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Note that: $$ Z_tZ_{t+h} = \left(\sum_{i=0}^{n}a_iX_{t-i}\right) \left(\sum_{j=0}^{n}a_jX_{t+h-j}\right) \not =\sum_{i=0}^{n}a_i^2X_{t-i}X_{t+h-i} $$ Yes, as the expectation operator is linear, all we need is for: $$ E[X_{t-i}]=E[X_{s-i}]$$ and $$ E[X_{t-i}X_{t+h-j}] = E[X_{s-i}X_{s+h-j}] $$ to hold for all $h$, $i$, and $j$.


Other than the binomial expansion, what is the relationship between $B$ and $w$? As I understand correctly, there is no immediate relationship between the lag operator $B$ and the weights $w$. However, going from the fractional differencing operator, $(1-B)^d$ to $\hat{X}_t$ can be done as follows: \begin{align} (1-B)^d X_t &= \sum_{k=0}^{\infty} \begin{...


First, note that $\epsilon_t \sim N(0,1)$ is a white noise process and the random variates are simulated from a standard normal distribution. Hence, it does not make sense for you to multiply ret[2] and ret[3] with the MA-parameters, in order to reproduce ret[4]. The source code reveals how to reproduce the simulation values of the ARMA process: From the ...

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