How to Simplify Yule Walker Expression for AR Coefficients?












2












$begingroup$


We have an AR(2) process (w/ intercept omitted):
$$
y _ { t } = a _ { 1 } y _ { t - 1 } + a _ { 2 } y _ { t - 2 } + varepsilon _ { t }
$$

We multiply the second order equation by $ y_{t-s} text{ for } s = 0, s = 1, s = 2 ldots $
$$
begin{aligned} E y _ { t } y _ { t } & = a _ { 1 } E y _ { t - 1 } y _ { t } + a _ { 2 } E y _ { t - 2 } y _ { t } + E varepsilon _ { t } y _ { t } \ E y _ { t } y _ { t - 1 } & = a _ { 1 } E y _ { t - 1 } y _ { t - 1 } + a _ { 2 } E y _ { t - 2 } y _ { t - 1 } + E varepsilon _ { t } y _ { t - 1 } \ E y _ { t } y _ { t - 2 } & = a _ { 1 } E y _ { t - 1 } y _ { t - 2 } + a _ { 2 } E y _ { t - 2 } y _ { t - 2 } + E varepsilon _ { t } y _ { t - 2 } \ cdots & \ & cdots \ E y _ { t } y _ { t - s } & = a _ { 1 } E y _ { t - 1 } y _ { t - s } + a _ { 2 } E y _ { t - 2 } y _ { t - s } + E varepsilon _ { t } y _ { t - s } end{aligned}
$$



By definition, the autocovariances of a stationary series are such that $E y _ { t } y _ { t - s } =$
$E y _ { t - s } y _ { t } = E y _ { t - k } y _ { t - k - s } = gamma _ { s } .$ We also know that $E varepsilon _ { t } y _ { t } = sigma ^ { 2 }$ and $E varepsilon _ { t } y _ { t - s } = 0 .$ Hence, we
can use the equations in $( 2.24 )$ to form



$$
gamma _ { 0 } = a _ { 1 } gamma _ { 1 } + a _ { 2 } gamma _ { 2 } + sigma ^ { 2 }
$$

$$
begin{aligned} gamma _ { 1 } & = a _ { 1 } gamma _ { 0 } + a _ { 2 } gamma _ { 1 } \ gamma _ { s } & = a _ { 1 } gamma _ { s - 1 } + a _ { 2 } gamma _ { s - 2 } \ text { Dividing by } gamma _ { 0 } text { yields } \ rho _ { 1 } & = a _ { 1 } rho _ { 0 } + a _ { 2 } rho _ { 1 } \ rho _ { s } & = a _ { 1 } rho _ { s - 1 } + a _ { 2 } rho _ { s - 2 } end{aligned}
$$



Question: How did $ gamma_1 / gamma_0 $ turn into the $rho_1$ function?



The simple algebra of this division isn't making sense. What happens to the Variance $ sigma^2 $ that's in $ gamma_0 $ but somehow eliminated in the $ gamma_1$ function?










share|cite|improve this question









$endgroup$












  • $begingroup$
    You forgot the 1st line with $gamma 0$
    $endgroup$
    – Damien
    Dec 24 '18 at 17:01










  • $begingroup$
    @Damien I think it's there? Just above the line for $ gamma_{1} $ and centered. Maybe I'm misunderstanding?
    $endgroup$
    – AIS91
    Dec 24 '18 at 17:05










  • $begingroup$
    Yes this line. Don't forget it!
    $endgroup$
    – Damien
    Dec 24 '18 at 17:32










  • $begingroup$
    I realized the first small caveat that helps me solve the problem. $E varepsilon _ { t } y _ { t } = sigma^2 $ but this has to be 0 to eliminate serial correlation.
    $endgroup$
    – AIS91
    Dec 24 '18 at 17:49










  • $begingroup$
    What do you mean with eliminate ? This term is very inportant ...
    $endgroup$
    – Damien
    Dec 24 '18 at 18:18
















2












$begingroup$


We have an AR(2) process (w/ intercept omitted):
$$
y _ { t } = a _ { 1 } y _ { t - 1 } + a _ { 2 } y _ { t - 2 } + varepsilon _ { t }
$$

