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I'm trying to generate a very simple AR(1) Process with *Mathematica* using the `ARProcess[]`

function. The process must have the following format:

$Y_t = \alpha + \beta Y_{t−1} + ϵ_t$

However, after reading the *Mathematica* help file and some posts about `ARProcess[]`

at MMA.SE, I couldn't find a way to correctly generate the process. I seems that the `ARProcess[]`

function cannot deal with the $\alpha$ term of the process.

For instance, consider the process described below:

$Y_t = 10 + .6Y_{t−1} + ϵ_t$ , where $ϵ_t$ is Normally distributed with mean $0$ and variance $4$ (i.e., $ϵ_t \sim N(0,2)$).

If I use `ARProcess[{.6},4]`

I get

```
Mean[ARProcess[{.6}, 4][t]]
Variance[ARProcess[{.6}, 4][t]]
```

Out[1]= 0

Out[2]= 6.25

which are consistent with an AR(1) process of the form $Y_t = 0 + .6Y_{t−1} + ϵ_t$

I would like to be able to, though the `ARProcess[]`

function, calculate the **mean** and the **variance** of the process $Y_t = 10 + .6Y_{t−1} + ϵ_t$ in order to get a **Mean** of $25$ and a **Variance** of $6.25$.

Is there any way to achieve this in *Mathematica*?

@Fred it does't work... what you've suggested works like an AR(2) Process of the form $Y_t=0+10 Y_{t-1} + .6 Y_{t-2} + \epsilon_t$, which is not stationary and, as such, the mean cannot be computed. – Rod – 2013-05-20T01:26:01.180

Sorry, I'm deleting the comment and will delete this one later. – Fred Kline – 2013-05-20T01:44:21.840

Don't worry @Fred... I think it's somehow a bug in

Mathematica... – Rod – 2013-05-20T01:47:17.520One other thought, is there a way to get the $10$ into the function? – Fred Kline – 2013-05-20T02:04:10.947

Ahah... That's

exactlywhat I'm trying to figure out! I think it's not possible and, as such, it could be a bug inMathematica... – Rod – 2013-05-20T02:05:22.503