ARMA(1,1) Processes

For an ARMA(1, 1) process

Now let’s suppose that |φ₁| < 1. We show how to create the MA(∞) representation as follows:

Thus

where ψ₀ = 1 and for j > 0

The MA(∞) representation is therefore

If |φ₁| < 1, then this ARMA(1,1) process is stationary. It also turns out that when |θ₁| < 1, the process is invertible.

Example 1: Find the MA(∞) form of the ARMA(1, 1) process y_i =.4y_i-1 +ε_i – .2ε_i-1

etc.

We get the same result using the Real Statistics PSICoeff array function as shown in Figure 1.

Figure 1 – Finding MA(∞) Coefficients

Property 1: The following is true for an ARMA(1,1) process

Proof: See ARMA Proofs

Property 2: The following is true for an ARMA(1,1) process

and for k > 1

Proof: See ARMA Proofs

Observation: Since z₁ = 1/φ₁ is the root of the characteristic polynomial φ(z) = 1 – φ₁z, we can express the second term in Property 2 as

It turns out that this can be generalized to other ARMA processes.

Property 3: The following is true for an ARMA(1, 1) process

and for k > 1

Proof: See ARMA Proofs

Observation: When θ₁ = –φ₁ for an ARMA(1, 1) process, we note that γ₀ = σ² and ρ_k = 0 for all k > 1, which are the characteristics of white noise.

In fact, the white noise process with zero mean takes the form

y_i = ε_i

We see that φ₁ y_i – φ₁ ε_i  = 0  for all i, and in particular φ₁ y_i-1 – φ₁ ε_i-1  = 0. Thus, the process takes the form

which is an ARMA(1,1) process with θ₁ = –φ₁.

Example 2: Simulate a sample of 105 elements from the ARMA(1,1) process

where ε_i ∼ N(0, 1) and calculate the ACF.

This is done in Figure 2 by placing the formula =NORM.S.INV(RAND()) in cell B7 and the formula =0.7*C6+B7-0.2*B6 in cell C7, highlighting the range B7:C111 and pressing Ctrl-D (only the first 16 elements of the simulation is shown in Figure 2).

Figure 2 – Simulated ARMA(1,1) process

The ACF values are shown in Figure 3 where the theoretical values are based on Property 3.

Figure 3 – ACF for ARMA(1,1) Process

Cell M6 contains the formula =ACF($C$12:$C$111,L6), and similarly, for the other cells in column M, cell N6 contains the formula =(Q5+Q6)*(1+Q5*Q6)/(1+2*Q5*Q6+Q5^2) and cell N7 contains the formula =N6*Q$5, and similarly for rest of the cells in column N.

Since |φ₁| = .7 < 1 and |θ₁| = .2 < 1, this process is both stationary and invertible.

Example 3: Simulate a sample of 105 elements from the ARMA(1,1) process

where ε_i ∼ N(0, 1) and calculate the ACF.

The only difference between this process and the one in Example 2 is the constant term. The approach is identical, except that this time you place the formula =3+0.7*C6+B7-0.2*B6 in cell C7. The result is shown in Figure 4. Note that the ACF values are identical to those in Figure 3.

Figure 4 – Simulated ARMA(1,1) Process with non-zero mean