Sum of Gaussians and Uncorrelatedness vs. Independence

Random Variables · Easy · Free problem

Suppose $X$ and $Y$ are independent standard Gaussian random variables ($X, Y \sim N(0,1)$).

  1. Is $X + Y$ necessarily Gaussian? What if $X$ and $Y$ are each marginally $N(0,1)$ but not independent?
  1. Explain the relationship between uncorrelatedness and independence for general random variables. Then explain why, for jointly Gaussian random variables, uncorrelatedness *does* imply independence.

Hints

  1. Think about what makes two variables *jointly* Gaussian versus each being *marginally* Gaussian. The sum inherits properties from the joint distribution, not the marginals.
  2. For a counterexample to marginal-implies-joint, try $Y = X \cdot Z$ where $Z$ is an independent random sign. Check that $Y$ is $N(0,1)$ but $X + Y$ is not Gaussian.
  3. For the uncorrelated-implies-independent claim in the Gaussian case, write out the bivariate normal density with $\rho = 0$ and observe that it factors into a product of marginals.

Worked Solution

How to Think About It: This problem tests a distinction that trips up many candidates: the difference between *marginal* normality and *joint* normality. Two variables can each be perfectly Gaussian on their own, yet their joint distribution can be weird enough that their sum is not Gaussian. The key is that sums inherit nice properties from the *joint* distribution, not from the marginals. Separately, the uncorrelated-vs-independent question hinges on whether zero covariance tells you everything about the joint distribution -- for Gaussians it does, for general distributions it does not.

Quick Estimate: For part (1), if $X$ and $Y$ are independent $N(0,1)$, then $X + Y \sim N(0, 2)$ -- this is immediate from the MGF or characteristic function. The interesting case is when they are only marginally normal.

Formal Solution:

Part 1: Is $X + Y$ Gaussian?

Case 1 -- Independent (or jointly Gaussian): If $X \sim N(0,1)$ and $Y \sim N(0,1)$ are independent, then $X + Y$ is Gaussian. The MGF of the sum is the product of MGFs:

$M_{X+Y}(t) = M_X(t) \cdot M_Y(t) = e^{t^2/2} \cdot e^{t^2/2} = e^{t^2}$

which is the MGF of $N(0, 2)$. More generally, any linear combination of jointly Gaussian variables is Gaussian.

Case 2 -- Marginally Gaussian but not jointly Gaussian: $X + Y$ may *not* be Gaussian.

Counterexample: Let $X \sim N(0,1)$ and let $Z$ be an independent Rademacher variable ($P(Z = 1) = P(Z = -1) = 1/2$). Define $Y = XZ$. Then: - $Y$ is marginally $N(0,1)$ (since $|Y| = |X|$ and the sign is random). - But $X + Y = X(1 + Z)$, which equals

X$ with probability 1/2 and $0$ with probability 1/2. This is a 50-50 mixture of $N(0, 4)$ and a point mass at 0 -- clearly not Gaussian.

The takeaway: the sum of two marginally normal variables is only guaranteed to be normal if they are jointly normal.

Part 2: Uncorrelatedness vs. Independence

General case -- uncorrelated does NOT imply independent:

Uncorrelatedness means $\text{Cov}(X, Y) = E[XY] - E[X]E[Y] = 0$. This only constrains the *linear* relationship. Nonlinear dependence can be invisible to covariance.

Classic example: Let $X \sim N(0,1)$ and $Y = X^2$. Then: - $\text{Cov}(X, Y) = E[X^3] - E[X]E[X^2] = 0 - 0 = 0$ (by symmetry of the normal distribution, all odd moments are zero). - But $Y$ is a deterministic function of $X$, so they are maximally dependent.

Jointly Gaussian case -- uncorrelated DOES imply independent:

If $(X, Y)$ is jointly Gaussian, the joint density is:

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

where $\rho = \text{Corr}(X,Y)$. If $\rho = 0$ (uncorrelated), the cross term vanishes and the density factors:

$f(x,y) = \frac{1}{\sqrt{2\pi}\sigma_X}e^{-\frac{(x-\mu_X)^2}{2\sigma_X^2}} \cdot \frac{1}{\sqrt{2\pi}\sigma_Y}e^{-\frac{(y-\mu_Y)^2}{2\sigma_Y^2}} = f_X(x) \cdot f_Y(y)$

Since the joint density factors into the product of marginals, $X$ and $Y$ are independent. The key reason this works for Gaussians but not in general: the Gaussian joint distribution is completely determined by its first two moments (means, variances, and covariance). So zero covariance pins down the entire joint structure.

Answer:

  1. $X + Y$ is Gaussian when $X$ and $Y$ are jointly Gaussian (in particular, when independent). If they are only marginally normal, $X + Y$ can fail to be Gaussian.
  1. In general, uncorrelatedness does not imply independence because covariance only captures linear dependence. For jointly Gaussian variables, uncorrelatedness implies independence because the joint distribution is fully characterized by its first two moments.

Intuition

The deeper lesson is about the gap between *marginal* and *joint* properties. Knowing each variable's distribution individually tells you nothing about their joint behavior -- you need the full joint distribution (or at least joint normality) to make conclusions about sums or independence. This is why, in portfolio theory, knowing each asset's return distribution is not enough; you need the copula or joint distribution to understand portfolio-level risk.

The Gaussian distribution is special precisely because it is determined by its first two moments. This is why correlation is such a powerful tool in Gaussian models (like mean-variance portfolio optimization) but dangerously misleading outside that setting. In practice, financial returns have fat tails and asymmetric dependence, which means uncorrelated assets can still crash together -- a lesson painfully learned in every tail event.

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