We multiply the second order equation by $ y_{t-s} text{ for } s = 0, s = 1, s = 2 ldots $
$$
begin{aligned} E y _ { t } y _ { t } & = a _ { 1 } E y _ { t - 1 } y _ { t } + a _ { 2 } E y _ { t - 2 } y _ { t } + E varepsilon _ { t } y _ { t } \ E y _ { t } y _ { t - 1 } & = a _ { 1 } E y _ { t - 1 } y _ { t - 1 } + a _ { 2 } E y _ { t - 2 } y _ { t - 1 } + E varepsilon _ { t } y _ { t - 1 } \ E y _ { t } y _ { t - 2 } & = a _ { 1 } E y _ { t - 1 } y _ { t - 2 } + a _ { 2 } E y _ { t - 2 } y _ { t - 2 } + E varepsilon _ { t } y _ { t - 2 } \ cdots & \ & cdots \ E y _ { t } y _ { t - s } & = a _ { 1 } E y _ { t - 1 } y _ { t - s } + a _ { 2 } E y _ { t - 2 } y _ { t - s } + E varepsilon _ { t } y _ { t - s } end{aligned}
$$



By definition, the autocovariances of a stationary series are such that $E y _ { t } y _ { t - s } =$
$E y _ { t - s } y _ { t } = E y _ { t - k } y _ { t - k - s } = gamma _ { s } .$ We also know that $E varepsilon _ { t } y _ { t } = sigma ^ { 2 }$ and $E varepsilon _ { t } y _ { t - s } = 0 .$ Hence, we
can use the equations in $( 2.24 )$ to form



$$
gamma _ { 0 } = a _ { 1 } gamma _ { 1 } + a _ { 2 } gamma _ { 2 } + sigma ^ { 2 }
$$

$$
begin{aligned} gamma _ { 1 } & = a _ { 1 } gamma _ { 0 } + a _ { 2 } gamma _ { 1 } \ gamma _ { s } & = a _ { 1 } gamma _ { s - 1 } + a _ { 2 } gamma _ { s - 2 } \ text { Dividing by } gamma _ { 0 } text { yields } \ rho _ { 1 } & = a _ { 1 } rho _ { 0 } + a _ { 2 } rho _ { 1 } \ rho _ { s } & = a _ { 1 } rho _ { s - 1 } + a _ { 2 } rho _ { s - 2 } end{aligned}
$$



Question: How did $ gamma_1 / gamma_0 $ turn into the $rho_1$ function?



The simple algebra of this division isn't making sense. What happens to the Variance $ sigma^2 $ that's in $ gamma_0 $ but somehow eliminated in the $ gamma_1$ function?










share|cite|improve this question









$endgroup$












  • $begingroup$
    You forgot the 1st line with $gamma 0$
    $endgroup$
    – Damien
    Dec 24 '18 at 17:01










  • $begingroup$
    @Damien I think it's there? Just above the line for $ gamma_{1} $ and centered. Maybe I'm misunderstanding?
    $endgroup$
    – AIS91
    Dec 24 '18 at 17:05










  • $begingroup$
    Yes this line. Don't forget it!
    $endgroup$
    – Damien
    Dec 24 '18 at 17:32










  • $begingroup$
    I realized the first small caveat that helps me solve the problem. $E varepsilon _ { t } y _ { t } = sigma^2 $ but this has to be 0 to eliminate serial correlation.
    $endgroup$
    – AIS91
    Dec 24 '18 at 17:49










  • $begingroup$
    What do you mean with eliminate ? This term is very inportant ...
    $endgroup$
    – Damien
    Dec 24 '18 at 18:18














2












2








2





$begingroup$


We have an AR(2) process (w/ intercept omitted):
$$
y _ { t } = a _ { 1 } y _ { t - 1 } + a _ { 2 } y _ { t - 2 } + varepsilon _ { t }
$$

We multiply the second order equation by $ y_{t-s} text{ for } s = 0, s = 1, s = 2 ldots $
$$
begin{aligned} E y _ { t } y _ { t } & = a _ { 1 } E y _ { t - 1 } y _ { t } + a _ { 2 } E y _ { t - 2 } y _ { t } + E varepsilon _ { t } y _ { t } \ E y _ { t } y _ { t - 1 } & = a _ { 1 } E y _ { t - 1 } y _ { t - 1 } + a _ { 2 } E y _ { t - 2 } y _ { t - 1 } + E varepsilon _ { t } y _ { t - 1 } \ E y _ { t } y _ { t - 2 } & = a _ { 1 } E y _ { t - 1 } y _ { t - 2 } + a _ { 2 } E y _ { t - 2 } y _ { t - 2 } + E varepsilon _ { t } y _ { t - 2 } \ cdots & \ & cdots \ E y _ { t } y _ { t - s } & = a _ { 1 } E y _ { t - 1 } y _ { t - s } + a _ { 2 } E y _ { t - 2 } y _ { t - s } + E varepsilon _ { t } y _ { t - s } end{aligned}
$$



