3
$\begingroup$

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

$\endgroup$

1 Answer 1

5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.