Index: The Book of Statistical ProofsProbability DistributionsUnivariate continuous distributionsNormal distribution ▷ Normal and uncorrelated does not imply independent

Theorem: Consider two random variables X and Y. If each of them is normally distributed and both are uncorrelated, then X and Y are not necessarily independent.

Proof: As an example, let V follow a Bernoulli distribution with success probability 1/2 and let W be defined as a transformation of V:

\label{eq:V-W} \begin{split} V &\sim \mathrm{Bern}\left( \frac{1}{2} \right) \\ W &= 2V-1 \; . \end{split}

By definition of the Bernoulli distribution, it follows that

\label{eq:V-W-dist} p(V=0) = p(V=1) = \frac{1}{2} \quad \Rightarrow \quad p(W=-1) = p(W=+1) = \frac{1}{2} \; .

Moreover, let X follow a standard normal distribution and let Y be defined as a combination of X and W:

\label{eq:X-Y} \begin{split} X &\sim \mathcal{N}(0,1) \\ Y &= WX \; . \end{split}

Then, by the nature of the random variable W, it follows that

\label{eq:X-Y-dist} p(W=-1) = p(W=+1) = \frac{1}{2} \quad \Rightarrow \quad p(Y=-X) = p(Y=+X) = \frac{1}{2} \; .

Since the negative of a standard normal random variable is also standard normally distributed,

\label{eq:X-dist} X \sim \mathcal{N}(0,1) \quad \Rightarrow \quad -X \sim \mathcal{N}(0,1) \; ,

we can calculate the probability density function belonging to the mixture distribution of Y as follows:

\label{eq:Y-pdf} \begin{split} p(y) &= p(y|Y=-X) \cdot p(Y=-X) + p(y|Y=+X) \cdot p(Y=+X) \\ &\overset{\eqref{eq:X-Y-dist}}{=} \mathcal{N}(y; 0, 1) \cdot \frac{1}{2} + \mathcal{N}(y; 0, 1) \cdot \frac{1}{2} \\ &= \mathcal{N}(y; 0, 1) \end{split}

where we have used the law of marginal probability in the first line and \mathcal{N}(x; \mu, \sigma^2) denotes the probability density function of the normal distribution. Thus, Y is also standard normally distributed:

\label{eq:Y-dist} Y \sim \mathcal{N}(0,1) \; .

This means that both X and Y have expected value zero:

\label{eq:X-Y-mean} \mathrm{E}(X) = \mathrm{E}(Y) = 0 \; .

With that, we can start to work out the covariance of X and Y:

\label{eq:X-Y-cov-s1} \begin{split} \mathrm{Cov}(X,Y) &= \mathrm{E}\left[ \left( X-\mathrm{E}(X) \right) \left( Y-\mathrm{E}(Y) \right) \right] \\ &\overset{\eqref{eq:X-Y-mean}}{=} \mathrm{E}\left[ XY \right] \\ &\overset{\eqref{eq:X-Y}}{=} \mathrm{E}\left[ XWX \right] \\ &= \mathrm{E}\left[ WX^2 \right] \; . \end{split}

Since W and X are independent by construction, their expected values factorize:

\label{eq:X-Y-cov-s2} \begin{split} \mathrm{Cov}(X,Y) &= \mathrm{E}[W] \cdot \mathrm{E}[X^2] \\ &= \left( (-1) \cdot p(W=-1) + (+1) \cdot p(W=+1) \right) \cdot \mathrm{E}[X^2] \\ &\overset{\eqref{eq:V-W-dist}}{=} \left( (-1) \cdot \frac{1}{2} + (+1) \cdot \frac{1}{2} \right) \cdot \mathrm{E}[X^2] \\ &= 0 \cdot \mathrm{E}[X^2] \\ &= 0 \; . \end{split}

Thus, X and Y are uncorrelated:

\label{eq:X-Y-corr} \mathrm{Corr}(X,Y) = \frac{\mathrm{Cov}(X,Y)}{\sqrt{\mathrm{Var}(X)} \sqrt{\mathrm{Var}(Y)}} = 0 \; .

Yet, X and Y are not independent, since the marginal density of Y is

\label{eq:Y-dist-marg} p(y) = \mathcal{N}(y; 0, 1) \; ,

but the conditional density of Y given X is

\label{eq:Y-dist-cond} p(y|x) = \left\{ \begin{array}{rl} 1/2 \; , & \text{if} \; y = -x \\ 1/2 \; , & \text{if} \; y = +x \\ 0 \; , & \text{otherwise} \end{array} \right. \; ,

thus violating the behavior of probability under independence:

\label{eq:X-Y-dep} p(Y) \neq p(Y|X) \; .

Therefore, X and Y defined by \eqref{eq:X-Y} and \eqref{eq:V-W} constitute an example for two random variables that are normally distributed and uncorrelated, but not independent.

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Metadata: ID: P473 | shortcut: norm-corrind | author: JoramSoch | date: 2024-10-04, 10:59.