Zero Correlation and Independence

Why Zero Correlation Needs a Model

Correlation measures linear association: it summarizes how two variables vary together on a standardized scale. Independence is stronger. It says that learning the value of one variable does not change the probability distribution of the other. Independence always implies zero correlation when the variables have finite, positive variances, but zero correlation does not generally imply independence.

The bivariate normal model is an important exception. Within that model, zero correlation and independence are equivalent. The model condition matters: a conclusion that is valid for bivariate normal variables need not hold for other distributions.

Definition: A pair $(X,Y)$ is bivariate normal if every linear combination $aX+bY$, for constants $a$ and $b$, has a normal distribution, with a constant combination allowed to be degenerate. Suppose the marginal variances are positive, so that $\sigma_X>0$ and $\sigma_Y>0$. The correlation is $$ \rho=\frac{\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y}. $$ For a bivariate normal pair with positive marginal variances, $X$ and $Y$ are independent if and only if $\rho=0$.

This statement includes the boundary cases $\rho=1$ and $\rho=-1$: neither is zero, and the two components are dependent. The usual bivariate normal density formula below applies only when $|\rho|<1$. If a marginal variance is zero, the correlation is undefined, so the statement about zero correlation does not apply.

Why Zero Correlation Gives Independence in the Normal Model

For $|\rho|<1$, a bivariate normal pair with means $\mu_X,\mu_Y$, standard deviations $\sigma_X,\sigma_Y$, and correlation $\rho$ has joint density

f_{X,Y}(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} -2\rho\frac{(x-\mu_X)(y-\mu_Y)}{\sigma_X\sigma_Y} +\frac{(y-\mu_Y)^2}{\sigma_Y^2} \right) \right].

Set $\rho=0$. The cross-product term disappears, the square-root factor becomes 1, and the exponent separates into a term in $x$ and a term in $y$. Thus

f_{X,Y}(x,y) = \left[ \frac{1}{\sqrt{2\pi}\sigma_X} \exp\!\left(-\frac{(x-\mu_X)^2}{2\sigma_X^2}\right) \right] \left[ \frac{1}{\sqrt{2\pi}\sigma_Y} \exp\!\left(-\frac{(y-\mu_Y)^2}{2\sigma_Y^2}\right) \right] =f_X(x)f_Y(y).

The bracketed factors are the marginal normal densities of $X$ and $Y$. A joint density that factors as the product of its marginal densities gives independence: for any suitable sets $A$ and $B$, integrating the joint density over $A\times B$ gives $P(X\in A,Y\in B)=P(X\in A)P(Y\in B)$. This proves that zero correlation implies independence in the bivariate normal model.

For the reverse implication, independence gives $E[XY]=E[X]E[Y]$ when the expectations exist. Therefore

\operatorname{Cov}(X,Y) =E[XY]-E[X]E[Y]=0.

Normal variables have finite variances. With positive marginal variances, dividing this covariance by $\sigma_X\sigma_Y$ gives $\rho=0$. This reverse direction does not require normality; it is the forward direction that depends on the bivariate normal model.

Worked Examples

Worked Example: Using Zero Correlation to Factor an Event

Suppose $(X,Y)$ is bivariate normal with $X\sim N(12,9)$, $Y\sim N(40,25)$, and $\operatorname{Cov}(X,Y)=0$. Find $P(X>15,\ Y\leq 40)$. State why multiplying the marginal probabilities is valid.

Step 1: Find the correlation. The marginal standard deviations are $\sigma_X=\sqrt{9}=3$ and $\sigma_Y=\sqrt{25}=5$. Hence $$ \rho=\frac{0}{(3)(5)}=0. $$

Step 2: Apply the model result. The pair is bivariate normal and both marginal variances are positive. Therefore zero correlation implies that $X$ and $Y$ are independent. In particular, the two events in the question are independent.

Step 3: Calculate each marginal probability. For $X$, $$ P(X>15)=P\!\left(\frac{X-12}{3}>\frac{15-12}{3}\right) =1-\Phi(1)\approx 0.1587. $$ Since $Y$ is continuous and has mean 40, its symmetric normal distribution gives $$ P(Y\leq 40)=0.5000. $$

Step 4: Multiply and check. $$ P(X>15,\ Y\leq 40)=(0.1587)(0.5000)\approx 0.0793. $$ The probability is reported to four decimal places. The multiplication is justified by bivariate normality and zero correlation, not by zero correlation alone.

Method and trap avoided: Verify the joint model before using a product of probabilities. Treating any two zero-correlation variables as independent would be an invalid shortcut.

Worked Example: Nonzero Correlation Rules Out Independence

Suppose $(X,Y)$ is bivariate normal with $X\sim N(5,4)$, $Y\sim N(8,25)$, and $\operatorname{Cov}(X,Y)=6$. Determine whether the variables are independent. As a check, find the conditional mean of $Y$ given $X=7$.

Step 1: Calculate the correlation. The standard deviations are $\sigma_X=2$ and $\sigma_Y=5$, so $$ \rho=\frac{6}{(2)(5)}=0.6. $$ Since $\rho\ne 0$, the bivariate normal equivalence shows that the variables are not independent.

Step 2: Check using the conditional mean. In a bivariate normal model, the conditional mean is $$ E[Y\mid X=x]=\mu_Y+\rho\frac{\sigma_Y}{\sigma_X}(x-\mu_X). $$ Substitute $x=7$: $$ E[Y\mid X=7] =8+(0.6)\frac{5}{2}(7-5) =8+(0.6)(2.5)(2) =8+3 =11. $$

The conditional mean 11 differs from the marginal mean $E[Y]=8$, confirming that the conditional distribution changes with the observed value of $X$. The calculation is consistent with dependence. This check uses the bivariate normal conditional-mean formula; the independence conclusion already follows from the nonzero correlation.

