Mixing
What is the probability distribution of this AR(1) function?
$begingroup$
I'm preparing the exam for "stochastic models" and I encountered this exercise which is giving me a lot of problems:
Let $X_t sim AR(1)$, with
$$X_t=-0.8X_{t-1}+ epsilon_t, ~~~~~~~~~~epsilon_t sim WN(0,4)$$
1) Compute the autocovariance and the autocorrelation functions of $X_t$ at lags $h ge 0$
2) Assuming that $epsilon_t sim GWN(0,4)$, what is the probability distribution of $X_t$ for all $t$?
1) No problem for this point but I want to show you how I did it:
Autocovariance function:
$$gamma(1)=operatorname{Cov}(X_t,X_{t-1})~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$=operatorname{Cov}(-0.8X_{t-1}+epsilon_t, X_{t-1})$$
$$~~~~~~~~~~~~~~~~~~~=operatorname{Cov}(-0.8X_{t-1}, X_{t-1})+operatorname{Cov}(epsilon_t, X_{t-1})$$
$$=-0.8 cdot gamma(0)~~~~~~~~~~~~~~~~~~~~~~~~~~$$
Similarly,
$$gamma(2)=operatorname{Cov}(X_t,X_{t-2})~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$=operatorname{Cov}(-0.8X_{t-1}+epsilon_t, X_{t-2})$$
$$~~~~~~~~~~~~~~~~~~~=operatorname{Cov}(-0.8X_{t-1}, X_{t-2})+operatorname{Cov}(epsilon_t, X_{t-2})$$
$$=(-0.8) cdot gamma(1)~~~~~~~~~~~~~~~~~~~~~~~$$
$$=(-0.8)^2 cdot gamma(0)~~~~~~~~~~~~~~~~~~~~~$$
$$vdots$$
$$gamma(h)=gammaleft(0right)cdotphi^{left|hright|}$$
Autocorrelation function:
$$rho = frac{gamma(h)}{gamma(0)}$$
which
$$gamma(0)=Var(X_t)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=Var(-0.8X_{t-1})+Var(epsilon_t)+2operatorname{Cov}(X_{t-1},epsilon_t)$$
$$=(-0.8)^2 Var(X_t) + 4 + 0$$
$$=11.bar{11}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
Point 2 is my main problem because I don't know where to begin in order to solve it. One sure thing is that the Gaussian White Noise with zero-mean has a normal distribution. Then? How can I use the data that I have in order to find this probability distribution?
Any help would be appreciated
probability probability-distributions normal-distribution self-learning time-series
asked Jan 20 at 13:26
Martina MartyMartina Marty
345
$endgroup$
add a comment |
$begingroup$
I'm preparing the exam for "stochastic models" and I encountered this exercise which is giving me a lot of problems:
Let $X_t sim AR(1)$, with
$$X_t=-0.8X_{t-1}+ epsilon_t, ~~~~~~~~~~epsilon_t sim WN(0,4)$$
1) Compute the autocovariance and the autocorrelation functions of $X_t$ at lags $h ge 0$
2) Assuming that $epsilon_t sim GWN(0,4)$, what is the probability distribution of $X_t$ for all $t$?
1) No problem for this point but I want to show you how I did it:
Autocovariance function:
$$gamma(1)=operatorname{Cov}(X_t,X_{t-1})~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$=operatorname{Cov}(-0.8X_{t-1}+epsilon_t, X_{t-1})$$
$$~~~~~~~~~~~~~~~~~~~=operatorname{Cov}(-0.8X_{t-1}, X_{t-1})+operatorname{Cov}(epsilon_t, X_{t-1})$$
$$=-0.8 cdot gamma(0)~~~~~~~~~~~~~~~~~~~~~~~~~~$$
Similarly,
$$gamma(2)=operatorname{Cov}(X_t,X_{t-2})~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$=operatorname{Cov}(-0.8X_{t-1}+epsilon_t, X_{t-2})$$
$$~~~~~~~~~~~~~~~~~~~=operatorname{Cov}(-0.8X_{t-1}, X_{t-2})+operatorname{Cov}(epsilon_t, X_{t-2})$$
$$=(-0.8) cdot gamma(1)~~~~~~~~~~~~~~~~~~~~~~~$$
$$=(-0.8)^2 cdot gamma(0)~~~~~~~~~~~~~~~~~~~~~$$
$$vdots$$
$$gamma(h)=gammaleft(0right)cdotphi^{left|hright|}$$
Autocorrelation function:
$$rho = frac{gamma(h)}{gamma(0)}$$
which
$$gamma(0)=Var(X_t)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=Var(-0.8X_{t-1})+Var(epsilon_t)+2operatorname{Cov}(X_{t-1},epsilon_t)$$
$$=(-0.8)^2 Var(X_t) + 4 + 0$$
$$=11.bar{11}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
Point 2 is my main problem because I don't know where to begin in order to solve it. One sure thing is that the Gaussian White Noise with zero-mean has a normal distribution. Then? How can I use the data that I have in order to find this probability distribution?
