Time-series and autocorrelation inequality 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 
$-1\leq \rho_X(1)\leq 1$
$|\rho_X(1)|\leq 1$
$2\rho_X(1)^2\leq2$
$2\rho_X(1)^2-1\leq 1$
and 
$\rho_X(j)=\frac{\gamma_X(j)}{\sigma^2}$
where $\gamma_X(j)=cov(X_t, X_{t+j})$.
However, I am unsure how to proceed. Any help would be appreciated.
 A: 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:
$−1≤\rho_X≤1$
Subtract 1 from all of this and you get:
$−2≤\rho_X-1≤0$
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.