Method and trap avoided: A nonzero covariance with finite positive marginal variances gives nonzero correlation, so independence is impossible. Do not confuse the converse direction: zero correlation would not establish independence without the normal-model condition.

Worked Example: Zero Correlation Without Independence

Let $X$ be uniform on $(-2,2)$, and define $Y=X^2$. Calculate the covariance, then compare $P(X>1,\ Y>1)$ with $P(X>1)P(Y>1)$. Use the comparison to decide whether $X$ and $Y$ are independent.

Step 1: Calculate the moments. The density of $X$ is $1/4$ on $(-2,2)$, and symmetry gives $$ E[X]=\frac{1}{4}\int_{-2}^{2}x\,dx=0, \qquad E[X^3]=\frac{1}{4}\int_{-2}^{2}x^3\,dx=0. $$ Also, $$ E[X^2]=\frac{1}{4}\int_{-2}^{2}x^2\,dx =\frac{1}{4}\left[\frac{x^3}{3}\right]_{-2}^{2} =\frac{1}{4}\left(\frac{8}{3}-\left(-\frac{8}{3}\right)\right) =\frac{4}{3}. $$ Because $Y=X^2$, $E[Y]=4/3$ and $E[XY]=E[X^3]=0$. Therefore $$ \operatorname{Cov}(X,Y) =E[XY]-E[X]E[Y] =0-(0)\left(\frac{4}{3}\right) =0. $$

Step 2: Calculate the event probabilities. The event $X>1$ corresponds to the interval $(1,2)$, whose length is 1 out of a total interval length of 4: $$ P(X>1)=\frac{2-1}{4}=\frac{1}{4}. $$ The condition $Y>1$ means $X^2>1$, or $|X|>1$. This consists of $(-2,-1)$ and $(1,2)$, total length 2, so $$ P(Y>1)=\frac{(2-1)+((-1)-(-2))}{4} =\frac{1+1}{4} =\frac{1}{2}. $$ When $X>1$, $Y=X^2>1$ automatically. Hence the intersection is just the event $X>1$: $$ P(X>1,\ Y>1)=\frac{1}{4}. $$

Step 3: Compare with the product. $$ P(X>1)P(Y>1) =\left(\frac{1}{4}\right)\left(\frac{1}{2}\right) =\frac{1}{8}. $$ Since $1/4\ne 1/8$, the events are not independent, and therefore $X$ and $Y$ are dependent despite having zero covariance.

As a separate probability check, let $C=\{|X|>1/2\}$. Its total interval length is $3$, so $P(C)=3/4$. Then $P(X>1)P(C)=(1/4)(3/4)=3/16=0.1875$, whereas $P(X>1,C)=1/4$, because $X>1$ implies $C$. This is a different event comparison; for the original events, the product is $1/8$, not $0.1875$.

Method and trap avoided: First translate each event into a set of $X$-values, then use interval lengths. In particular, $Y>1$ includes both tails, while the intersection with $X>1$ includes only the right tail. A zero covariance is not a substitute for checking independence in a general model.

What the Result Does—and Does Not—Say

The bivariate normal equivalence is useful because it turns a potentially difficult independence question into a parameter check. If a pair is known to be jointly normal, compute the covariance or correlation: zero means independence, and a nonzero value means dependence. The pair must be jointly bivariate normal; knowing only that each marginal distribution is normal is not enough to invoke the result.

Outside the bivariate normal model, zero correlation only says that the covariance vanishes. It does not rule out nonlinear dependence. In the uniform example, $Y$ is completely determined by $X$, yet the positive and negative values of $X$ cancel in the covariance calculation. The event comparison exposes the dependence that covariance misses.

Key takeaway: For a bivariate normal pair with positive marginal variances, $X$ and $Y$ are independent exactly when $\rho=0$. For general random variables, independence implies zero correlation when correlation is defined, but zero correlation alone does not imply independence.

Check Your Understanding

For each problem, identify whether the bivariate normal equivalence applies before drawing a conclusion.

A bivariate normal pair has $X\sim N(3,4)$, $Y\sim N(-1,9)$, and covariance 0. Determine whether the components are independent and find $P(X\leq 3,\ Y\leq -1)$. (Answer: Independent; probability $0.2500$.)
Let $X$ be uniform on $(-1,1)$ and $Y=X^2$. Find the covariance, then compare $P(X>1/2,\ Y>1/4)$ with the product of the marginal event probabilities. (Answer: Covariance $0$; intersection $1/4$, product $1/8$; dependent.)
A bivariate normal pair has marginal variances 16 and 9, and covariance $-6$. Calculate the correlation and decide whether the components are independent. (Answer: $\rho=-6/(4\cdot3)=-0.5$; not independent.)
A bivariate normal pair has $X\sim N(10,4)$, $Y\sim N(6,16)$, and correlation 0. Find $P(X\leq10,\ Y>6)$. (Answer: Independent; $(0.5)(0.5)=0.2500$.)
For a nonnormal pair, suppose $X$ is uniform on $(-2,2)$ and $Y=X^2$. Compare $P(X>1)$ and $P(Y>1)$ with their intersection to test independence. (Answer: Marginals $1/4$ and $1/2$, product $1/8$, intersection $1/4$; dependent despite zero covariance.)

What You'll Learn