions](/notes/01-foundation-bsma1004-stats-2-week05-15-joint-continuous-distributions)), Normal ([Week 4.2: Common Continuous Distributions](/notes/01-foundation-bsma1004-stats-2-week04-13-common-cont...

Week 5.3: Bivariate Normal Distribution

Prerequisites: Joint Continuous (Week 5.1: Joint Continuous Distributions), Normal (Week 4.2: Common Continuous Distributions) Cross-links: BSMA3012 (Linear Stat Models) — multivariate normal Core question: What is the most important joint continuous distribution?

1. Intuition: The "Correlated Normal"

The bivariate normal is the natural joint distribution for two normally-distributed variables that may be correlated. Examples:
  • Height and weight of individuals
  • Exam scores in two subjects
  • Stock returns on consecutive days

2. Definition

(X,Y)N(μX,μY,σX2,σY2,ρ)(X,Y) \sim \mathcal{N}(\mu_X, \mu_Y, \sigma_X^2, \sigma_Y^2, \rho) has PDF:
fXY(x,y)=12πσXσY1ρ2exp(12(1ρ2)Q(x,y)),f_{XY}(x,y) = \frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)} Q(x,y) \right),
where
Q(x,y)=(xμX)2σX22ρ(xμX)(yμY)σXσY+(yμY)2σY2.Q(x,y) = \frac{(x-\mu_X)^2}{\sigma_X^2} - 2\rho\frac{(x-\mu_X)(y-\mu_Y)}{\sigma_X\sigma_Y} + \frac{(y-\mu_Y)^2}{\sigma_Y^2}.

3. Key Properties

  1. Marginals are normal: XN(μX,σX2)X \sim \mathcal{N}(\mu_X, \sigma_X^2), YN(μY,σY2)Y \sim \mathcal{N}(\mu_Y, \sigma_Y^2).
  2. ρ\rho is the correlation coefficient between XX and YY.
  3. If ρ=0\rho = 0, XX and YY are independent (unique to normal!).
  4. Conditional distributions are normal:
YX=xN(μY+ρσYσX(xμX),  σY2(1ρ2)).Y \mid X = x \sim \mathcal{N}\left( \mu_Y + \rho\frac{\sigma_Y}{\sigma_X}(x-\mu_X),\; \sigma_Y^2(1-\rho^2) \right).
The conditional mean is linear in xx — this is the foundation of linear regression.

4. Visualising the Bivariate Normal

(Diagram)

5. Practice Questions

Q1 (Easy)

(X,Y)(X,Y) bivariate normal with μX=50\mu_X=50, μY=60\mu_Y=60, σX=10\sigma_X=10, σY=12\sigma_Y=12, ρ=0.6\rho=0.6. Find E[YX=70]\mathbb{E}[Y \mid X=70].
Full Solution
E[YX=70]=60+0.61210(7050)=60+0.61.220=60+14.4=74.4\mathbb{E}[Y \mid X=70] = 60 + 0.6 \cdot \frac{12}{10}(70-50) = 60 + 0.6 \cdot 1.2 \cdot 20 = 60 + 14.4 = 74.4.

Q2 (Medium)

For the same distribution, find Var(YX=70)\operatorname{Var}(Y \mid X=70).
Full Solution
Var(YX=70)=σY2(1ρ2)=144(10.36)=1440.64=92.16\operatorname{Var}(Y \mid X=70) = \sigma_Y^2(1-\rho^2) = 144(1-0.36) = 144 \cdot 0.64 = 92.16.

Q3 (Hard)

If XX and YY are bivariate normal with ρ=0\rho=0, show they are independent.
Full Solution
If ρ=0\rho=0, then Q(x,y)=(xμX)2σX2+(yμY)2σY2Q(x,y) = \frac{(x-\mu_X)^2}{\sigma_X^2} + \frac{(y-\mu_Y)^2}{\sigma_Y^2}, and 1ρ2=1\sqrt{1-\rho^2}=1.
fXY(x,y)=12πσXσYexp((xμX)22σX2(yμY)22σY2)=fX(x)fY(y)f_{XY}(x,y) = \frac{1}{2\pi\sigma_X\sigma_Y} \exp\left(-\frac{(x-\mu_X)^2}{2\sigma_X^2} - \frac{(y-\mu_Y)^2}{2\sigma_Y^2}\right) = f_X(x) \cdot f_Y(y).
Thus XX and YY are independent.

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