By definition, the autocovariances of a stationary series are such that $E y _ { t } y _ { t - s } =$
$E y _ { t - s } y _ { t } = E y _ { t - k } y _ { t - k - s } = gamma _ { s } .$ We also know that $E varepsilon _ { t } y _ { t } = sigma ^ { 2 }$ and $E varepsilon _ { t } y _ { t - s } = 0 .$ Hence, we
can use the equations in $( 2.24 )$ to form



$$
gamma _ { 0 } = a _ { 1 } gamma _ { 1 } + a _ { 2 } gamma _ { 2 } + sigma ^ { 2 }
$$

$$
begin{aligned} gamma _ { 1 } & = a _ { 1 } gamma _ { 0 } + a _ { 2 } gamma _ { 1 } \ gamma _ { s } & = a _ { 1 } gamma _ { s - 1 } + a _ { 2 } gamma _ { s - 2 } \ text { Dividing by } gamma _ { 0 } text { yields } \ rho _ { 1 } & = a _ { 1 } rho _ { 0 } + a _ { 2 } rho _ { 1 } \ rho _ { s } & = a _ { 1 } rho _ { s - 1 } + a _ { 2 } rho _ { s - 2 } end{aligned}
$$



Question: How did $ gamma_1 / gamma_0 $ turn into the $rho_1$ function?



The simple algebra of this division isn't making sense. What happens to the Variance $ sigma^2 $ that's in $ gamma_0 $ but somehow eliminated in the $ gamma_1$ function?










share|cite|improve this question









$endgroup$




We have an AR(2) process (w/ intercept omitted):
$$
y _ { t } = a _ { 1 } y _ { t - 1 } + a _ { 2 } y _ { t - 2 } + varepsilon _ { t }
$$

We multiply the second order equation by $ y_{t-s} text{ for } s = 0, s = 1, s = 2 ldots $
$$
begin{aligned} E y _ { t } y _ { t } & = a _ { 1 } E y _ { t - 1 } y _ { t } + a _ { 2 } E y _ { t - 2 } y _ { t } + E varepsilon _ { t } y _ { t } \ E y _ { t } y _ { t - 1 } & = a _ { 1 } E y _ { t - 1 } y _ { t - 1 } + a _ { 2 } E y _ { t - 2 } y _ { t - 1 } + E varepsilon _ { t } y _ { t - 1 } \ E y _ { t } y _ { t - 2 } & = a _ { 1 } E y _ { t - 1 } y _ { t - 2 } + a _ { 2 } E y _ { t - 2 } y _ { t - 2 } + E varepsilon _ { t } y _ { t - 2 } \ cdots & \ & cdots \ E y _ { t } y _ { t - s } & = a _ { 1 } E y _ { t - 1 } y _ { t - s } + a _ { 2 } E y _ { t - 2 } y _ { t - s } + E varepsilon _ { t } y _ { t - s } end{aligned}
$$



By definition, the autocovariances of a stationary series are such that $E y _ { t } y _ { t - s } =$
$E y _ { t - s } y _ { t } = E y _ { t - k } y _ { t - k - s } = gamma _ { s } .$ We also know that $E varepsilon _ { t } y _ { t } = sigma ^ { 2 }$ and $E varepsilon _ { t } y _ { t - s } = 0 .$ Hence, we
can use the equations in $( 2.24 )$ to form



$$
gamma _ { 0 } = a _ { 1 } gamma _ { 1 } + a _ { 2 } gamma _ { 2 } + sigma ^ { 2 }
$$

$$
begin{aligned} gamma _ { 1 } & = a _ { 1 } gamma _ { 0 } + a _ { 2 } gamma _ { 1 } \ gamma _ { s } & = a _ { 1 } gamma _ { s - 1 } + a _ { 2 } gamma _ { s - 2 } \ text { Dividing by } gamma _ { 0 } text { yields } \ rho _ { 1 } & = a _ { 1 } rho _ { 0 } + a _ { 2 } rho _ { 1 } \ rho _ { s } & = a _ { 1 } rho _ { s - 1 } + a _ { 2 } rho _ { s - 2 } end{aligned}
$$



Question: How did $ gamma_1 / gamma_0 $ turn into the $rho_1$ function?



