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Gaussian distribution of AR(1) model
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This is very basic, but I have been stuck here for a while.
Consider an AR(1) model $Y_t = c+phi Y_t-1 +epsilon_t$, where $c$ is a constant. If $epsilon_t sim i.i.d. N(0, sigma^2),$ then $Y_1, dots, Y_T$ are also Gaussian, where $Y_1$ is the first observation in the sample.
I don't quite understand how we have each single realization of $Y_t$ Gaussian. It seems that the conditional distribution $Y_t|Y_t-1$ is Gaussian, but why $Y_t$ is Gaussian unconditionally?
regression normal-distribution gaussian-process autoregressive
asked 3 hours ago
Vivian Miller
183
 |Â
show 2 more comments
up vote
2
down vote
favorite
This is very basic, but I have been stuck here for a while.
Consider an AR(1) model $Y_t = c+phi Y_t-1 +epsilon_t$, where $c$ is a constant. If $epsilon_t sim i.i.d. N(0, sigma^2),$ then $Y_1, dots, Y_T$ are also Gaussian, where $Y_1$ is the first observation in the sample.
I don't quite understand how we have each single realization of $Y_t$ Gaussian. It seems that the conditional distribution $Y_t|Y_t-1$ is Gaussian, but why $Y_t$ is Gaussian unconditionally?
regression normal-distribution gaussian-process autoregressive
asked 3 hours ago
Vivian Miller
183
For last question: Because $Y_t+1$ and $epsilon_t} are normal distributed. See here for proof. mathworld.wolfram.com/NormalSumDistribution.html
– a_statistician
2 hours ago
@a_statistician Wait, why $Y_t+1$ is Gaussian?
– Vivian Miller
2 hours ago
typo. Should be $Y_t-1$ and $epsilon_t$ are normal, so their linear combination (one of them is $Y_t$) is normal according to linked site.
– a_statistician
2 hours ago
@a_statistician How do you know $Y_t-1$ is normal?
– Vivian Miller
1 hour ago
Think from beginning when $t=1$. Then $t=2$ ... $t=t-1$
– a_statistician
1 hour ago
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This is very basic, but I have been stuck here for a while.
Consider an AR(1) model $Y_t = c+phi Y_t-1 +epsilon_t$, where $c$ is a constant. If $epsilon_t sim i.i.d. N(0, sigma^2),$ then $Y_1, dots, Y_T$ are also Gaussian, where $Y_1$ is the first observation in the sample.
I don't quite understand how we have each single realization of $Y_t$ Gaussian. It seems that the conditional distribution $Y_t|Y_t-1$ is Gaussian, but why $Y_t$ is Gaussian unconditionally?
regression normal-distribution gaussian-process autoregressive
asked 3 hours ago
Vivian Miller
183
This is very basic, but I have been stuck here for a while.
Consider an AR(1) model $Y_t = c+phi Y_t-1 +epsilon_t$, where $c$ is a constant. If $epsilon_t sim i.i.d. N(0, sigma^2),$ then $Y_1, dots, Y_T$ are also Gaussian, where $Y_1$ is the first observation in the sample.
I don't quite understand how we have each single realization of $Y_t$ Gaussian. It seems that the conditional distribution $Y_t|Y_t-1$ is Gaussian, but why $Y_t$ is Gaussian unconditionally?
regression normal-distribution gaussian-process autoregressive
regression normal-distribution gaussian-process autoregressive
asked 3 hours ago
Vivian Miller
183
asked 3 hours ago
Vivian Miller
183
asked 3 hours ago
Vivian Miller
183
asked 3 hours ago
Vivian Miller
183
asked 3 hours ago
Vivian Miller
183
183
For last question: Because $Y_t+1$ and $epsilon_t} are normal distributed. See here for proof. mathworld.wolfram.com/NormalSumDistribution.html
– a_statistician
2 hours ago
@a_statistician Wait, why $Y_t+1$ is Gaussian?
– Vivian Miller
2 hours ago
typo. Should be $Y_t-1$ and $epsilon_t$ are normal, so their linear combination (one of them is $Y_t$) is normal according to linked site.
– a_statistician
2 hours ago
@a_statistician How do you know $Y_t-1$ is normal?
– Vivian Miller
1 hour ago
Think from beginning when $t=1$. Then $t=2$ ... $t=t-1$
– a_statistician
1 hour ago
 |Â
show 2 more comments
For last question: Because $Y_t+1$ and $epsilon_t} are normal distributed. See here for proof. mathworld.wolfram.com/NormalSumDistribution.html
– a_statistician
2 hours ago
@a_statistician Wait, why $Y_t+1$ is Gaussian?
– Vivian Miller
2 hours ago
typo. Should be $Y_t-1$ and $epsilon_t$ are normal, so their linear combination (one of them is $Y_t$) is normal according to linked site.
