In order to prove asymptotic normality of the time series OLS estimator (this context is not important), I have done the following:


• $\mathrm{var}\left(y_{t}\right)=\gamma_{0}\quad\forall t$

• $ \mathrm{cov}\left(y_{t},y_{t-j}\right)=\gamma_{j}\in\mathbb{R}^{k}\quad\forall j$

• $\underset{j\rightarrow\infty}{\lim}\mathrm{cov}\left(y_{t},y_{t-j}\right)=\gamma_{j}=0$

• ${\sum}\left|\gamma_{j}\right|<\infty$


$$\begin{align} L&= \underset{{\scriptstyle T\rightarrow\infty}}{\lim}\mathrm{\mathrm{var}}\left(\sqrt{T}\left(T^{-1}\sum_{t}y_{t}-\mu\right)\right) \\ &= \underset{{\scriptstyle T\rightarrow\infty}}{\lim}\mathrm{\mathrm{var}}\left(T^{-1/2}\sum_{t}y_{t}\right) \\ &=\underset{{\scriptstyle T\rightarrow\infty}}{\lim}T^{-1}\mathrm{\mathrm{var}}\left(\sum_{t}y_{t}\right) \\ &=\underset{{\scriptstyle T\rightarrow\infty}}{\lim}T^{-1}\sum_{t}\sum_{s}\mathrm{cov}\left(y_{t},y_{s}\right) \\ &=\underset{{\scriptstyle T\rightarrow\infty}}{\lim}T^{-1}\left[T\cdot\gamma_{0}+2\cdot\left(T-1\right)\cdot\gamma_{1}+2\cdot\left(T-2\right)\cdot\gamma_{2}+...+2\cdot\gamma_{T-1}\right] \\ &=\underset{{\scriptstyle T\rightarrow\infty}}{\lim}T^{-1}\left[T\gamma_{0}+2\sum_{1\leq j\leq T-1}\left(T-j\right)\gamma_{j}\right] \\ &=\underset{{\scriptstyle T\rightarrow\infty}}{\lim}\gamma_{0}+2\sum_{1\leq j\leq T-1}\left(1-\dfrac{j}{T}\right)\gamma_{j} \end{align}$$ I'm assuming the last step equals: $\gamma_{0}+2\sum_{j=1}^{\infty}\gamma_{j}$

since concluding from here is rather easy. The problem being that I do not think it's the case, since j also converges to infinity.

Am I wrong? If so, why?

  • $\begingroup$ It's definitely true if $\sum j\gamma_j$ converges. $\endgroup$ – David Kleiman Feb 2 '16 at 15:35
  • $\begingroup$ Said sum converges (as $\sum{|\gamma_{j}|}$ does), as assumed. Could you please expose why? $\endgroup$ – rsm Feb 2 '16 at 15:37
  • $\begingroup$ The sum does not necessarily converge. Take $\gamma_j = 1/j^2$ . (In this case it your statement is true anyway). $\endgroup$ – David Kleiman Feb 2 '16 at 15:57
  • $\begingroup$ If you just multiply out the terms, you get $2(\sum\gamma_j \gamma_j - T^{-1}\sum j\gamma_j)$, so all you need is that $T^{-1}\sum j\gamma_j = 0$. $\endgroup$ – David Kleiman Feb 2 '16 at 15:59
  • $\begingroup$ So what you really need is that $\sum j\gamma_j$ is asymptotically less than $T$. $\endgroup$ – David Kleiman Feb 2 '16 at 16:01

Ok, so after a long battle with this one I think what you have written is true. Note that, since $\sum|\gamma_j|$ converges, we can find $N\in\mathbb{N}$ such that:

$$ \sum_{j=N}^T j| \gamma_j| \leq \sum_{j=N}^T j \frac cj = cT \;\;\;\; \forall c\in \mathbb{R} \setminus \{0\}. $$

This gives that

$$ \lim_{T\to \infty}T^{-1}\sum_{j=N}^T j |\gamma_j| \leq c\;\;\;\; \forall c\in \mathbb{R}\setminus \{0\}, $$

which implies that

$$ \lim_{T\to \infty}T^{-1}\sum_{j=1}^T j |\gamma_j| = 0. $$


$$ \underset{{\scriptstyle T\rightarrow\infty}}{\lim}\gamma_{0}+2\sum_{1\leq j\leq T-1}\left(1-\dfrac{j}{T}\right)\gamma_{j} = {\lim_{T\to\infty}}\gamma_{0}+2\sum_{j=1}^\infty\gamma_j - 2T^{-1}\sum_{j=1}^\infty j\gamma_j \\ ={\lim_{T\to\infty}}\gamma_{0}+2\sum_{j=1}^\infty\gamma_j $$

  • $\begingroup$ Perfect! Thank you. Notice that you meant that c is non-zero, if you want to edit. $\endgroup$ – rsm Feb 2 '16 at 17:37
  • $\begingroup$ Yes thank you. I forgot to mention that. $\endgroup$ – David Kleiman Feb 2 '16 at 17:38
  • $\begingroup$ Can you explain how did you get your first inequality. Thanks $\endgroup$ – chandu1729 Feb 2 '16 at 18:21
  • $\begingroup$ @chandu1729 I edited my answer to fix up some details. Since the series $\sum |\gamma_j|$ converges, its tail is less than any constant $c$ times the tail of the harmonic series (which diverges). $\endgroup$ – David Kleiman Feb 2 '16 at 19:19
  • $\begingroup$ But how does that imply your inequality. Its not clear. I feel your arguments are wrong. Please be more rigorous $\endgroup$ – chandu1729 Feb 2 '16 at 19:30

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