The simple algebra of this division isn't making sense. What happens to the Variance $ sigma^2 $ that's in $ gamma_0 $ but somehow eliminated in the $ gamma_1$ function?







sequences-and-series functions stochastic-processes systems-of-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 24 '18 at 16:49









AIS91AIS91

183




183












  • $begingroup$
    You forgot the 1st line with $gamma 0$
    $endgroup$
    – Damien
    Dec 24 '18 at 17:01










  • $begingroup$
    @Damien I think it's there? Just above the line for $ gamma_{1} $ and centered. Maybe I'm misunderstanding?
    $endgroup$
    – AIS91
    Dec 24 '18 at 17:05










  • $begingroup$
    Yes this line. Don't forget it!
    $endgroup$
    – Damien
    Dec 24 '18 at 17:32










  • $begingroup$
    I realized the first small caveat that helps me solve the problem. $E varepsilon _ { t } y _ { t } = sigma^2 $ but this has to be 0 to eliminate serial correlation.
    $endgroup$
    – AIS91
    Dec 24 '18 at 17:49










  • $begingroup$
    What do you mean with eliminate ? This term is very inportant ...
    $endgroup$
    – Damien
    Dec 24 '18 at 18:18


















  • $begingroup$
    You forgot the 1st line with $gamma 0$
    $endgroup$
    – Damien
    Dec 24 '18 at 17:01










  • $begingroup$
    @Damien I think it's there? Just above the line for $ gamma_{1} $ and centered. Maybe I'm misunderstanding?
    $endgroup$
    – AIS91
    Dec 24 '18 at 17:05










  • $begingroup$
    Yes this line. Don't forget it!
    $endgroup$
    – Damien
    Dec 24 '18 at 17:32










  • $begingroup$
    I realized the first small caveat that helps me solve the problem. $E varepsilon _ { t } y _ { t } = sigma^2 $ but this has to be 0 to eliminate serial correlation.
    $endgroup$
    – AIS91
    Dec 24 '18 at 17:49










  • $begingroup$
    What do you mean with eliminate ? This term is very inportant ...
    $endgroup$
    – Damien
    Dec 24 '18 at 18:18
















$begingroup$
You forgot the 1st line with $gamma 0$
$endgroup$
– Damien
Dec 24 '18 at 17:01




$begingroup$
You forgot the 1st line with $gamma 0$
$endgroup$
– Damien
Dec 24 '18 at 17:01












$begingroup$
@Damien I think it's there? Just above the line for $ gamma_{1} $ and centered. Maybe I'm misunderstanding?
$endgroup$
– AIS91
Dec 24 '18 at 17:05




$begingroup$
@Damien I think it's there? Just above the line for $ gamma_{1} $ and centered. Maybe I'm misunderstanding?
$endgroup$
– AIS91
Dec 24 '18 at 17:05












$begingroup$
Yes this line. Don't forget it!
$endgroup$
– Damien
Dec 24 '18 at 17:32




$begingroup$
Yes this line. Don't forget it!
$endgroup$
– Damien
Dec 24 '18 at 17:32












$begingroup$
I realized the first small caveat that helps me solve the problem. $E varepsilon _ { t } y _ { t } = sigma^2 $ but this has to be 0 to eliminate serial correlation.
$endgroup$
– AIS91
Dec 24 '18 at 17:49




$begingroup$
I realized the first small caveat that helps me solve the problem. $E varepsilon _ { t } y _ { t } = sigma^2 $ but this has to be 0 to eliminate serial correlation.
$endgroup$
– AIS91
Dec 24 '18 at 17:49












$begingroup$
What do you mean with eliminate ? This term is very inportant ...
$endgroup$
– Damien
Dec 24 '18 at 18:18




$begingroup$
What do you mean with eliminate ? This term is very inportant ...
$endgroup$
– Damien
Dec 24 '18 at 18:18










1 Answer
1






active

oldest

votes


















1












$begingroup$

Suppose we have an AR$(2)$ process



$$X_t=frac{1}{3}X_{t-1}+frac{1}{2}X_{t-2}+Z_t$$



From properties of the lag operator:



$$L(B)=1-frac{1}{3}B-frac{1}{2}B^2$$



This has real roots, both are outside the unit circle. This implies this process is weakly stationary.




Remark: A stochastic process is said to be weakly stationary when its mean and autocovariance does not vary with respect to time. Also,
the second moment is finite at all times. For more details see this




Next multiply both sides by $X_{t-k}$ and take expectation to obtain $E(X_{t-k}X_t)=gamma(k)$. (now the part you are interested in). Since $mu=0$, assuming $gamma(k)=0$, then:



$$gamma(-k)=frac{1}{3}gamma(-k+1)+frac{1}{2}gamma(-k+2)$$



Here note $gamma(k)=gamma(-k)$ for any $k$.



$$gamma(k)=frac{1}{3}gamma(k-1)+frac{1}{2}gamma(k-2)$$



Dividing both sides by $gamma(0)=sigma^2$ gives:



$$rho(k)=frac{1}{3}rho(k-1)+frac{1}{2}rho(k-2)$$



Notice $Var(X)=(E(X))^2-E(XY)$. See the above assumption and for stationary time series $E(X)=mu=0$. The definition of autocorrelation is given here.



