backshift operator notation
$begingroup$
- Original equation:
$$begin{equation}
z_t = phi z_{t-1} + z_{t-1} - phi z_{t-2} + omega_t
end{equation}$$
- Rewrite the equation, re-arrange terms, and factorize them:
$$begin{align}
z_t - z_{t-1} &= phi (z_{t-1} - z_{t-2}) + omega_t \
(z_t - z_{t-1}) - phi (z_{t-1} - z_{t-2}) &= omega_t \
(1 - phi B)(z_t - z_{t-1}) &= omega_t \
(1 - phi B) bigtriangledown z_t &= omega_t \
(1 - phi B)(1 - B)z_t &= omega_t
end{align}$$
BACKGROUND:
I'm trying to understand backshift operator notation in the context of this ARIMA example from machine learning time series https://github.com/etcrago/Tutorial-Arima-w-jeffrey-yau (lecture 2).
QUESTION:
In the 3rd line he somehow jumps to $(1-B)$, but my understanding is the $phi$
part of the equation should be factored to $(B-B^2)$, not $(1-B)$. How does the author jump to that $(1-B)$ factorization ?
notation time-series
asked Dec 23 '18 at 14:28
alan dalan d
213
$endgroup$
add a comment |
$begingroup$
- Original equation:
$$begin{equation}
z_t = phi z_{t-1} + z_{t-1} - phi z_{t-2} + omega_t
end{equation}$$
- Rewrite the equation, re-arrange terms, and factorize them:
$$begin{align}
z_t - z_{t-1} &= phi (z_{t-1} - z_{t-2}) + omega_t \
(z_t - z_{t-1}) - phi (z_{t-1} - z_{t-2}) &= omega_t \
(1 - phi B)(z_t - z_{t-1}) &= omega_t \
(1 - phi B) bigtriangledown z_t &= omega_t \
(1 - phi B)(1 - B)z_t &= omega_t
end{align}$$
BACKGROUND:
I'm trying to understand backshift operator notation in the context of this ARIMA example from machine learning time series https://github.com/etcrago/Tutorial-Arima-w-jeffrey-yau (lecture 2).
QUESTION:
In the 3rd line he somehow jumps to $(1-B)$, but my understanding is the $phi$
part of the equation should be factored to $(B-B^2)$, not $(1-B)$. How does the author jump to that $(1-B)$ factorization ?
notation time-series
asked Dec 23 '18 at 14:28
alan dalan d
213
$endgroup$
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 14:36
2
$begingroup$
DONE. Thank you.
$endgroup$
– alan d
Dec 23 '18 at 15:28
add a comment |
$begingroup$
- Original equation:
$$begin{equation}
z_t = phi z_{t-1} + z_{t-1} - phi z_{t-2} + omega_t
end{equation}$$
- Rewrite the equation, re-arrange terms, and factorize them:
$$begin{align}
z_t - z_{t-1} &= phi (z_{t-1} - z_{t-2}) + omega_t \
(z_t - z_{t-1}) - phi (z_{t-1} - z_{t-2}) &= omega_t \
(1 - phi B)(z_t - z_{t-1}) &= omega_t \
(1 - phi B) bigtriangledown z_t &= omega_t \
(1 - phi B)(1 - B)z_t &= omega_t
end{align}$$
BACKGROUND:
I'm trying to understand backshift operator notation in the context of this ARIMA example from machine learning time series https://github.com/etcrago/Tutorial-Arima-w-jeffrey-yau (lecture 2).
QUESTION:
In the 3rd line he somehow jumps to $(1-B)$, but my understanding is the $phi$
part of the equation should be factored to $(B-B^2)$, not $(1-B)$. How does the author jump to that $(1-B)$ factorization ?
notation time-series
asked Dec 23 '18 at 14:28
alan dalan d
213
$endgroup$
- Original equation:
$$begin{equation}
z_t = phi z_{t-1} + z_{t-1} - phi z_{t-2} + omega_t
end{equation}$$
- Rewrite the equation, re-arrange terms, and factorize them:
$$begin{align}
z_t - z_{t-1} &= phi (z_{t-1} - z_{t-2}) + omega_t \
(z_t - z_{t-1}) - phi (z_{t-1} - z_{t-2}) &= omega_t \
(1 - phi B)(z_t - z_{t-1}) &= omega_t \
(1 - phi B) bigtriangledown z_t &= omega_t \
(1 - phi B)(1 - B)z_t &= omega_t
end{align}$$
BACKGROUND:
I'm trying to understand backshift operator notation in the context of this ARIMA example from machine learning time series https://github.com/etcrago/Tutorial-Arima-w-jeffrey-yau (lecture 2).
QUESTION:
In the 3rd line he somehow jumps to $(1-B)$, but my understanding is the $phi$
part of the equation should be factored to $(B-B^2)$, not $(1-B)$. How does the author jump to that $(1-B)$ factorization ?
notation time-series
notation time-series
asked Dec 23 '18 at 14:28
alan dalan d
213
asked Dec 23 '18 at 14:28
alan dalan d
213
edited Dec 23 '18 at 15:38
alan d
asked Dec 23 '18 at 14:28
alan dalan d
213
asked Dec 23 '18 at 14:28
alan dalan d
213
asked Dec 23 '18 at 14:28
alan dalan d
213
213
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 14:36
2
$begingroup$
DONE. Thank you.
$endgroup$
– alan d
Dec 23 '18 at 15:28
add a comment |
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 14:36
2
$begingroup$
DONE. Thank you.
$endgroup$
– alan d
Dec 23 '18 at 15:28
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 14:36
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 14:36
2
2
$begingroup$
DONE. Thank you.
$endgroup$
– alan d
Dec 23 '18 at 15:28
$begingroup$
DONE. Thank you.
$endgroup$
– alan d
Dec 23 '18 at 15:28
add a comment |
1 Answer
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$begingroup$
I misread this problem. The better way to look at it is that the author factored $(z_t−z_{t−1})$ out such that the equations do work when multiplied to $(1-phi B)$.
answered Dec 23 '18 at 17:18
alan dalan d
213
$endgroup$
add a comment |
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$begingroup$
I misread this problem. The better way to look at it is that the author factored $(z_t−z_{t−1})$ out such that the equations do work when multiplied to $(1-phi B)$.
answered Dec 23 '18 at 17:18
alan dalan d
213
$endgroup$
add a comment |
$begingroup$
I misread this problem. The better way to look at it is that the author factored $(z_t−z_{t−1})$ out such that the equations do work when multiplied to $(1-phi B)$.
answered Dec 23 '18 at 17:18
alan dalan d
213
$endgroup$
add a comment |
$begingroup$
I misread this problem. The better way to look at it is that the author factored $(z_t−z_{t−1})$ out such that the equations do work when multiplied to $(1-phi B)$.
answered Dec 23 '18 at 17:18
alan dalan d
213
$endgroup$
I misread this problem. The better way to look at it is that the author factored $(z_t−z_{t−1})$ out such that the equations do work when multiplied to $(1-phi B)$.
answered Dec 23 '18 at 17:18
alan dalan d
213
answered Dec 23 '18 at 17:18
alan dalan d
213
answered Dec 23 '18 at 17:18
alan dalan d
213
answered Dec 23 '18 at 17:18
alan dalan d
213
213
add a comment |
add a comment |
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$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 14:36
2
$begingroup$
DONE. Thank you.
$endgroup$
– alan d
Dec 23 '18 at 15:28