Let ${X_t} = \mu + \sum\limits_{j = - \infty }^{ + \infty } {{\psi _j}{\varepsilon _{t - j}}}$ with $\varepsilon$ is a white noise iid with variance $\sigma^2$ , $\sum\limits_{j = - \infty }^{ + \infty } {\left| {{\psi _j}} \right|} < \infty $ and $\sum\limits_{j = - \infty }^{ + \infty } {{\psi _j}} \ne 0$

I want to show the asymptotic distribution for the mean $\mu$: $\sqrt n ({\bar X_n} - \mu ) \sim AN( {0,\sum\limits_{h = - \infty }^{ + \infty } {\gamma (h)} } )$

this is what I did:

Let ${X_{tm}} = \mu + \sum\limits_{j = - m}^m {{\psi _j}{\varepsilon _{t - j}}}$ and $W_{nm}={\overline X _{tm}} = \frac{1}{n}\sum\limits_{t = 1}^n {{X_{tm}}}$

Since $X_{tm}$ is $m$-dependent, $\sqrt n ({W_{nm}} - \mu ) \stackrel{D}{\rightarrow} {W_m}$ with ${W_m} \sim N( {0,\sum\limits_{j = - m}^m {\gamma (j)} } )$ hence, ${W_m}\xrightarrow{D}W$ with $W \sim N( {0,\sum\limits_{j = - \infty }^\infty {\gamma (j)} } )$

now, it left to show that $Var\left( {\sqrt n \left( {{{\bar X}_n} - {W_{nm}}} \right)} \right) \to 0$ as $m$,$n\rightarrow +\infty$

I compute:

$Var( {\sqrt n ( {{{\bar X}_n} - {W_{nm}}} )} ) = nVar( {\frac{1}{n}\sum\limits_{t = 1}^n {\sum\limits_{\left| j \right| > m} {{\psi _j}{\varepsilon _{t - j}}} } } )$

but I am stuck on proving that $nVar( {\frac{1}{n}\sum\limits_{t = 1}^n {\sum\limits_{\left| j \right| > m} {{\psi _j}{\varepsilon _{t - j}}} } } ) \to 0$ as $m$,$n\rightarrow +\infty$

$\gamma(h)$ is the autocovariance function of X

Some help would be appreciated


1 Answer 1


You may show that: $$n\mathrm{Var}\left(n^{-1}\sum_{t = 1}^n \sum_{|j| > m}\psi_j\varepsilon_{t - j}\right) \to \left(\sum_{|j| > m}\psi_j\right)^2\sigma^2$$ as $n \to \infty$ by expanding the variane of sums. Then the result follows from the condition of the absolute convergence of $\psi$ series. I also want to point out that the wording "as $m, n \to \infty$" is sloppy, the rigorous statement is: $$\lim_{m \to \infty}\limsup_{n \to \infty}n\mathrm{Var}\left(n^{-1}\sum_{t = 1}^n \sum_{|j| > m}\psi_j\varepsilon_{t - j}\right) = 0.$$

In other words, the order of $m$ and $n$ approach to $\infty$ matters. Also, to check variance converges to zero is a sufficient but not necessary condition, see, for example, Proposition 6.3.9 of Time Series: Theory and Methods (2nd edition) by P. J. Brockwell and R. A. Davis.


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