Joint and Conditional Distributions - Key Proof

We prove that for bivariate normal distributions, zero correlation implies independence—a property unique to the normal distribution.

Independence from Zero Correlation

Theorem

Let $(X,Y)$ have bivariate normal distribution with correlation $\rho$ . Then: $\rho = 0 \iff X \text{ and } Y \text{ are independent}$

Proof

Direction ( $\Leftarrow$ ): If $X$ and $Y$ are independent, then: $\text{Cov}(X,Y) = E[XY] - E[X]E[Y] = E[X]E[Y] - E[X]E[Y] = 0$

Therefore $\rho = \frac{\text{Cov}(X,Y)}{\sigma_X\sigma_Y} = 0$ . ✓

Direction ( $\Rightarrow$ ): This is the interesting direction. Assume $\rho = 0$ .

The bivariate normal PDF is: $f(x,y) = \frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}} \exp\left(-\frac{Q}{2(1-\rho^2)}\right)$

where: $Q = \left(\frac{x-\mu_X}{\sigma_X}\right)^2 - 2\rho\frac{(x-\mu_X)(y-\mu_Y)}{\sigma_X\sigma_Y} + \left(\frac{y-\mu_Y}{\sigma_Y}\right)^2$

When $\rho = 0$ : $Q = \left(\frac{x-\mu_X}{\sigma_X}\right)^2 + \left(\frac{y-\mu_Y}{\sigma_Y}\right)^2$

Substituting: $f(x,y) = \frac{1}{2\pi\sigma_X\sigma_Y} \exp\left(-\frac{1}{2}\left[\left(\frac{x-\mu_X}{\sigma_X}\right)^2 + \left(\frac{y-\mu_Y}{\sigma_Y}\right)^2\right]\right)$

Factoring the exponential: $= \frac{1}{\sqrt{2\pi}\sigma_X} \exp\left(-\frac{(x-\mu_X)^2}{2\sigma_X^2}\right) \cdot \frac{1}{\sqrt{2\pi}\sigma_Y} \exp\left(-\frac{(y-\mu_Y)^2}{2\sigma_Y^2}\right)$

$= f_X(x) \cdot f_Y(y)$

Since the joint PDF factors as the product of marginal PDFs, $X$ and $Y$ are independent. □

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Counterexample for Non-Normal

Example

Let $(X,Y)$ be uniformly distributed on the circle $x^2 + y^2 = 1$ .

By symmetry: $E[X] = E[Y] = 0$ , $E[XY] = 0$ , so $\rho = 0$ .

But $X$ and $Y$ are clearly dependent: knowing $X = 0.5$ implies $Y = \pm\sqrt{0.75}$ , not any value!

Zero correlation $\not\Rightarrow$ independence in general.

Implications

This result explains why multivariate normal distributions are so tractable:

Covariance matrix $\boldsymbol{\Sigma}$ completely characterizes dependence
If $\boldsymbol{\Sigma}$ is diagonal, components are independent
Correlation analysis suffices for detecting all dependencies

Remark

The equivalence of zero correlation and independence for bivariate (and multivariate) normal distributions is exceptional. For non-normal distributions, uncorrelated variables can be highly dependent. This distinction is crucial in data analysis—correlation only captures linear relationships.