# Gaussian distribution of AR(1) model

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?

• For last question: Because $Y_{t+1}$ and $\epsilon_t} are normal distributed. See here for proof. mathworld.wolfram.com/NormalSumDistribution.html – user158565 Oct 21 '18 at 20:14 • @a_statistician Wait, why$Y_{t+1}$is Gaussian? – Vivian Miller Oct 21 '18 at 20:38 • typo. Should be$Y_{t-1}$and$\epsilon_t$are normal, so their linear combination (one of them is$Y_t$) is normal according to linked site. – user158565 Oct 21 '18 at 20:54 • @a_statistician How do you know$Y_{t-1}$is normal? – Vivian Miller Oct 21 '18 at 21:05 • Think from beginning when$t=1$. Then$t=2$...$t=t-1$– user158565 Oct 21 '18 at 21:12 ## 1 Answer 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. • 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 Oct 21 '18 at 22:40
• Good point - edited. – Ben Oct 21 '18 at 23:14