Uncorrelated

In probability theory and statistics, two real-valued random variables are said to be uncorrelated if their covariance is zero. Uncorrelatedness is by definition pairwise; i.e. to say that more than two random variables are uncorrelated simply means that any two of them are uncorrelated.

Uncorrelated random variables have a correlation coefficient of zero, except in the trivial case when either variable has zero variance (is a constant). In this case the correlation is undefined.

In general, uncorrelatedness is not the same as orthogonality, except in the special case where either X or Y has an expected value of 0. In this case, the covariance is the expectation of the product, and X and Y are uncorrelated if and only if E(XY) = 0.

If X and Y are independent, then they are uncorrelated. However, not all uncorrelated variables are independent. For example, if X is a continuous random variable uniformly distributed on [−1, 1] and Y = X², then X and Y are uncorrelated even though X determines Y and a particular value of Y can be produced by only one or two values of X.

A set of two or more random variables is called uncorrelated if each pair of them are uncorrelated.

Another example to prove that uncorrelated random variables are not necessarily independent

• Let X be a random variable that takes the value 0 with probability 1/2, and takes the value 1 with probability 1/2.
• Let Z be a random variable that takes the value -1 with probability 1/2, and takes the value 1 with probability 1/2.
• Let U be a random variable constructed as U=XZ.

The claim is that U and X have zero covariance (and thus are uncorrelated), but are not independent.

Proof:

First note:

• $E[U] = 0\times1/2 + 1\times1/4 + (-1)\times 1/4 = 0$
• $E[X] = 0\times1/2+1\times1/2 = 1/2$

Now, by definition $\mathrm{cov}(U,X) = E[(U-E[U])(X-E[X])] = E[ U (X-1/2)] = E[X^2Z - (1/2)XZ] = E[X^2Z] - (1/2)E[XZ]$

• $E[X^2Z] = 0\times1/2 + 1\times1/4 + (-1)\times 1/4 = 0$
• $E[XZ] = 0\times1/2 + 1\times1/4 + (-1)\times 1/4 = 0$

Therefore $\mathrm{cov}(U,X)=0$

A necessary condition for showing that U and X are independent is showing that for any number a and b, $Pr(U=a|X=b) = Pr(U=a)$. We prove that this is not true. PIck a=1 and b=0.

• $Pr(U=1|X=0) = Pr(XZ=1|X=0) = 0$
• $Pr(U=1) = Pr(XZ=1) = 1/4$

Thus $Pr(U=1|X=0)\ne Pr(U=1)$ so U and X are not independent.

Q.E.D.

When uncorrelatedness implies independence

There are cases in which uncorrelatedness does imply independence. One of these cases is when both random variables are two-valued (which reduces to binomial distributions with n=1). See Binomial_distribution#Covariance_between_two_binomials for more information. Further, two jointly normally distributed random variables are independent if they are uncorrelated, although this does not hold for variables whose marginal distributions are normal and uncorrelated but whose joint distribution is not joint normal: see Normally distributed and uncorrelated does not imply independent.

See also

• Correlation and dependence

References

• Galen R. Shorack, Probability for Statisticians, Birkhäuser, 2000, p127.