Imagine we have given a stationary process $\{X_t\}_{t \in \mathbb Z}$;

Now I want to show that

$$\operatorname{Var}(X_1+\cdots+X_T) \to \infty,$$ for $T \to \infty$.

We have that the process if $\phi$-mixing with $\phi(n) \leq n^{-(1+\varepsilon)}$ for some $\varepsilon >0$.

For those who don't know the meaning of $\phi$-mixing: You can see it here: https://www.encyclopediaofmath.org/images/a/a7/Strong_mixing_conditions.pdf, Page 2;

Now I don't have a clear plan; I could say:

$$\operatorname{Var}(X_1+\cdots+X_T)=\sum_{i=1}^T \sum_{j=1}^T \operatorname{Cov}(X_i,X_j)$$

The variance of each element is the same and strictly greater than zero since the process is not constant; Now I somehow should use the mixing coefficient to show that the variance of the sum really goes to infinity;

One idea is using $\rho(n)\leq 2\sqrt{\phi(n)}$, where $\rho$ is defined again as in:

https://www.encyclopediaofmath.org/images/a/a7/Strong_mixing_conditions.pdf, Page 2

(since $\operatorname{Cov}(X_1,X_j)\geq -\rho(\vert i-j\vert)\operatorname{Var}(X_1)$)

However then the decay-rate of the $\phi$-mixing coefficient is not enough; Do you have any other idea to show the statement?

  • 2
    $\begingroup$ The conclusion $\mathsf{Var}(X_1+\cdots+X_T)\to\infty$ may not be true, for example, $\varepsilon_n, n\ge 1$ is a sequence of $N(0,1)$ and i.i.d. random variables and $X_n=\varepsilon_n-\varepsilon_{n+1}$. $\endgroup$ – JGWang Aug 11 '17 at 7:18
  • $\begingroup$ Thank you very much - What could be a property that my statement still holds? $\endgroup$ – user299124 Aug 11 '17 at 18:35
  • 1
    $\begingroup$ If $\{X_n,n\in\mathbb{Z}\}$ exists spectral density $f(\lambda)$ and $f$ is continuous at $\lambda=0$, then $\mathsf{var}[\sum\limits_{j=1}^nX_j]=2\pi f(0)n+o(n)$. $\endgroup$ – JGWang Aug 12 '17 at 2:30
  • 1
    $\begingroup$ If $\{X_n,n\in\mathbb{Z}\}$ is $\phi$-mixing stationary process with spectral function $F$, then $\lim\limits_{n\to\infty}\mathsf{Var}[\sum\limits_{j=1}^nX_j]=\frac12\int_{-\pi}^\pi\text{cosec}^2(\frac\lambda2)dF(\lambda)$. $\endgroup$ – JGWang Aug 12 '17 at 2:49
  • $\begingroup$ Okay, that seems to be helpful - thank you $\endgroup$ – user299124 Aug 13 '17 at 15:03

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