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Extend autocorrelation

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@ -179,8 +179,8 @@ mathematically formulated by strict stationarity.
$Cov[X_t,X_{t+h}] = \gamma_h$ & autocovariance depends only on lag $h$ \\
\end{tabular}
\subsubsection{Weak}
It is impossible to "prove" the theoretical concept of stationarity from data. We can only search for evidence in favor or against it. \\
\subsubsection{Weak} \label{weak-stationarity}
It is impossible to "prove" the theoretical concept of stationarity from data. We can only search for evidence in favor or against it. \\
\vspace{0.1cm}
However, with strict stationarity, even finding evidence only is too difficult. We thus resort to the concept of weak stationarity.
@ -393,7 +393,7 @@ fit <- gam(log(maine) ~ s(tnum) + mm)
\end{lstlisting}
\section{Autocorrelation}
For most of the rest of this course, we will deal with (weakly) stationary time series. See \ref{}
For most of the rest of this course, we will deal with (weakly) stationary time series. See \ref{weak-stationarity} \\
\vspace{.2cm}
Definition of autocorrelation at lag $k$
$$Cor(X_{t+k},X_t) = \frac{Cov(X_{k+t},X_t)}{\sqrt{Var(X_{k+t})\cdot Var(X_t)}} = \rho(k)$$
@ -410,25 +410,134 @@ We assume $\rho(k) = 0.7$
\item From this we can also conclude that any $\rho(k) < 0.4$ is not a strong association, i.e. has a small effect on the next observation only.
\end{itemize}
\subsection{Lagged scatterplot approach}
Create a plot of $(x_t, x_{t+k}) \, \forall \, t = 1,...,n-k$ and compute the canonical Pearson correlation coefficient of these pairs and use it as an estimation for the autocorrelation $\tilde{\rho}(k)$
\begin{lstlisting}[language=R]
lag.plot(wave, do.lines=FALSE, pch=20)
\end{lstlisting}
\begin{figure}[H]
\centering
\includegraphics[width=.25\textwidth]{lagged-scatterplot.png}
\caption{Lagged scatterplot example for $k=1$}
\label{fig:lagged-scatterplot}
\end{figure}
\subsection{Plug-in estimation}
Plug-in estimation relies on the canonical covariance estimator:
$$\hat{\rho}(k) = \frac{Cov(X_t,X_{t+k})}{Var(X_t)}$$
Plug-in estimates are biased, i.e. shrunken towards zero for large lags $k$. Nevertheless, they are generally more reliable and precise.
\begin{figure}[H]
\centering
\includegraphics[width=.25\textwidth]{lagged-scatterplot-vs-plug-in.png}
\caption{Lagged scatterplot estimation vs. plug-in estimation}
\label{fig:lagged-scatterplot-vs-plug-in}
\end{figure}
\subsection{Important points on ACF estimation}
\begin{itemize}
\item Correlations measure linear association and usually fail if there are non-linear associations between the variables.
\item The bigger the lag $k$ for which $\rho(k)$ is estimated, the fewer data pairs remain. Hence the higher the lag, the bigger the variability in $\hat{\rho}(k)$ .
\item To avoid spurious autocorrelation, the plug-in approach shrinks $\hat{\rho}(k)$ for large $k$ towards zero. This creates a bias, but pays off in terms of mean squared error.
\item Autocorrelations are only computed and inspected for lags up to $10 \log_{10}(n)$, where they have less bias/variance
\end{itemize}
\subsection{Correlogram}
\begin{lstlisting}[language=R]
acf(wave, ylim=c(-1,1))
\end{lstlisting}
\begin{figure}[H]
\centering
\includegraphics[width=.25\textwidth]{correlogram.png}
\caption{Example correlogram}
\label{fig:correlogram}
\end{figure}
\subsubsection{Confidence Bands}
Even for an i.i.d. series $X_t$ without autocorrelation, i.e. $\rho(k) = 0 \, \forall \, k$, the estimates will be different from zero: $\hat{\rho}(k) \neq 0$ \\
\textbf{Question}: Which $\hat{\rho}(k)$ are significantly different from zero?
$$\hat{\rho}(k) \sim N(0,1/n), \; \mathrm{for \, large} \, n$$
\begin{itemize}
\item Under the null hypothesis of an i.i.d. series, a 95\% acceptance region for the null is given by the interval $\pm 1.96 / \sqrt{n}$
\item For any stationary series, $\hat{\rho}(k)$ within the confidence bands are considered to be different from 0 only by chance, while those outside are considered to be truly different from zero.