The example is from the course by the state university of New York






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a ton for this! Are you assuming this process is white noise? Hence why you're able to define $mu = frac { 1 } { 3 } mu + frac { 1 } { 2 } mu$ as being 0? Or is this an assumption for stationarity? I would think the former?
    $endgroup$
    – AIS91
    Dec 26 '18 at 21:13












  • $begingroup$
    No, see the update.
    $endgroup$
    – Waqas
    Dec 26 '18 at 21:22










  • $begingroup$
    $$ begin{array} { l } { E left( y _ { t } right) = E left( y _ { t - s } right) = mu } \ { E left[ left( y _ { t } - mu right) ^ { 2 } right] = E left[ left( y _ { t - s } - mu right) ^ { 2 } right] = sigma _ { y } ^ { 2 } } end{array} $$ and $$ E left[ left( y _ { t } - mu right) left( y _ { t - s } - mu right) right] = E left[ left( y _ { t - j } - mu right) left( y _ { t - j - s } - mu right) right] = gamma _ { s } $$. Where $ mu, sigma^2, gamma_s $ constant. But how does that make $ mu = 0 $ from your equation?
    $endgroup$
    – AIS91
    Dec 26 '18 at 21:28






  • 1




    $begingroup$
    Yes, here $epsilon_t$ is indeed white noise. Upvote :)
    $endgroup$
    – Waqas
    Dec 26 '18 at 21:42






  • 1




    $begingroup$
    When you consider $gamma(k)=E(X_{t-k},X_t)$ for the given time series you get $gamma(-k)$, which is equal to $gamma(k)$. There is a lot of material available on the web regarding Time series modelling; you may find that helpful.
    $endgroup$
    – Waqas
    Dec 27 '18 at 22:23











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Suppose we have an AR$(2)$ process



$$X_t=frac{1}{3}X_{t-1}+frac{1}{2}X_{t-2}+Z_t$$



From properties of the lag operator:



$$L(B)=1-frac{1}{3}B-frac{1}{2}B^2$$



This has real roots, both are outside the unit circle. This implies this process is weakly stationary.




Remark: A stochastic process is said to be weakly stationary when its mean and autocovariance does not vary with respect to time. Also,
the second moment is finite at all times. For more details see this




Next multiply both sides by $X_{t-k}$ and take expectation to obtain $E(X_{t-k}X_t)=gamma(k)$. (now the part you are interested in). Since $mu=0$, assuming $gamma(k)=0$, then:



$$gamma(-k)=frac{1}{3}gamma(-k+1)+frac{1}{2}gamma(-k+2)$$



Here note $gamma(k)=gamma(-k)$ for any $k$.



$$gamma(k)=frac{1}{3}gamma(k-1)+frac{1}{2}gamma(k-2)$$



Dividing both sides by $gamma(0)=sigma^2$ gives:



$$rho(k)=frac{1}{3}rho(k-1)+frac{1}{2}rho(k-2)$$



Notice $Var(X)=(E(X))^2-E(XY)$. See the above assumption and for stationary time series $E(X)=mu=0$. The definition of autocorrelation is given here.



The example is from the course by the state university of New York






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a ton for this! Are you assuming this process is white noise? Hence why you're able to define $mu = frac { 1 } { 3 } mu + frac { 1 } { 2 } mu$ as being 0? Or is this an assumption for stationarity? I would think the former?
    $endgroup$
    – AIS91
    Dec 26 '18 at 21:13












  • $begingroup$
    No, see the update.
    $endgroup$
    – Waqas
    Dec 26 '18 at 21:22










  • $begingroup$
    $$ begin{array} { l } { E left( y _ { t } right) = E left( y _ { t - s } right) = mu } \ { E left[ left( y _ { t } - mu right) ^ { 2 } right] = E left[ left( y _ { t - s } - mu right) ^ { 2 } right] = sigma _ { y } ^ { 2 } } end{array} $$ and $$ E left[ left( y _ { t } - mu right) left( y _ { t - s } - mu right) right] = E left[ left( y _ { t - j } - mu right) left( y _ { t - j - s } - mu right) right] = gamma _ { s } $$. Where $ mu, sigma^2, gamma_s $ constant. But how does that make $ mu = 0 $ from your equation?
    $endgroup$
    – AIS91
    Dec 26 '18 at 21:28