Any help would be appreciated
probability probability-distributions normal-distribution self-learning time-series
asked Jan 20 at 13:26
Martina MartyMartina Marty
345
$endgroup$
add a comment |
$begingroup$
I'm preparing the exam for "stochastic models" and I encountered this exercise which is giving me a lot of problems:
Let $X_t sim AR(1)$, with
$$X_t=-0.8X_{t-1}+ epsilon_t, ~~~~~~~~~~epsilon_t sim WN(0,4)$$
1) Compute the autocovariance and the autocorrelation functions of $X_t$ at lags $h ge 0$
2) Assuming that $epsilon_t sim GWN(0,4)$, what is the probability distribution of $X_t$ for all $t$?
1) No problem for this point but I want to show you how I did it:
Autocovariance function:
$$gamma(1)=operatorname{Cov}(X_t,X_{t-1})~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$=operatorname{Cov}(-0.8X_{t-1}+epsilon_t, X_{t-1})$$
$$~~~~~~~~~~~~~~~~~~~=operatorname{Cov}(-0.8X_{t-1}, X_{t-1})+operatorname{Cov}(epsilon_t, X_{t-1})$$
$$=-0.8 cdot gamma(0)~~~~~~~~~~~~~~~~~~~~~~~~~~$$
Similarly,
$$gamma(2)=operatorname{Cov}(X_t,X_{t-2})~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$=operatorname{Cov}(-0.8X_{t-1}+epsilon_t, X_{t-2})$$
$$~~~~~~~~~~~~~~~~~~~=operatorname{Cov}(-0.8X_{t-1}, X_{t-2})+operatorname{Cov}(epsilon_t, X_{t-2})$$
$$=(-0.8) cdot gamma(1)~~~~~~~~~~~~~~~~~~~~~~~$$
$$=(-0.8)^2 cdot gamma(0)~~~~~~~~~~~~~~~~~~~~~$$
$$vdots$$
$$gamma(h)=gammaleft(0right)cdotphi^{left|hright|}$$
Autocorrelation function:
$$rho = frac{gamma(h)}{gamma(0)}$$
which
$$gamma(0)=Var(X_t)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=Var(-0.8X_{t-1})+Var(epsilon_t)+2operatorname{Cov}(X_{t-1},epsilon_t)$$
$$=(-0.8)^2 Var(X_t) + 4 + 0$$
$$=11.bar{11}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
Point 2 is my main problem because I don't know where to begin in order to solve it. One sure thing is that the Gaussian White Noise with zero-mean has a normal distribution. Then? How can I use the data that I have in order to find this probability distribution?
Any help would be appreciated
probability probability-distributions normal-distribution self-learning time-series
asked Jan 20 at 13:26
Martina MartyMartina Marty
345
$endgroup$
I'm preparing the exam for "stochastic models" and I encountered this exercise which is giving me a lot of problems:
Let $X_t sim AR(1)$, with
$$X_t=-0.8X_{t-1}+ epsilon_t, ~~~~~~~~~~epsilon_t sim WN(0,4)$$
1) Compute the autocovariance and the autocorrelation functions of $X_t$ at lags $h ge 0$
2) Assuming that $epsilon_t sim GWN(0,4)$, what is the probability distribution of $X_t$ for all $t$?