– a_statistician
2 hours ago
@a_statistician How do you know $Y_t-1$ is normal?
– Vivian Miller
1 hour ago
Think from beginning when $t=1$. Then $t=2$ ... $t=t-1$
– a_statistician
1 hour ago
For last question: Because $Y_t+1$ and $epsilon_t} are normal distributed. See here for proof. mathworld.wolfram.com/NormalSumDistribution.html
– a_statistician
2 hours ago
For last question: Because $Y_t+1$ and $epsilon_t} are normal distributed. See here for proof. mathworld.wolfram.com/NormalSumDistribution.html
– a_statistician
2 hours ago
@a_statistician Wait, why $Y_t+1$ is Gaussian?
– Vivian Miller
2 hours ago
@a_statistician Wait, why $Y_t+1$ is Gaussian?
– Vivian Miller
2 hours ago
typo. Should be $Y_t-1$ and $epsilon_t$ are normal, so their linear combination (one of them is $Y_t$) is normal according to linked site.
– a_statistician
2 hours ago
typo. Should be $Y_t-1$ and $epsilon_t$ are normal, so their linear combination (one of them is $Y_t$) is normal according to linked site.
– a_statistician
2 hours ago
@a_statistician How do you know $Y_t-1$ is normal?
– Vivian Miller
1 hour ago
@a_statistician How do you know $Y_t-1$ is normal?
– Vivian Miller
1 hour ago
Think from beginning when $t=1$. Then $t=2$ ... $t=t-1$
– a_statistician
1 hour ago
Think from beginning when $t=1$. Then $t=2$ ... $t=t-1$
– a_statistician
1 hour ago
 |Â
show 2 more comments
1 Answer
1
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oldest
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up vote
2
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You are right to be confused here. Strictly speaking, the asserted conclusion in the highlighted statement is a non sequiter. (Can you please add the source of the statement?) The element $Y_1$ is defined recursively in terms of $Y_0$ in the specified recursive equation. Since there is no specification of the distribution of $Y_0$, the distributions of the observable values is not determined.
What they should have specified is that $Y_0 sim textN.$ Unfortunately, people are notoriously sloppy in setting up time-series models, and it is commonly the case for the model to not be properly specified. With a bit of practice you get used to "reading between the lines" to figure out what was intended.
answered 44 mins ago
Ben
16.4k12185
Actually, specifying that $Y_0$ is Gaussian without also saying that $Y_0$ is independent of all the $varepsilon_i$'s (or at least jointly Gaussian with all the $varepsilon_i$'s) is also sloppy because without such a clause, $Y_1 = c + phi Y_0 + varepsilon_1$ is not necessarily Gaussian: the sum of two marginally Gaussian random variables is not Gaussian unless joint Gaussianity is also assumed. Perhaps it is simpler to set $Y_0=0$ and avoid the extra complications.
– Dilip Sarwate
16 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You are right to be confused here. Strictly speaking, the asserted conclusion in the highlighted statement is a non sequiter. (Can you please add the source of the statement?) The element $Y_1$ is defined recursively in terms of $Y_0$ in the specified recursive equation. Since there is no specification of the distribution of $Y_0$, the distributions of the observable values is not determined.
What they should have specified is that $Y_0 sim textN.$ Unfortunately, people are notoriously sloppy in setting up time-series models, and it is commonly the case for the model to not be properly specified. With a bit of practice you get used to "reading between the lines" to figure out what was intended.
answered 44 mins ago
Ben
16.4k12185
Actually, specifying that $Y_0$ is Gaussian without also saying that $Y_0$ is independent of all the $varepsilon_i$'s (or at least jointly Gaussian with all the $varepsilon_i$'s) is also sloppy because without such a clause, $Y_1 = c + phi Y_0 + varepsilon_1$ is not necessarily Gaussian: the sum of two marginally Gaussian random variables is not Gaussian unless joint Gaussianity is also assumed. Perhaps it is simpler to set $Y_0=0$ and avoid the extra complications.
– Dilip Sarwate
16 mins ago
add a comment |Â
up vote
2
down vote
You are right to be confused here. Strictly speaking, the asserted conclusion in the highlighted statement is a non sequiter. (Can you please add the source of the statement?) The element $Y_1$ is defined recursively in terms of $Y_0$ in the specified recursive equation. Since there is no specification of the distribution of $Y_0$, the distributions of the observable values is not determined.