  • 1




    $begingroup$
    Yes, here $epsilon_t$ is indeed white noise. Upvote :)
    $endgroup$
    – Waqas
    Dec 26 '18 at 21:42






  • 1




    $begingroup$
    When you consider $gamma(k)=E(X_{t-k},X_t)$ for the given time series you get $gamma(-k)$, which is equal to $gamma(k)$. There is a lot of material available on the web regarding Time series modelling; you may find that helpful.
    $endgroup$
    – Waqas
    Dec 27 '18 at 22:23
















1












$begingroup$

Suppose we have an AR$(2)$ process



$$X_t=frac{1}{3}X_{t-1}+frac{1}{2}X_{t-2}+Z_t$$



From properties of the lag operator:



$$L(B)=1-frac{1}{3}B-frac{1}{2}B^2$$



This has real roots, both are outside the unit circle. This implies this process is weakly stationary.




Remark: A stochastic process is said to be weakly stationary when its mean and autocovariance does not vary with respect to time. Also,
the second moment is finite at all times. For more details see this




Next multiply both sides by $X_{t-k}$ and take expectation to obtain $E(X_{t-k}X_t)=gamma(k)$. (now the part you are interested in). Since $mu=0$, assuming $gamma(k)=0$, then:



$$gamma(-k)=frac{1}{3}gamma(-k+1)+frac{1}{2}gamma(-k+2)$$



Here note $gamma(k)=gamma(-k)$ for any $k$.



$$gamma(k)=frac{1}{3}gamma(k-1)+frac{1}{2}gamma(k-2)$$



Dividing both sides by $gamma(0)=sigma^2$ gives:



$$rho(k)=frac{1}{3}rho(k-1)+frac{1}{2}rho(k-2)$$



Notice $Var(X)=(E(X))^2-E(XY)$. See the above assumption and for stationary time series $E(X)=mu=0$. The definition of autocorrelation is given here.



The example is from the course by the state university of New York






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a ton for this! Are you assuming this process is white noise? Hence why you're able to define $mu = frac { 1 } { 3 } mu + frac { 1 } { 2 } mu$ as being 0? Or is this an assumption for stationarity? I would think the former?
    $endgroup$
    – AIS91
    Dec 26 '18 at 21:13












  • $begingroup$
    No, see the update.
    $endgroup$
    – Waqas
    Dec 26 '18 at 21:22










  • $begingroup$
    $$ begin{array} { l } { E left( y _ { t } right) = E left( y _ { t - s } right) = mu } \ { E left[ left( y _ { t } - mu right) ^ { 2 } right] = E left[ left( y _ { t - s } - mu right) ^ { 2 } right] = sigma _ { y } ^ { 2 } } end{array} $$ and $$ E left[ left( y _ { t } - mu right) left( y _ { t - s } - mu right) right] = E left[ left( y _ { t - j } - mu right) left( y _ { t - j - s } - mu right) right] = gamma _ { s } $$. Where $ mu, sigma^2, gamma_s $ constant. But how does that make $ mu = 0 $ from your equation?
    $endgroup$
    – AIS91
    Dec 26 '18 at 21:28






  • 1




    $begingroup$
    Yes, here $epsilon_t$ is indeed white noise. Upvote :)
    $endgroup$
    – Waqas
    Dec 26 '18 at 21:42






  • 1




    $begingroup$
    When you consider $gamma(k)=E(X_{t-k},X_t)$ for the given time series you get $gamma(-k)$, which is equal to $gamma(k)$. There is a lot of material available on the web regarding Time series modelling; you may find that helpful.
    $endgroup$
    – Waqas
    Dec 27 '18 at 22:23














1












1








1





$begingroup$

Suppose we have an AR$(2)$ process



$$X_t=frac{1}{3}X_{t-1}+frac{1}{2}X_{t-2}+Z_t$$



From properties of the lag operator:



$$L(B)=1-frac{1}{3}B-frac{1}{2}B^2$$



This has real roots, both are outside the unit circle. This implies this process is weakly stationary.