1) No problem for this point but I want to show you how I did it:
Autocovariance function:
$$gamma(1)=operatorname{Cov}(X_t,X_{t-1})~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$=operatorname{Cov}(-0.8X_{t-1}+epsilon_t, X_{t-1})$$
$$~~~~~~~~~~~~~~~~~~~=operatorname{Cov}(-0.8X_{t-1}, X_{t-1})+operatorname{Cov}(epsilon_t, X_{t-1})$$
$$=-0.8 cdot gamma(0)~~~~~~~~~~~~~~~~~~~~~~~~~~$$
Similarly,
$$gamma(2)=operatorname{Cov}(X_t,X_{t-2})~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$=operatorname{Cov}(-0.8X_{t-1}+epsilon_t, X_{t-2})$$
$$~~~~~~~~~~~~~~~~~~~=operatorname{Cov}(-0.8X_{t-1}, X_{t-2})+operatorname{Cov}(epsilon_t, X_{t-2})$$
$$=(-0.8) cdot gamma(1)~~~~~~~~~~~~~~~~~~~~~~~$$
$$=(-0.8)^2 cdot gamma(0)~~~~~~~~~~~~~~~~~~~~~$$
$$vdots$$
$$gamma(h)=gammaleft(0right)cdotphi^{left|hright|}$$
Autocorrelation function:
$$rho = frac{gamma(h)}{gamma(0)}$$
which
$$gamma(0)=Var(X_t)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=Var(-0.8X_{t-1})+Var(epsilon_t)+2operatorname{Cov}(X_{t-1},epsilon_t)$$
$$=(-0.8)^2 Var(X_t) + 4 + 0$$
$$=11.bar{11}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
Point 2 is my main problem because I don't know where to begin in order to solve it. One sure thing is that the Gaussian White Noise with zero-mean has a normal distribution. Then? How can I use the data that I have in order to find this probability distribution?
Any help would be appreciated
probability probability-distributions normal-distribution self-learning time-series
probability probability-distributions normal-distribution self-learning time-series
asked Jan 20 at 13:26
Martina MartyMartina Marty
345
asked Jan 20 at 13:26
Martina MartyMartina Marty
345
asked Jan 20 at 13:26
Martina MartyMartina Marty
345
asked Jan 20 at 13:26
Martina MartyMartina Marty
345
asked Jan 20 at 13:26
Martina MartyMartina Marty
345
345
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $L$ denote the lag operator, i.e. $LX_{t}=X_{t-1}$ and $Lepsilon_t =epsilon_{t-1}$. The given equation can be written as
$$
X_t = -.8LX_t +epsilon_t
$$ or equivalently
$$
(1+.8L)X_t = epsilon_t.
$$ By inverting $1+.8L$, we get
$$
X_t =alpha+(1+.8L)^{-1}epsilon_t =alpha+sum_{j=0}^infty (-.8L)^j epsilon_t=alpha+sum_{j=0}^infty (-.8)^j epsilon_{t-j}.
$$ Now, we claim that $sum_{j=0}^infty (-.8)^j epsilon_{t-j}$ is normally distributed with mean $0$ and variance equal to $4sum_{j=0}^infty (.64)^j=frac{100}{9}.$ The mean-square convergence of the series can be easily seen from the fact that
$$
Bbb Eleft[left|sum_{nle jle m}(-.8)^j epsilon_{t-j}right|^2right]=4sum_{nle jle m}(.64)^j to 0
$$as $n,mtoinfty$. By examining its characteristic function, we get
$$begin{eqnarray}
Bbb Eleft[e^{itsumlimits_{j=0}^infty (-.8)^j epsilon_{t-j}}right]&=&prod_{j=0}^infty Bbb Eleft[e^{it (-.8)^j epsilon_{t-j}}right]\
&=&prod_{j=0}^infty expleft[-frac{t^2 text{Var}left[(-.8)^j epsilon_{t-j}right]}{2}right]\&=&prod_{j=0}^infty expleft[-frac{4t^2cdot(.64)^j}{2}right]\
&=&expleft[-frac{4t^2cdotsum_{j=0}^infty(.64)^j}{2}right]=expleft[-frac{t^2cdotfrac{100}{9}}{2}right].
end{eqnarray}$$ Since the characteristic function of $mathcal{N}(mu,sigma^2)$ is $expleft(itmu -frac{sigma^2t^2}{2}right)$, it follows
$$
X_t sim_d mathcal{N}left(alpha,left(frac{10}{3}right)^2right)
$$ for some $alpha$. ($alpha$ is not determined.)
answered Jan 20 at 13:50
SongSong
16.5k1741
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add a comment |
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1 Answer
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$begingroup$
Let $L$ denote the lag operator, i.e. $LX_{t}=X_{t-1}$ and $Lepsilon_t =epsilon_{t-1}$. The given equation can be written as
$$
X_t = -.8LX_t +epsilon_t
$$ or equivalently
$$
(1+.8L)X_t = epsilon_t.
