Jointly Normal and Uncorrelated Implies Independent
Part 1: Prove that if $(X, Y)$ are jointly normal and uncorrelated, then $X$ and $Y$ are independent.
Part 2: Give an example of two random variables that are each marginally normal and uncorrelated, but are not independent. Explain why this does not contradict Part 1.
Hints
- For Part 1, write out the bivariate normal density with $\sigma_{XY} = 0$ and observe that it factors into two independent univariate normal densities.
- Alternatively, show the joint MGF factors when the off-diagonal of $\Sigma$ is zero. Factorization of the MGF is equivalent to independence.
- For Part 2, you need normal marginals but a non-Gaussian joint distribution. Try modifying $Z \sim N(0,1)$ by flipping its sign in a region chosen to zero out the covariance.
Worked Solution
How to Think About It: The statement "uncorrelated implies independent" is false in general -- correlation only captures linear dependence. But for jointly normal random variables, the entire distribution is parameterized by the mean vector and covariance matrix. Setting the off-diagonal covariance to zero forces the joint pdf to factor, which is independence. The catch is the word "jointly" -- it is doing all the heavy lifting. If the pair is not jointly normal (even if each marginal is normal), the covariance matrix no longer determines the distribution, and the implication breaks.
Formal Solution:
Part 1: Proof that jointly normal + uncorrelated $\Rightarrow$ independent.
Let $(X, Y) \sim N(\boldsymbol{\mu}, \Sigma)$ where $\boldsymbol{\mu} = (\mu_X, \mu_Y)^T$ and:
$$\Sigma = \begin{pmatrix} \sigma_X^2 & \sigma_{XY} \\\ \sigma_{XY} & \sigma_Y^2 \end{pmatrix}$$
Uncorrelated means $\sigma_{XY} = \text{Cov}(X, Y) = 0$, so $\Sigma$ is diagonal:
$$\Sigma = \begin{pmatrix} \sigma_X^2 & 0 \\\ 0 & \sigma_Y^2 \end{pmatrix}$$
The joint density of $(X, Y)$ is:
$$f_{X,Y}(x, y) = \frac{1}{2\pi |\Sigma|^{1/2}} \exp\!\left(-\frac{1}{2} (\mathbf{z} - \boldsymbol{\mu})^T \Sigma^{-1} (\mathbf{z} - \boldsymbol{\mu})\right)$$
With $\Sigma$ diagonal, $|\Sigma| = \sigma_X^2 \sigma_Y^2$ and $\Sigma^{-1} = \text{diag}(1/\sigma_X^2,\, 1/\sigma_Y^2)$. The exponent becomes:
$$-\frac{(x - \mu_X)^2}{2\sigma_X^2} - \frac{(y - \mu_Y)^2}{2\sigma_Y^2}$$
So the joint density factors:
$$f_{X,Y}(x, y) = \underbrace{\frac{1}{\sqrt{2\pi}\sigma_X} \exp\!\left(-\frac{(x - \mu_X)^2}{2\sigma_X^2}\right)}_{f_X(x)} \cdot \underbrace{\frac{1}{\sqrt{2\pi}\sigma_Y} \exp\!\left(-\frac{(y - \mu_Y)^2}{2\sigma_Y^2}\right)}_{f_Y(y)}$$
Since $f_{X,Y}(x, y) = f_X(x) \cdot f_Y(y)$ for all $(x, y)$, the variables $X$ and $Y$ are independent. $\square$
Alternative proof via MGFs: The joint MGF of $(X, Y) \sim N(\boldsymbol{\mu}, \Sigma)$ is:
$$M(t_1, t_2) = \exp\!\left(\boldsymbol{\mu}^T \mathbf{t} + \frac{1}{2} \mathbf{t}^T \Sigma \mathbf{t}\right)$$
With $\sigma_{XY} = 0$, the cross term $2t_1 t_2 \sigma_{XY}$ vanishes, and the MGF factors: $M(t_1, t_2) = M_X(t_1) \cdot M_Y(t_2)$. Factorization of the MGF is equivalent to independence.
Part 2: Counterexample -- normal marginals, uncorrelated, but dependent.
Let $Z \sim N(0, 1)$ and define:
$$W = \begin{cases} Z & \text{if } |Z| \le c \\\ -Z & \text{if } |Z| > c \end{cases}$$
where $c > 0$ is chosen so that $E[ZW] = 0$.
$W$ is standard normal: Since $|W| = |Z|$ and the sign of $W$ is symmetric (the density of $Z$ is symmetric about 0, and the sign-flipping preserves this symmetry), $W \sim N(0, 1)$.
$Z$ and $W$ are uncorrelated: We need:
$$E[ZW] = E[Z^2 \cdot \mathbf{1}_{|Z| \le c}] - E[Z^2 \cdot \mathbf{1}_{|Z| > c}] = 0$$
This requires $E[Z^2 \cdot \mathbf{1}_{|Z| \le c}] = E[Z^2 \cdot \mathbf{1}_{|Z| > c}] = 1/2$ (since $E[Z^2] = 1$). The function $g(c) = E[Z^2 \cdot \mathbf{1}_{|Z| \le c}]$ is continuous, equals 0 at $c = 0$ and 1 at $c = \infty$, so by the intermediate value theorem there exists a $c$ where $g(c) = 1/2$. (Numerically, $c \approx 1.54$.)
$Z$ and $W$ are dependent: They are clearly dependent because $|W| = |Z|$ always. Knowing $|Z| = 2$ tells you $|W| = 2$ with certainty.
Why no contradiction: This does not contradict Part 1 because $(Z, W)$ is not jointly normal. The joint distribution is concentrated on the union of lines $w = z$ and $w = -z$, which is not a bivariate Gaussian.
Answer:
- Part 1: For jointly normal $(X, Y)$, zero covariance makes $\Sigma$ diagonal, which forces the joint density to factor into the product of marginals. Hence $X \perp Y$.
- Part 2: Let $Z \sim N(0,1)$ and $W = Z \cdot \text{sign}(c - |Z|)$ with appropriate $c$. Both marginals are $N(0,1)$ and $\text{Cov}(Z, W) = 0$, but $(Z, W)$ is not jointly normal, so the pair is dependent despite being uncorrelated.
Intuition
This result highlights a very special property of the Gaussian distribution: it is completely determined by its first two moments (mean and covariance). For any other distribution family, the covariance matrix captures only linear relationships and misses everything else. But for jointly normal variables, there is nothing else -- the entire dependence structure lives in the covariance matrix. Setting the covariance to zero kills all dependence, not just linear dependence.
The counterexample in Part 2 is essential to internalize. In practice, when someone says "$X$ and $Y$ are uncorrelated," you should not assume independence unless you have a strong reason to believe the joint distribution is Gaussian. In finance, asset returns are famously non-Gaussian (heavy tails, asymmetric dependence in crashes), so zero correlation emphatically does not mean independence. This is one of the most commonly tested conceptual points in quant interviews.