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Show that the autocovariance function of stationary process {${X_t}$} with mean $\mu_X$ and variance $\gamma_X (0) > 0$ is positive definite, i.e.,

$\begin{equation} \sum^n_{t=1} \sum^n_{t'=1} a_ta_{t'}\gamma_X(|t-t'|) >0 \end{equation}$

in which {$a_i$} can be any sequence of real numbers.

My work so far, I was able to prove by stating $\mu_X = 0$ and doing so,

$a \in ℝ^n$ and $ \begin{pmatrix} X_1 \\ \vdots \\ X_n \end{pmatrix}$ with $E[X] = 0$,

$Cov(X) = \Gamma_n$ with $(\Gamma_n)_{t,t'} = \gamma(t-t')$,

Then $0 \leq E[(a^tX)^2] = Cov(a^tX)=a^t\Gamma_na=\sum^n_{t=1}\sum^n_{t'=1} \gamma(t-t')a_ta_{t'}$

but what changes when it comes to $\mu_X$ instead of 0?

motipai
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  • If $\mu_X \neq 0$, then your $0 \leq E[(a^tX)^2] = Cov(a^tX)$ needs to be replaced by $$0 \leq E[(a^t(X-\mu))^2] = Cov(a^tX)$$ and you are back in business. – Dilip Sarwate Oct 14 '19 at 18:23
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    Your proof already works for all cases. If you define $Y_t = X_t - \mu_X$, then all your calculations works to show that the autocovariance function of $Y$ is positive definite. But $\mbox{Cov}(Y_t, Y_{t'}) = \mbox{Cov}(X_t - \mu_X, X_{t'} - \mu_X) = \mbox{Cov}(X_t, X_{t'})$, hence they have the same autocovariance function. – Lucas Prates Oct 14 '19 at 18:43
  • but with a slight difference, in the formula on the question it does not include 0 and also there is a modulus inside de autocovariance – motipai Oct 14 '19 at 21:42

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