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I'm having a problem calculating the best linear predictor of a time series. I'm using the book Brockwell-Davis 2016 - Introduction to Time Series and Forecasts. First let me write down one notational convention, one proposition and one prerequisite problem:

  1. The best linear predictor in terms of $1,X_n,...,X_1$ is denoted by $P_nX_{n+h}$ and clearly has the form $P_nX_{n+h}=a_0+a_1X_n+...+a_nX_1$
  2. Let $(X_t, t\in\mathbb{Z})$ be a timeseries with $\text{Var}(X_t)<\infty$ for all $t\in \mathbb{Z}$ and $X^n := (X_{t_1},...,X_{t_n})$ a collection of random variables of the time series at $n$ different times. Then the best linear predictor of $X_t$ is given $P_nX_{n+h}$ as above in point 1. The coefficients $a_0,...,a_n$ are determined by the linear equations \begin{align} \mathbb{E}(X_t-P_nX_{n+h})&=0\\ \mathbb{E}(X_{t_j}(X_t-P_nX_{n+h})) &= 0, \quad \text{for all} \quad j=1,...,n. \tag1 \end{align}

The needed problem: Show that the process $X_t=A\cos(\omega t)+B\sin(\omega t), \ t=0,\pm1,...$ where $A$ and $B$ are uncorrelated random variables with mean 0 and variance 1 and $\omega$ a fixed frequency in $[0,\pi]$, is stationary and find its mean and autocovariance function.

Solving the above I came up with the answers $\mu_X=0$ and $\gamma_X(h)=\cos(\omega h)$ which are verified correct. Now the task is the following:

Let $\{X_t\}$ be the process defined in the problem above. Find $P_1X_2.$

Attempt:

According to the above we have $P_1X_2 = a_0+a_1X_1 = a_0+a_1A\cos(\omega)+a_1B\sin(\omega)$. The first equation in $(1)$ gives

\begin{align} \mathbb{E}[X_2-P_1X_2] &= \mathbb{E}[X_2-a_0-a_1A\cos(2\omega)-a_1B\sin(2\omega)]\\ &=\mathbb{E}[X_2]-a_0=0+a_0 = 0 \Longleftrightarrow a_0=0. \end{align}

The second equation in $(1)$ gives

\begin{align} \mathbb{E}[X_1(X_2-P_1X_2)] &= \mathbb{E}[X_1X_2] - \mathbb{E}[(X_1(a_0 + a_1A\cos(2\omega)+a_1B\sin(2\omega)))] \\ &= \mathbb{E}[2A^2\cos^3(\omega)-A^2\cos(\omega)+2B^2\cos(\omega)-2B^2\cos^3(\omega)]\\ &-a_1\mathbb{E}[2B^2\cos^3(\omega)-A^2\cos(\omega)+2B^2\cos(\omega)-2B^2cos(\omega)]\\ &= \mathbb{E}[(2B^2-A^2)\cos(\omega)]-a_1\mathbb{E}[(2B^2-A^2)\cos(\omega)]\\ &= \cos(\omega) - a_1\cos(\omega) = 0\Longleftrightarrow a_1 = 1. \end{align}

after plugging in $X_1$ and using that $\mathbb{E}[A^2]=\text{Var}[A]=1 =\mathbb{E}[B^2]=\text{Var}[B]$ as well as some trigonometric identities. Thus the best linear predictor of $X_2$ based on $1, X_1$ is $P_1X_2 = a_0 + a_1X_1 = X_1.$ Am I doing this correctly?

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So the question is asking: Let $X_t$ be the process defined in the problem above. Find $P_1X_2$.

$P_nX_{n+h}$ is the form of the linear predictor, I don't quite understand why you are assuming that they are asking you to predict $X_2$, as you are assuming that $n+h = t$.

They themselves state that $t = 0$ in the question statement above in the equation:

$X_t=A\cos(\omega t)+B\sin(\omega t), \ t=0,\pm1,...$

Let me know if this helps. I believe you may be incorrectly evaluating $X_t$ as $X_2$.

I'm not an expert and could very well be wrong however.

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  • $\begingroup$ Because $P_nX_{n+h}$ is the best linear predictor of $X_{n+h}$ in terms of $1, X_1,...,X_n$. So in my case $n=1$ and $h=2$ so we get $P_1X_2$ which is the best linear predictor of $X_2$ given $1, X_1$. $\endgroup$
    – Parseval
    May 24 at 22:08

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