$$ By inverting $1+.8L$, we get
$$
X_t =alpha+(1+.8L)^{-1}epsilon_t =alpha+sum_{j=0}^infty (-.8L)^j epsilon_t=alpha+sum_{j=0}^infty (-.8)^j epsilon_{t-j}.
$$ Now, we claim that $sum_{j=0}^infty (-.8)^j epsilon_{t-j}$ is normally distributed with mean $0$ and variance equal to $4sum_{j=0}^infty (.64)^j=frac{100}{9}.$ The mean-square convergence of the series can be easily seen from the fact that
$$
Bbb Eleft[left|sum_{nle jle m}(-.8)^j epsilon_{t-j}right|^2right]=4sum_{nle jle m}(.64)^j to 0
$$as $n,mtoinfty$. By examining its characteristic function, we get
$$begin{eqnarray}
Bbb Eleft[e^{itsumlimits_{j=0}^infty (-.8)^j epsilon_{t-j}}right]&=&prod_{j=0}^infty Bbb Eleft[e^{it (-.8)^j epsilon_{t-j}}right]\
&=&prod_{j=0}^infty expleft[-frac{t^2 text{Var}left[(-.8)^j epsilon_{t-j}right]}{2}right]\&=&prod_{j=0}^infty expleft[-frac{4t^2cdot(.64)^j}{2}right]\
&=&expleft[-frac{4t^2cdotsum_{j=0}^infty(.64)^j}{2}right]=expleft[-frac{t^2cdotfrac{100}{9}}{2}right].
end{eqnarray}$$ Since the characteristic function of $mathcal{N}(mu,sigma^2)$ is $expleft(itmu -frac{sigma^2t^2}{2}right)$, it follows
$$
X_t sim_d mathcal{N}left(alpha,left(frac{10}{3}right)^2right)
$$ for some $alpha$. ($alpha$ is not determined.)
answered Jan 20 at 13:50
SongSong
16.5k1741
$endgroup$
add a comment |
$begingroup$
Let $L$ denote the lag operator, i.e. $LX_{t}=X_{t-1}$ and $Lepsilon_t =epsilon_{t-1}$. The given equation can be written as
$$
X_t = -.8LX_t +epsilon_t
$$ or equivalently
$$
(1+.8L)X_t = epsilon_t.
$$ By inverting $1+.8L$, we get
$$
X_t =alpha+(1+.8L)^{-1}epsilon_t =alpha+sum_{j=0}^infty (-.8L)^j epsilon_t=alpha+sum_{j=0}^infty (-.8)^j epsilon_{t-j}.
$$ Now, we claim that $sum_{j=0}^infty (-.8)^j epsilon_{t-j}$ is normally distributed with mean $0$ and variance equal to $4sum_{j=0}^infty (.64)^j=frac{100}{9}.$ The mean-square convergence of the series can be easily seen from the fact that
$$
Bbb Eleft[left|sum_{nle jle m}(-.8)^j epsilon_{t-j}right|^2right]=4sum_{nle jle m}(.64)^j to 0
$$as $n,mtoinfty$. By examining its characteristic function, we get
$$begin{eqnarray}
Bbb Eleft[e^{itsumlimits_{j=0}^infty (-.8)^j epsilon_{t-j}}right]&=&prod_{j=0}^infty Bbb Eleft[e^{it (-.8)^j epsilon_{t-j}}right]\
&=&prod_{j=0}^infty expleft[-frac{t^2 text{Var}left[(-.8)^j epsilon_{t-j}right]}{2}right]\&=&prod_{j=0}^infty expleft[-frac{4t^2cdot(.64)^j}{2}right]\
&=&expleft[-frac{4t^2cdotsum_{j=0}^infty(.64)^j}{2}right]=expleft[-frac{t^2cdotfrac{100}{9}}{2}right].
end{eqnarray}$$ Since the characteristic function of $mathcal{N}(mu,sigma^2)$ is $expleft(itmu -frac{sigma^2t^2}{2}right)$, it follows
$$
X_t sim_d mathcal{N}left(alpha,left(frac{10}{3}right)^2right)
$$ for some $alpha$. ($alpha$ is not determined.)
answered Jan 20 at 13:50
SongSong
16.5k1741
$endgroup$
add a comment |
$begingroup$
Let $L$ denote the lag operator, i.e. $LX_{t}=X_{t-1}$ and $Lepsilon_t =epsilon_{t-1}$. The given equation can be written as
$$
X_t = -.8LX_t +epsilon_t
$$ or equivalently
$$
(1+.8L)X_t = epsilon_t.
