Def [ lag -l sample autocorrelation] is

\[\hat{\rho} = \frac{\sum_{t=l+1}^T (x_t - \bar{x})(x_{t-l}-\bar{x})}{\sum_{t=1}^T (x_t-\bar{x})^2}\]

consider two case:

  • Case (i): if $(x_t)$ are iid st. $\text{Var}(x_t) =\sigma^2$ and $\mathbb{E}[x_t^4] < \infty$. then

    \[\sqrt{T} \hat{\rho} = \frac{\frac{1}{\sqrt{T}}\sum_{l+1}^T (x_t - \bar{x})(x_{t+l}-\bar{x})}{\frac{1}{T}\sum_{t=1}^T(x_t-\bar{x})^2}\]

    replace $\bar{x}$ by $\mu$ because $x-\bar{x} = (x-\mu + \mu-\bar{x})$ and $\mu-\bar{x}$ is of higher order.


    \[\sqrt{T}\hat{\rho} = \frac{\frac{1}{\sqrt{T}}\sum_{t=l+1}^T(x_t-\mu)(x_{t-l}-\mu)}{\frac{1}{T}\sum_{t=1}^T (x_t-\mu)^2}\]

    The denominator is $\frac{1}{T}\sum_{t=1}^T (x_t-\mu)^2$ converges to $\mathbb{E}(x_t-\mu)^2=\sigma^2$ .

    Treat $y_t^{(l)} = (x_t-\mu)(x_{t-l}-\mu)$ and it is an $l$-dependent sequence with $\mathbb{E}y_{t}^{(l)}=0$. Then by central limit theorem,

    \[\frac{1}{\sqrt{T}} \sum_{t=l+1}^T y_t^{(l)} \rightarrow \mathcal{N}(0,\tau^2)\]


    \[\tau^2 =\gamma_0 + 2\gamma_1 + \dots + 2\gamma_l\]

    because $y^{(l)}$ is an $l$-dependent sequence. Then

    \[\begin{aligned} \gamma_m &= \text{Cov}(y_t^{(l)},y_{t+m}^{(l)}) \\ &= \text{Cov} (x_t-\mu)(x_{t-l}-\mu) (x_{t+m}-\mu)(x_{t+m-l}-\mu) \end{aligned}\]

    if $m\geq 1$ if $l\neq m$ $\gamma_m=0$. if $l=m$

    \[\gamma_m = \mathbb{E}[(x_t-\mu)^2(x_{t-l}-\mu)(x_-\mu)] = 0\]


    \[\begin{aligned} \tau^2 &= \gamma_0 = \text{Var}(y_t) = \mathbb{E}[y_t^2] \\ &=\mathbb{E}[(x_t-\mu)^2(x_{t-l}-\mu)^2] \\ &= \mathbb{E}[(x_t-\mu)^2] \mathbb{E}[(x_{t-l}-\mu)^2]=\sigma^2 \end{aligned}\]

    Thus, since the numerator is $\sigma^2$.

    \[\sqrt{T}\hat{\rho}\rightarrow \mathcal{N}(0,1)\]
  • Case(ii) $(x_t)$ is a linear time series.

    \[x_t = \mu + \epsilon_1 + \psi_1\epsilon_{t-1} + \dots + \psi_{q}\epsilon_{t-q}\]

    where $\epsilon_t$ are iid with mean zero and $\mathbb{E}[\epsilon^4]<\infty$ with $\psi_0=1$. Then

    \[\sqrt{T}\hat{\rho}_l \rightarrow \mathcal{N}(0,\tau^2)\]


    \[\tau^2 = 1 + 2\sum_{i=1}^q \rho_i^2 \quad \text{ for all } l>q\]

    Remark The asymptotic variance only depends on the second moments and no fourth moments show up. This is because the series is linear.

    General formula when $(x_t)$ is linear time series for $l\geq 1$.

    \[\sqrt{T} (\hat{\rho}_l - \rho_l) \rightarrow \mathcal{N}(0,\tau^2)\]


    \[\tau^2 = \sum_{h=1}^\infty (\rho_{h+l}+\rho_{h-l} - 2\rho_h\rho_l)^2\]

    this is called Bartlett’s formula.

    for lag $l>q$,

    \[\begin{aligned} \tau^2 &= \sum_{h=1}^\infty (\rho_{h+l} + \rho_{h-l} - 2\rho_h\rho_l) \\ &= \sum_{h=1}^\infty \rho_{h-l}^2 \\ &= \rho_{l-1}^2 + \rho_{l-2}^2 + \dots + \rho_0^2 + \rho_{-1}^2 + \dots \\ &= \rho_{q}^2 + \dots + \rho_1^2+\rho_0^2+\rho_{-1}^2+ \dots + \rho_{-q}^2 \\ &= 1 + 2\sum_{i=1}^q \rho_i^2 \end{aligned}\]

Testing for ACF

  • inidividual test: given $l\geq 1$, test $H_0 : \rho_l = 0$ vs $H_1: \rho_l \neq 0$,

    \[\frac{\sqrt{T}\hat{\rho}_l}{\tau} \rightarrow \mathcal{N}(0,1)\]


    \[\tau = \frac{1}{T} \left(1+2\sum_{i=1}^{l-1}\rho_i^2\right)\]

    Here we repalce the upper limit to $l-1$ because we assume $\rho_l$ decays with $l$ and if we assume $\rho_l =0$, we neglect other terms. This is a weaker form of the test. It underestimate the variance. It rejects more cases and the false negative error is lowered and thus the power is higher.

  • Portmanteau test:

    $H_0:$ $\rho_1 = \dots = \rho_m = 0$. vs $H_1:$ $\rho_i \neq 0$ for some $i\in{1,\dots,m}$.

    Box-Pierce : one of the portmanteau test. The statistic: \(Q^*(m) = T\sum_{i=1}^m \hat{\rho}_i^2\) If $x_t$ are iid with moment conditions then \(Q^*(m) \rightarrow \chi^2(m) ,\quad \text{ as } T\rightarrow \infty\)

    Ljung-Box: The statistic:

    \[Q(m) = T(T+2) \sum_{i=1}^m \frac{\hat{\rho}^2_i}{T-i}\]

    This gives more weights to the small legs estimators because there are more samples in the small legs estimation. This imporved power in finite sample.

    The choice of $m$ would be $m\sim \log(T)$.


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