Example 1: A non-mixing autoregressive process

The process (1, . . . , n) is simulated, according to the AR(1) equation:

k+1=

1

2(k+ ηk+1),

where 1 is uniformly distributed over [−1₂,1₂], and (ηi)i≥2 is a sequence of i.i.d. random variables,

independent of 1, such that P(ηi = −1₂) = P(ηi = 1₂) = 1₂. In this example, Fi = σ(ηk, k ≤ i), and the

σ-algebra F−∞is trivial.

The transition kernel of the chain (i)i≥1is:

K(f )(x) = 1 2  f x 2 + 1 4  + f x 2 − 1 4  ,

and the uniform distribution on [−1₂,1₂]is the unique invariant distribution by K. Hence, the chain (i)i≥1

is strictly stationary. Furthermore, it is not α-mixing in the sense of Rosenblatt [21], but it is ˜φ-dependent. Indeed, one can prove that the coefficient ˜φof the chain (i)i≥1decreases geometrically [39]:

˜

φ(k) ≤ 2−k.

Consequently Hannan’s conditions are satisfied and the Fisher tests can be corrected as indicated above.

The first model simulated with this error process is the following linear regression model, for all i in {1, ..., n}:

Yi= β0+ β1i + 10i.

The random variables iare multiplied by 10 to increase the variance. The coefficient β0is chosen equal to 3. We test the hypothesis H0: β1= 0, against the hypothesis H1: β16= 0.

The estimated level of the Fisher test will be studied for different choices of n and an, which is the

number of covariance terms considered. Under the hypothesis H0, the same Fisher test is carried out 2000times. Then we look at the frequency of rejection of the test when we are under H0, that is to say the estimated level of the test. Let us specify that we want an estimated level close to 5%.

• Case β1= 0and an = 0(no correction):

n 200 400 600 800 1000

Estimated level 0.2745 0.2655 0.2615 0.2845 0.2445

Here, since an = 0, we do not estimate any of the covariance terms. The result is that the estimated

levels are too large. This means that the test will reject the null hypothesis too often.

The quantities anmay be chosen by analyzing the graph of the empirical autocovariances, Figure 1.1,

Figure 1.1 – Empirical autocovariances for the first model of Example 1, n = 600. • Case β1= 0, an= 2:

n 200 400 600 800 1000

Estimated level 0.0805 0.086 0.0745 0.0675 0.077

As suggested by the graph of the empirical autocovariances, the choice an = 2 gives a better esti-

mated level than an= 0.

• Case β1= 0, an= 3:

n 200 400 600 800 1000

Estimated level 0.078 0.0725 0.074 0.059 0.0625

Here, we see that the choice an = 3works well also, and seems even slightly better than the choice

an = 2. If one increases the size of the samples n, and the number of estimated covariance terms an,

we are getting closer to the estimated level 5 %. If n = 5000 and an = 4, the estimated level is around

0.05.

• Case β1= 0.005, an = 3:

In this example, H0is not satisfied. We choose β1equal to 0.005, and we perform the same tests as above (N = 2000) to estimate the power of the test.

n 200 400 600 800 1000

As one can see, the estimated power is always greater than 0.05, as expected. Still as expected, the estimated power increases with the size of the samples. For n = 200, the power of the test is around 0.2255, and for n = 800, the power is around 1. As soon as n = 800, the test always rejects the H0- hypothesis.

The second model considered is the following linear regression model, for all i in {1, ..., n}:

Yi= β0+ β1i + β2i2+ 10i.

Here, we test the hypothesis H0: β1 = β2 = 0against H1: β1 6= 0 or β2 6= 0. The coefficient β0 is equal to 3, and we use the same simulation scheme as above.

• Case β1= β2= 0and an = 0(no correction):

n 200 400 600 800 1000

Estimated level 0.402 0.378 0.385 0.393 0.376 As for the first simulation, if an= 0the test will reject the null hypothesis too often.

As suggested by the graph of the estimated autocovariances Figure 1.2, the choice an = 4should

give a better result for the estimated level.

Figure 1.2 – Empirical autocovariances for the second model of Example 1, n = 600. • Case β1= β2= 0, an = 4:

n 200 400 600 800 1000

Here, we see that the choice an = 4works well. For n = 1000, the estimated level is around 0.06. If

n = 2000and an= 4, the estimated level is around 0.05.

• Case β1= 0.005, β2= 0, an = 4:

Now, we study the estimated power of the test. The coefficient β1is chosen equal to 0.005 and β2is zero.

n 200 400 600 800 1000

Estimated power 0.2145 0.634 0.9855 1 1

As expected, the estimated power increases with the size of the samples, and it is around 1 as soon as n = 800.

The third model that we consider is the following linear regression model, for all i in {1, ..., n}:

Yi= β0+ β1 √

i + β2log(i) + 10i.

We test again the hypothesis H0: β1 = β2 = 0against H1: β1 6= 0 or β2 6= 0. The coefficient β0 is equal to 3. The conditions of the simulation are the same as above except for the size of the samples. Indeed, for this model, the size of the samples n must be greater than previously to have an estimated level close to 5% with the correction.

• Case β1= β2= 0and an= 0(no correction):

n 500 1000 2000 3000 4000 5000

Estimated level 0.4435 0.4415 0.427 0.3925 0.397 0.4075

As for the first and second simulation, if an= 0the test will reject the null hypothesis too often.

As suggested by the graph of the estimated autocovariances Figure 1.3, the choice an = 4should

give a better result for the estimated level. • Case β1= β2= 0, an= 4:

n 500 1000 2000 3000 4000 5000 Estimated level 0.106 0.1 0.078 0.072 0.077 0.068

For an = 4and n = 5000, the estimated level is around 0.07. If n = 10000, it is around 5%.

Then, we study the estimated power of the test for β0or β1non equal to 0. • Case β1= 0, β2= 0.2, an = 4:

Figure 1.3 – Empirical autocovariances for the third model of Example 1, n = 2000.

n 500 1000 2000 3000 4000 5000

Estimated power 0.2505 0.317 0.4965 0.6005 0.725 0.801

As expected, the estimated power increases with the size of the samples, and it is around 0.8 as soon as n = 5000.