Normally distributed and uncorrelated does not imply independent

To say that the pair $(X,Y)$ of random variables has a bivariate normal distribution means that every linear combination $aX+bY$ of $X$ and $Y$ for constant (i.e. not random) coefficients $a$ and $b$ has a univariate normal distribution. In that case, if $X$ and $Y$ are uncorrelated then they are independent. However, it is possible for two random variables $X$ and $Y$ to be so distributed jointly that each one alone is marginally normally distributed, and they are uncorrelated, but they are not independent; examples are given below.

Examples

A symmetric example

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Joint range of $X$ and $Y$. Darker indicates higher value of the density function.

Suppose $X$ has a normal distribution with expected value 0 and variance 1. Let $W$ have the Rademacher distribution, so that $W=1$ or $W=-1$, each with probability 1/2, and assume $W$ is independent of $X$. Let $Y=WX$. Then

• $X$ and $Y$ are uncorrelated;
• both have the same normal distribution; and
• $X$ and $Y$ are not independent.

To see that $X$ and $Y$ are uncorrelated, one may consider the covariance $\mathrm{cov}(X,Y)$: by definition, it is

$\mathrm{cov}(X,Y)=\mathrm{E}(XY)-\mathrm{E}(X)\mathrm{E}(Y).$

Then by definition of the random variables $X$, $Y$, and $W$, and the independence of $W$ from $X$, one has

$\mathrm{cov}(X,Y)=\mathrm{E}(XY)-0=\mathrm{E}(X^{2}W)=\mathrm{E}(X^2)\mathrm{E}(W)=\mathrm{E}(X^2)\cdot 0=0.$

To see that $Y$ has the same normal distribution as $X$, consider

$\begin{array}{rl}\mathrm{Pr}(Y\le x)& =\mathrm{E}(\mathrm{Pr}(Y\le x\mid W))\\ & =\mathrm{Pr}(X\le x)\mathrm{Pr}(W=1)+\mathrm{Pr}(-X\le x)\mathrm{Pr}(W=-1)\\ & =\mathrm{\Phi }(x)\cdot \frac{1}{2}+\mathrm{\Phi }(x)\cdot \frac{1}{2}\end{array}$

(since $X$ and $-X$ both have the same normal distribution), where $\mathrm{\Phi }(x)$ is the cumulative distribution function of the normal distribution..

To see that $X$ and $Y$ are not independent, observe that $|Y|=|X|$ or that $\mathrm{Pr}(Y>1|X=1/2)=\mathrm{Pr}(X>1|X=1/2)=0$.

Finally, the distribution of the simple linear combination $X+Y$ concentrates positive probability at 0: $\mathrm{Pr}(X+Y=0)=1/2$. Therefore, the random variable $X+Y$ is not normally distributed, and so also $X$ and $Y$ are not jointly normally distributed (by the definition above).

An asymmetric example

The joint density of $X$ and $Y$. Darker indicates a higher value of the density.

Suppose $X$ has a normal distribution with expected value 0 and variance 1. Let

$Y=\{\begin{array}{cc}X& \text{if }\left|X\right|\le c\\ -X& \text{if }\left|X\right|>c\end{array}$

where $c$ is a positive number to be specified below. If $c$ is very small, then the correlation $\mathrm{corr}(X,Y)$ is near $-1$ if $c$ is very large, then $\mathrm{corr}(X,Y)$ is near 1. Since the correlation is a continuous function of $c$, the intermediate value theorem implies there is some particular value of $c$ that makes the correlation 0. That value is approximately 1.54. In that case, $X$ and $Y$ are uncorrelated, but they are clearly not independent, since $X$ completely determines $Y$.

To see that $Y$ is normally distributed—indeed, that its distribution is the same as that of $X$ —one may compute its cumulative distribution function:

$\begin{array}{rl}\mathrm{Pr}(Y\le x)& =\mathrm{Pr}(\{|X|\le c\text{ and }X\le x\}\text{ or }\{|X|>c\text{ and }-X\le x\})\\ & =\mathrm{Pr}(|X|\le c\text{ and }X\le x)+\mathrm{Pr}(|X|>c\text{ and }-X\le x)\\ & =\mathrm{Pr}(|X|\le c\text{ and }X\le x)+\mathrm{Pr}(|X|>c\text{ and }X\le x)\\ & =\mathrm{Pr}(X\le x),\end{array}$

where the next-to-last equality follows from the symmetry of the distribution of $X$ and the symmetry of the condition that $|X|\le c$.

In this example, the difference $X-Y$ is nowhere near being normally distributed, since it has a substantial probability (about 0.88) of it being equal to 0. By contrast, the normal distribution, being a continuous distribution, has no discrete part—that is, it does not concentrate more than zero probability at any single point. Consequently $X$ and $Y$ are not jointly normally distributed, even though they are separately normally distributed.

See also

• Correlation and dependence