I was given this problem to think about, but I do not feel it is well posed.

Consider $X_1, X_2, X_3$ with zero mean and unit variance. Consider the correlation between any two of the variables to be $\rho$. If $X_1^2 + X_2^2 + X_3^2 = 1$, what is the range of values that $\rho$ can realize?

I think this question is ill-posed because:

$\text{var}(X_i) = 1 = E[X_i^2] - E[X_i]^2 = E[X_i^2]$

Then $E[X_1^2 + X_2^2 + X_3^2] = 3 \neq 1$. Which doesn't seem to make sense?

If we ignore this fact, and continue with the problem, I think we can show the lower bound on $\rho$ is 0.

$$ \text{cov}(X_i, X_j) = E[X_iX_j] - E[X_i]E[X_j] = E[X_iX_j] $$ If we assume $X_i, X_j$ are independent, then we will get 0 for the covariance, hence proving the lower bound of 0 on the correlation coefficient.

If they were dependent, is it possible that $E[X_iX_j]$ could be negative (which would give us a lower bound)?

How do I go about getting the upper bound?


1 Answer 1


I think that the condition $X_1^2+X_2^2+X_3^2=\rm{const}$ does not add smth to the problem:

Let $X_1, X_2, X_3$ have zero mean and unit variance. Consider the correlation between any two of the variables to be $\rho$. What is the range of $\rho$?

In this statement the answer is: $\rho\in [-0.5,\,1]$. Indeed, $$(X_1+X_2+X_3)^2 = X_1^2+X_2^2+X_3^2 + 2X_1X_2+2X_2X_3+2X_1X_3,$$ and applying expectations obtain $$0\leq \mathbb E(X_1 + X_2+X_3)^2 = \mathbb EX_1^2+\mathbb EX_2^2+\mathbb EX_3^2 + 2\mathbb E(X_1X_2)+2\mathbb E(X_2X_3)+2\mathbb E(X_1X_3)$$ $$ =1+1+1+2\rho+2\rho+2\rho=3+6\rho, $$ so $\rho\geqslant -\frac12$. The right bound $\rho\leq 1$ can be achived in examples: say, if $X_1\equiv X_2\equiv X_3$ take values $\pm 1$ only with equal probabilities $$ \mathbb P(X_1=1)=\mathbb P(X_1=-1)=\frac12. $$

One can also construct examples when the values $-\frac12$ or any intermediate values of $\rho$ are reached. Stay with Rademacher r.v.'s with common distribution $$ \mathbb P((X_1,X_2,X_3) = (-1,-1,-1)) =\mathbb P(-1,-1,-1) = p, $$ $$ \mathbb P(1,1,-1) = \mathbb P(-1,1,1) = \mathbb P(1,-1,1) = q, $$ $$ \mathbb P(-1,1,-1) = \mathbb P(-1,-1,1) = \mathbb P(1,-1,-1) = \frac14-\frac{p+q}2, $$ $$ \mathbb P(1,1,1) = \frac14 - \frac{3q}{2}+\frac{p}{2} $$ Here $p$ and $q$ are non-negative with $2p-2q=\rho$, $p+q\leq \frac12$ and with some other inequalities so that all probabilities are nonnegative.

If we take $p=\frac18$, $q=\frac38$ then $\rho=-\frac12$. If $p=q=\frac18$ then all the probabilities are the same, the variables are independent and $\rho=0$. All the intermediate cases can be achieved too.

  • $\begingroup$ I understand this answer, if you're given a generic expression, say, $X+Y=2$. Shouldn't it be such that the expectation of the LHS is equal to the RHS? If so, how can we have an expresion such as $X^2+Y^2+Z^2 = 1$ if $X,Y,Z$ have zero mean with unit variance? It would seem to be a contradiction. $\endgroup$ Commented Aug 24, 2020 at 20:59
  • 1
    $\begingroup$ @user5965026 I understand and agreed with you that the statement of the problem containes contradiction. And you propose to ignore this fact. If you replace $1$ by $3$, there would be no contradictions. And this changes have no effects on the answer, $\endgroup$
    – NCh
    Commented Aug 26, 2020 at 13:10

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