$$ By inverting $1+.8L$, we get
$$
X_t =alpha+(1+.8L)^{-1}epsilon_t =alpha+sum_{j=0}^infty (-.8L)^j epsilon_t=alpha+sum_{j=0}^infty (-.8)^j epsilon_{t-j}.
$$ Now, we claim that $sum_{j=0}^infty (-.8)^j epsilon_{t-j}$ is normally distributed with mean $0$ and variance equal to $4sum_{j=0}^infty (.64)^j=frac{100}{9}.$ The mean-square convergence of the series can be easily seen from the fact that
$$
Bbb Eleft[left|sum_{nle jle m}(-.8)^j epsilon_{t-j}right|^2right]=4sum_{nle jle m}(.64)^j to 0
$$as $n,mtoinfty$. By examining its characteristic function, we get
$$begin{eqnarray}
Bbb Eleft[e^{itsumlimits_{j=0}^infty (-.8)^j epsilon_{t-j}}right]&=&prod_{j=0}^infty Bbb Eleft[e^{it (-.8)^j epsilon_{t-j}}right]\
&=&prod_{j=0}^infty expleft[-frac{t^2 text{Var}left[(-.8)^j epsilon_{t-j}right]}{2}right]\&=&prod_{j=0}^infty expleft[-frac{4t^2cdot(.64)^j}{2}right]\
&=&expleft[-frac{4t^2cdotsum_{j=0}^infty(.64)^j}{2}right]=expleft[-frac{t^2cdotfrac{100}{9}}{2}right].
end{eqnarray}$$ Since the characteristic function of $mathcal{N}(mu,sigma^2)$ is $expleft(itmu -frac{sigma^2t^2}{2}right)$, it follows
$$
X_t sim_d mathcal{N}left(alpha,left(frac{10}{3}right)^2right)
$$ for some $alpha$. ($alpha$ is not determined.)
answered Jan 20 at 13:50
SongSong
16.5k1741
$endgroup$
Let $L$ denote the lag operator, i.e. $LX_{t}=X_{t-1}$ and $Lepsilon_t =epsilon_{t-1}$. The given equation can be written as
$$
X_t = -.8LX_t +epsilon_t
$$ or equivalently
$$
(1+.8L)X_t = epsilon_t.
$$ By inverting $1+.8L$, we get
$$
X_t =alpha+(1+.8L)^{-1}epsilon_t =alpha+sum_{j=0}^infty (-.8L)^j epsilon_t=alpha+sum_{j=0}^infty (-.8)^j epsilon_{t-j}.
$$ Now, we claim that $sum_{j=0}^infty (-.8)^j epsilon_{t-j}$ is normally distributed with mean $0$ and variance equal to $4sum_{j=0}^infty (.64)^j=frac{100}{9}.$ The mean-square convergence of the series can be easily seen from the fact that
$$
Bbb Eleft[left|sum_{nle jle m}(-.8)^j epsilon_{t-j}right|^2right]=4sum_{nle jle m}(.64)^j to 0
$$as $n,mtoinfty$. By examining its characteristic function, we get
$$begin{eqnarray}
Bbb Eleft[e^{itsumlimits_{j=0}^infty (-.8)^j epsilon_{t-j}}right]&=&prod_{j=0}^infty Bbb Eleft[e^{it (-.8)^j epsilon_{t-j}}right]\
&=&prod_{j=0}^infty expleft[-frac{t^2 text{Var}left[(-.8)^j epsilon_{t-j}right]}{2}right]\&=&prod_{j=0}^infty expleft[-frac{4t^2cdot(.64)^j}{2}right]\
&=&expleft[-frac{4t^2cdotsum_{j=0}^infty(.64)^j}{2}right]=expleft[-frac{t^2cdotfrac{100}{9}}{2}right].
end{eqnarray}$$ Since the characteristic function of $mathcal{N}(mu,sigma^2)$ is $expleft(itmu -frac{sigma^2t^2}{2}right)$, it follows
$$
X_t sim_d mathcal{N}left(alpha,left(frac{10}{3}right)^2right)
$$ for some $alpha$. ($alpha$ is not determined.)
answered Jan 20 at 13:50
SongSong
16.5k1741
answered Jan 20 at 13:50
SongSong
16.5k1741
answered Jan 20 at 13:50
SongSong
16.5k1741
answered Jan 20 at 13:50
SongSong
16.5k1741
16.5k1741
add a comment |
add a comment |
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