Remark: A stochastic process is said to be weakly stationary when its mean and autocovariance does not vary with respect to time. Also,
the second moment is finite at all times. For more details see this




Next multiply both sides by $X_{t-k}$ and take expectation to obtain $E(X_{t-k}X_t)=gamma(k)$. (now the part you are interested in). Since $mu=0$, assuming $gamma(k)=0$, then:



$$gamma(-k)=frac{1}{3}gamma(-k+1)+frac{1}{2}gamma(-k+2)$$



Here note $gamma(k)=gamma(-k)$ for any $k$.



$$gamma(k)=frac{1}{3}gamma(k-1)+frac{1}{2}gamma(k-2)$$



Dividing both sides by $gamma(0)=sigma^2$ gives:



$$rho(k)=frac{1}{3}rho(k-1)+frac{1}{2}rho(k-2)$$



Notice $Var(X)=(E(X))^2-E(XY)$. See the above assumption and for stationary time series $E(X)=mu=0$. The definition of autocorrelation is given here.



The example is from the course by the state university of New York






share|cite|improve this answer











$endgroup$



Suppose we have an AR$(2)$ process



$$X_t=frac{1}{3}X_{t-1}+frac{1}{2}X_{t-2}+Z_t$$



From properties of the lag operator:



$$L(B)=1-frac{1}{3}B-frac{1}{2}B^2$$



This has real roots, both are outside the unit circle. This implies this process is weakly stationary.




Remark: A stochastic process is said to be weakly stationary when its mean and autocovariance does not vary with respect to time. Also,
the second moment is finite at all times. For more details see this




Next multiply both sides by $X_{t-k}$ and take expectation to obtain $E(X_{t-k}X_t)=gamma(k)$. (now the part you are interested in). Since $mu=0$, assuming $gamma(k)=0$, then:



$$gamma(-k)=frac{1}{3}gamma(-k+1)+frac{1}{2}gamma(-k+2)$$



Here note $gamma(k)=gamma(-k)$ for any $k$.



$$gamma(k)=frac{1}{3}gamma(k-1)+frac{1}{2}gamma(k-2)$$



Dividing both sides by $gamma(0)=sigma^2$ gives:



$$rho(k)=frac{1}{3}rho(k-1)+frac{1}{2}rho(k-2)$$



Notice $Var(X)=(E(X))^2-E(XY)$. See the above assumption and for stationary time series $E(X)=mu=0$. The definition of autocorrelation is given here.



The example is from the course by the state university of New York







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 26 '18 at 21:39

























answered Dec 25 '18 at 11:27









WaqasWaqas

15913




15913












  • $begingroup$
    Thanks a ton for this! Are you assuming this process is white noise? Hence why you're able to define $mu = frac { 1 } { 3 } mu + frac { 1 } { 2 } mu$ as being 0? Or is this an assumption for stationarity? I would think the former?
    $endgroup$
    – AIS91
    Dec 26 '18 at 21:13












  • $begingroup$
    No, see the update.
    $endgroup$
    – Waqas
    Dec 26 '18 at 21:22










  • $begingroup$
    $$ begin{array} { l } { E left( y _ { t } right) = E left( y _ { t - s } right) = mu } \ { E left[ left( y _ { t } - mu right) ^ { 2 } right] = E left[ left( y _ { t - s } - mu right) ^ { 2 } right] = sigma _ { y } ^ { 2 } } end{array} $$ and $$ E left[ left( y _ { t } - mu right) left( y _ { t - s } - mu right) right] = E left[ left( y _ { t - j } - mu right) left( y _ { t - j - s } - mu right) right] = gamma _ { s } $$. Where $ mu, sigma^2, gamma_s $ constant. But how does that make $ mu = 0 $ from your equation?
    $endgroup$
    – AIS91
    Dec 26 '18 at 21:28






  • 1




    $begingroup$
    Yes, here $epsilon_t$ is indeed white noise. Upvote :)
    $endgroup$
    – Waqas
    Dec 26 '18 at 21:42






  • 1




    $begingroup$
    When you consider $gamma(k)=E(X_{t-k},X_t)$ for the given time series you get $gamma(-k)$, which is equal to $gamma(k)$. There is a lot of material available on the web regarding Time series modelling; you may find that helpful.
    $endgroup$
    – Waqas
    Dec 27 '18 at 22:23


















  • $begingroup$
    Thanks a ton for this! Are you assuming this process is white noise? Hence why you're able to define $mu = frac { 1 } { 3 } mu + frac { 1 } { 2 } mu$ as being 0? Or is this an assumption for stationarity? I would think the former?
    $endgroup$
    – AIS91
    Dec 26 '18 at 21:13












  • $begingroup$
    No, see the update.
    $endgroup$
    – Waqas
    Dec 26 '18 at 21:22










