Cross Validated Asked on January 5, 2022

I don’t understand results of `ar()`

function in R.

I made up a very simple case, Fibonacci sequence:

```
x <- c(1,1,2,3,5,8,13,21,34,55)
ar(x)
```

result is

```
Coefficients:
1
0.5531
```

I would expect result of 1,1 – for the model `x(n) = 1* x(n-1) + 1 * x(n-2)`

Can anyone explain me please why I don’t get expected result?

You may use `ar.ols`

to estimate this non-stationary series.

From the docs,

ar.ols fits the general AR model to a possibly non-stationary and/or multivariate system of series x.

Example with your data:

```
ar.ols(x,demean=FALSE)
Call:
ar.ols(x = x, demean = FALSE)
Coefficients:
1 2
1 1
```

Answered by A. Webb on January 5, 2022

As I said in my comment, you are going toward a wrong direction.

`ar`

is assuming `x`

as a stationary process `AR(p)`

. The default estimation method `"yule-walker"`

is a moment estimator. Please see Yule-Walker equations of autoregressive process for more.

`ar`

selects order `p`

by minimizing AIC. For you example Fibonacci sequence `x`

, it has selected `p = 1`

. The resulting coefficient, by Yule-Walker equations, matches the sample ACF at lag 1:

```
z <- acf(x, lag.max = 1)
# 0 1
#1.000 0.553
```

Since model assumption is wrong, you definitely can not get `c(1, 1)`

as the answer. A crude way to get you to the right estimation is using least squares linear regression:

```
N <- length(x)
y <- x[3:N]
x1 <- x[2:(N-1)] # lag-1
x2 <- x[1:(N-2)] # lag-2
lm(y ~ x1 + x2 - 1) ## drop intercept (as you know it for sure)
#Coefficients:
#x1 x2
# 1 1
```

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