What they should have specified is that $Y_0 sim textN.$ Unfortunately, people are notoriously sloppy in setting up time-series models, and it is commonly the case for the model to not be properly specified. With a bit of practice you get used to "reading between the lines" to figure out what was intended.
answered 44 mins ago
Ben
16.4k12185
Actually, specifying that $Y_0$ is Gaussian without also saying that $Y_0$ is independent of all the $varepsilon_i$'s (or at least jointly Gaussian with all the $varepsilon_i$'s) is also sloppy because without such a clause, $Y_1 = c + phi Y_0 + varepsilon_1$ is not necessarily Gaussian: the sum of two marginally Gaussian random variables is not Gaussian unless joint Gaussianity is also assumed. Perhaps it is simpler to set $Y_0=0$ and avoid the extra complications.
– Dilip Sarwate
16 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You are right to be confused here. Strictly speaking, the asserted conclusion in the highlighted statement is a non sequiter. (Can you please add the source of the statement?) The element $Y_1$ is defined recursively in terms of $Y_0$ in the specified recursive equation. Since there is no specification of the distribution of $Y_0$, the distributions of the observable values is not determined.
What they should have specified is that $Y_0 sim textN.$ Unfortunately, people are notoriously sloppy in setting up time-series models, and it is commonly the case for the model to not be properly specified. With a bit of practice you get used to "reading between the lines" to figure out what was intended.
answered 44 mins ago
Ben
16.4k12185
You are right to be confused here. Strictly speaking, the asserted conclusion in the highlighted statement is a non sequiter. (Can you please add the source of the statement?) The element $Y_1$ is defined recursively in terms of $Y_0$ in the specified recursive equation. Since there is no specification of the distribution of $Y_0$, the distributions of the observable values is not determined.
What they should have specified is that $Y_0 sim textN.$ Unfortunately, people are notoriously sloppy in setting up time-series models, and it is commonly the case for the model to not be properly specified. With a bit of practice you get used to "reading between the lines" to figure out what was intended.
answered 44 mins ago
Ben
16.4k12185
answered 44 mins ago
Ben
16.4k12185
answered 44 mins ago
Ben
16.4k12185
answered 44 mins ago
Ben
16.4k12185
16.4k12185
Actually, specifying that $Y_0$ is Gaussian without also saying that $Y_0$ is independent of all the $varepsilon_i$'s (or at least jointly Gaussian with all the $varepsilon_i$'s) is also sloppy because without such a clause, $Y_1 = c + phi Y_0 + varepsilon_1$ is not necessarily Gaussian: the sum of two marginally Gaussian random variables is not Gaussian unless joint Gaussianity is also assumed. Perhaps it is simpler to set $Y_0=0$ and avoid the extra complications.
– Dilip Sarwate
16 mins ago
add a comment |Â
Actually, specifying that $Y_0$ is Gaussian without also saying that $Y_0$ is independent of all the $varepsilon_i$'s (or at least jointly Gaussian with all the $varepsilon_i$'s) is also sloppy because without such a clause, $Y_1 = c + phi Y_0 + varepsilon_1$ is not necessarily Gaussian: the sum of two marginally Gaussian random variables is not Gaussian unless joint Gaussianity is also assumed. Perhaps it is simpler to set $Y_0=0$ and avoid the extra complications.
– Dilip Sarwate
16 mins ago
Actually, specifying that $Y_0$ is Gaussian without also saying that $Y_0$ is independent of all the $varepsilon_i$'s (or at least jointly Gaussian with all the $varepsilon_i$'s) is also sloppy because without such a clause, $Y_1 = c + phi Y_0 + varepsilon_1$ is not necessarily Gaussian: the sum of two marginally Gaussian random variables is not Gaussian unless joint Gaussianity is also assumed. Perhaps it is simpler to set $Y_0=0$ and avoid the extra complications.
– Dilip Sarwate
16 mins ago
Actually, specifying that $Y_0$ is Gaussian without also saying that $Y_0$ is independent of all the $varepsilon_i$'s (or at least jointly Gaussian with all the $varepsilon_i$'s) is also sloppy because without such a clause, $Y_1 = c + phi Y_0 + varepsilon_1$ is not necessarily Gaussian: the sum of two marginally Gaussian random variables is not Gaussian unless joint Gaussianity is also assumed. Perhaps it is simpler to set $Y_0=0$ and avoid the extra complications.
– Dilip Sarwate
16 mins ago
add a comment |Â
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For last question: Because $Y_t+1$ and $epsilon_t} are normal distributed. See here for proof. mathworld.wolfram.com/NormalSumDistribution.html
– a_statistician
2 hours ago
@a_statistician Wait, why $Y_t+1$ is Gaussian?
– Vivian Miller
2 hours ago
typo. Should be $Y_t-1$ and $epsilon_t$ are normal, so their linear combination (one of them is $Y_t$) is normal according to linked site.
– a_statistician
2 hours ago
@a_statistician How do you know $Y_t-1$ is normal?
– Vivian Miller
1 hour ago
Think from beginning when $t=1$. Then $t=2$ ... $t=t-1$
– a_statistician
1 hour ago