  • $begingroup$
    $$ begin{array} { l } { E left( y _ { t } right) = E left( y _ { t - s } right) = mu } \ { E left[ left( y _ { t } - mu right) ^ { 2 } right] = E left[ left( y _ { t - s } - mu right) ^ { 2 } right] = sigma _ { y } ^ { 2 } } end{array} $$ and $$ E left[ left( y _ { t } - mu right) left( y _ { t - s } - mu right) right] = E left[ left( y _ { t - j } - mu right) left( y _ { t - j - s } - mu right) right] = gamma _ { s } $$. Where $ mu, sigma^2, gamma_s $ constant. But how does that make $ mu = 0 $ from your equation?
    $endgroup$
    – AIS91
    Dec 26 '18 at 21:28






  • 1




    $begingroup$
    Yes, here $epsilon_t$ is indeed white noise. Upvote :)
    $endgroup$
    – Waqas
    Dec 26 '18 at 21:42






  • 1




    $begingroup$
    When you consider $gamma(k)=E(X_{t-k},X_t)$ for the given time series you get $gamma(-k)$, which is equal to $gamma(k)$. There is a lot of material available on the web regarding Time series modelling; you may find that helpful.
    $endgroup$
    – Waqas
    Dec 27 '18 at 22:23
















$begingroup$
Thanks a ton for this! Are you assuming this process is white noise? Hence why you're able to define $mu = frac { 1 } { 3 } mu + frac { 1 } { 2 } mu$ as being 0? Or is this an assumption for stationarity? I would think the former?
$endgroup$
– AIS91
Dec 26 '18 at 21:13






$begingroup$
Thanks a ton for this! Are you assuming this process is white noise? Hence why you're able to define $mu = frac { 1 } { 3 } mu + frac { 1 } { 2 } mu$ as being 0? Or is this an assumption for stationarity? I would think the former?
$endgroup$
– AIS91
Dec 26 '18 at 21:13














$begingroup$
No, see the update.
$endgroup$
– Waqas
Dec 26 '18 at 21:22




$begingroup$
No, see the update.
$endgroup$
– Waqas
Dec 26 '18 at 21:22












$begingroup$
$$ begin{array} { l } { E left( y _ { t } right) = E left( y _ { t - s } right) = mu } \ { E left[ left( y _ { t } - mu right) ^ { 2 } right] = E left[ left( y _ { t - s } - mu right) ^ { 2 } right] = sigma _ { y } ^ { 2 } } end{array} $$ and $$ E left[ left( y _ { t } - mu right) left( y _ { t - s } - mu right) right] = E left[ left( y _ { t - j } - mu right) left( y _ { t - j - s } - mu right) right] = gamma _ { s } $$. Where $ mu, sigma^2, gamma_s $ constant. But how does that make $ mu = 0 $ from your equation?
$endgroup$
– AIS91
Dec 26 '18 at 21:28




$begingroup$
$$ begin{array} { l } { E left( y _ { t } right) = E left( y _ { t - s } right) = mu } \ { E left[ left( y _ { t } - mu right) ^ { 2 } right] = E left[ left( y _ { t - s } - mu right) ^ { 2 } right] = sigma _ { y } ^ { 2 } } end{array} $$ and $$ E left[ left( y _ { t } - mu right) left( y _ { t - s } - mu right) right] = E left[ left( y _ { t - j } - mu right) left( y _ { t - j - s } - mu right) right] = gamma _ { s } $$. Where $ mu, sigma^2, gamma_s $ constant. But how does that make $ mu = 0 $ from your equation?
$endgroup$
– AIS91
Dec 26 '18 at 21:28




1




1




$begingroup$
Yes, here $epsilon_t$ is indeed white noise. Upvote :)
$endgroup$
– Waqas
Dec 26 '18 at 21:42




$begingroup$
Yes, here $epsilon_t$ is indeed white noise. Upvote :)
$endgroup$
– Waqas
Dec 26 '18 at 21:42




1




1




$begingroup$
When you consider $gamma(k)=E(X_{t-k},X_t)$ for the given time series you get $gamma(-k)$, which is equal to $gamma(k)$. There is a lot of material available on the web regarding Time series modelling; you may find that helpful.
$endgroup$
– Waqas
Dec 27 '18 at 22:23




$begingroup$
When you consider $gamma(k)=E(X_{t-k},X_t)$ for the given time series you get $gamma(-k)$, which is equal to $gamma(k)$. There is a lot of material available on the web regarding Time series modelling; you may find that helpful.
$endgroup$
– Waqas
Dec 27 '18 at 22:23


















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