I am having problems proving for a weakly stationary process $\{X_t : t\in T\}$:

$\rho_X(2)\geq 2 (\rho_X(1))^2-1$

where $\rho_X(j)=corr(X_t, X_{t+j})$.

So far I have shown that

$-1\leq \rho_X(1)\leq 1$

$|\rho_X(1)|\leq 1$


$2\rho_X(1)^2-1\leq 1$



where $\gamma_X(j)=cov(X_t, X_{t+j})$.

However, I am unsure how to proceed. Any help would be appreciated.

  • 1
    $\begingroup$ Hint: The correlation matrix for $(X_t, X_{t+1}, X_{t+2})$ must be positive semidefinite. In particular, its determinant must be nonnegative. $\endgroup$
    – whuber
    Commented Feb 7, 2015 at 17:58
  • $\begingroup$ $\det\pmatrix{1&&\rho_X(1)&&\rho_X(2)\\\rho_X(1)&&1&&\rho_X(1)\\\rho_X(2)&&\rho_X(1) &&1}\geq0$. $\endgroup$ Commented Feb 7, 2015 at 18:35
  • $\begingroup$ $[1-\rho_X^2(1)]-\rho_X(1)[\rho_X(1)-\rho_X(1)\rho_X(2)]+\rho_X(2)[\rho_X^2(1)- \rho_X(2)]\geq0$ $1-2\rho_X^2(1)+2\rho_X^2(1)\rho_X(2)-\rho_X^2(2)\geq0$ But still cannot show the inequality in my original question holds? $\endgroup$ Commented Feb 7, 2015 at 20:23

1 Answer 1


The correlation matrix, $\Gamma$ of your weakly stationary time series $X$ has elements given by:

$\Gamma_{i,j} = \rho_X(|t_i - t_j|) = corr(X_{t_i}, X_{t_j})$

This matrix is symmetric semi-positive definite (SSPD). This means that any principal submatrix you take also has to be semi positive definite. As was suggested by whuber, you can take the principal submatrix corresponding to $(X_t, X_{t+1}, X_{t+2})$, since any principal submatrix of a SSPD matrix also has to be SSPD, this is an SSPD matrix. One property of SSPD matrices that their determinant is greater than or equal to zero. So you take the determinant of this principal submatrix:

$\det\pmatrix{1&&\rho_X(1)&&\rho_X(2)\\\rho_X(1)&&1&&\rho_X(1)\\\rho_X(2)&&\rho_X(1) &&1}$

Which you calculated correctly as:

$[1-\rho_X^2(1)]-\rho_X(1)[\rho_X(1)-\rho_X(1)\rho_X(2)]+\rho_X(2)[\rho_X^2(1)- \rho_X(2)]$

All you missed is that you can factor this into:

$[\rho_X(2) - 1][2\rho_X(1)^2 - \rho_X(2) - 1]$

Since $\rho_X$ is a correlation:


Subtract 1 from all of this and you get:


So in particular $\rho_X-1≤0$. Then, since $[\rho_X(2) - 1][2\rho_X(1)^2 - \rho_X(2) - 1] \geq 0$, the second factor must also be $\leq 0$ (one's nonpositive, the whole thing's positive, so the other thing's nonpositive).

Rearranging that gives you the desired inequality.

  • $\begingroup$ thank-you for pointing out the final couple of steps which I completely overlooked! $\endgroup$ Commented Feb 7, 2015 at 23:45
  • $\begingroup$ No worries. At that point it's just algebra. :) $\endgroup$
    – zanda
    Commented Feb 8, 2015 at 0:17

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