\end{itemize}
\textbf{Type I Errors} \\
For iid series, we need to expect 5\% of type I errors, i.e. $\hat{\rho}(k)$ that go beyond the confidence bands by chance. \\
\textbf{Non i.i.d. series} \\
The confidence bands are asymptotic for i.i.d. series. Real finite length non-i.i.d. series have different (unknown) properties.
\subsection{Ljung-box test}
The Ljung-Box approach tests the null hypothesis that a number of autocorrelation coefficients are simultaneously equal to zero. \\
Thus, it tests for significant autocorrelation in a series. The test statistic is:
$$Q(h) = n(n+2)\sum_{k=1}^h \frac{\hat{\rho}^2}{n-k} \sim \chi_h^2$$
\begin{lstlisting}[language=R]
Box.test(wave, lag=10, type="Ljung-Box")
\end{lstlisting}
\subsection{ACF and outliers}
The estimates $\hat{\rho}(k)$ are sensitive to outliers. They can be diagnosed using the lagged scatterplot, where every single outlier appears twice. \\
\vspace{.2cm}
\textbf{Some basic strategies for dealing with outliers}
\begin{itemize}
\item if it is bad data point: delete the observation
\item most (if not all) R functions can deal with missing data
\item if complete data are required, replace missing values with
\begin{itemize}
\item global mean of the series
\item local mean of the series, e.g. $\pm 3$ observations
\item fit a time series model and predict the missing value
\end{itemize}
\end{itemize}
\subsection{Properties of estimated ACF}
\begin{itemize}
\item Appearance of the series $\Rightarrow$ Appearance of the ACF \\ Appearance of the series $\nLeftarrow$ Appearance of the ACF
\item The compensation issue: \\ $\sum_{k=1}^{n-1}\hat{\rho}(k) = -1/2$ \\ All estimable autocorrelation coefficients sum up to -1/2
\item For large lags $k$ , there are only few data pairs for estimating $\rho(k)$. This leads to higher variability and hence the plug-in estimates are shrunken towards zero.
\end{itemize}
\subsection{Application: Variance of the arithmetic mean}
We need to estimate the mean of a realized/observed time series. We would like to attach a standard error
\begin{itemize}
\item If we estimate the mean of a time series without taking into account the dependency, the standard error will be flawed.
\item This leads to misinterpretation of tests and confidence intervals and therefore needs to be corrected.
\item The standard error of the mean can both be over-, but also underestimated. This depends on the ACF of the series.
\end{itemize}
\subsubsection{Confidence interval}
For a 95\% CI:
$$\hat{\mu} \pm 1.96 \sqrt{\frac{\gamma(0)}{n^2} \bigg(n + 2 \cdot \sum_{k=1}^{10log_{10}(n)}(n-k)\rho(k) \bigg)}$$
In R we can use
\begin{lstlisting}[language=R]
n <- length(b)
var.ts <- 1/n^2*acf(b,lag=0,type="cov")$acf[1]*(n+2*sum(((n-1):(n-10))*acf(b,10)$acf[-1]))
mean(b) + c(-1.96,1.96)*sqrt(var.ts)
\end{lstlisting}
\scriptsize
\section*{Copyleft}
\doclicenseImage \\
Dieses Dokument ist unter (CC BY-SA 3.0) freigegeben \\
\section*{Copyright}
Nearly everything is copy paste from the slides or the script. Copyright belongs to M. Dettling \\
\faGlobeEurope \kern 1em \url{https://n.ethz.ch/~jannisp/ats-zf} \\
\faGit \kern 0.88em \url{https://git.thisfro.ch/thisfro/ats-zf} \\
Jannis Portmann, FS21
\section*{Referenzen}
\section*{References}
\begin{enumerate}
\item Skript
\item ATSA\_Script\_v219219.docx, M. Dettling
\item ATSA\_Slides\_v219219.pptx, M. Dettling
\end{enumerate}
\section*{Bildquellen}
\begin{itemize}
\item Bild
\end{itemize}
\section*{Image sources}
All pictures are taken from the slides or the script mentioned above.
\end{multicols*}