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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?

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  • $\begingroup$ For last question: Because $Y_{t+1}$ and $\epsilon_t} are normal distributed. See here for proof. mathworld.wolfram.com/NormalSumDistribution.html $\endgroup$
    – user158565
    Oct 21, 2018 at 20:14
  • $\begingroup$ @a_statistician Wait, why $Y_{t+1}$ is Gaussian? $\endgroup$
    – Vivian Miller
    Oct 21, 2018 at 20:38
  • $\begingroup$ 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. $\endgroup$
    – user158565
    Oct 21, 2018 at 20:54
  • $\begingroup$ @a_statistician How do you know $Y_{t-1}$ is normal? $\endgroup$
    – Vivian Miller
    Oct 21, 2018 at 21:05
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    $\begingroup$ Think from beginning when $t=1$. Then $t=2$ ... $t=t-1$ $\endgroup$
    – user158565
    Oct 21, 2018 at 21:12

1 Answer 1

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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 $Y_0 | \boldsymbol{\epsilon} \sim \text{N}.$ 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.

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    $\begingroup$ 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. $\endgroup$
    – Dilip Sarwate
    Oct 21, 2018 at 22:40
  • $\begingroup$ Good point - edited. $\endgroup$
    – Ben
    Oct 21, 2018 at 23:14
  • $\begingroup$ Also, specifying $Y_0=0$ has the drawback that you lose stationarity, so that is not particularly desirable. $\endgroup$
    – Ben
    Nov 5, 2019 at 7:16

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