When will the ARMA(1,1) model (𝑟𝑟𝑡𝑡 =φ0 +φ1 𝑟𝑟𝑡𝑡−1 + 𝑎𝑎𝑡𝑡 −θ1 𝑎𝑎𝑡𝑡−1) be weakly stationary?a. always weakly stationaryb. when |φ1| < 1c. when |θ1| < 1d. when |φ1| < 1 and |θ1| < 1e. none of the above is correct

The ARMA(1,1) model will be weakly stationary when |φ1| < 1. So, the correct answer is b. when |φ1| < 1.

Here's why:

A weakly stationary process, also known as covariance stationary process, has a time invariant mean and autocovariance. For an ARMA(1,1) model to be weakly stationary, the root of its characteristic equation must lie outside the unit circle.

The characteristic equation of the ARMA(1,1) model is 1- φ1B = 0, where B is the backshift operator. The root of this equation is 1/φ1. For the process to be weakly stationary, the absolute value of this root must be greater than 1, which implies that |φ1| must be less than 1.

The parameter θ1 in the model does not affect the stationarity of the process, so it does not matter whether |θ1| < 1 or not.

The ARMA(1,1) model will be weakly stationary when |φ1| < 1. So, the correct answer is b. when |φ1| < 1.

Here's why:

A weakly stationary process, also known as covariance stationary process, has a time invariant mean and autocovariance. For an ARMA(1,1) model to be weakly stationary, the root of its characteristic equation must lie outside the unit circle.

The characteristic equation of an ARMA(1,1) model is 1- φ1B = 0, where B is the backshift operator. The root of this equation is 1/φ1. For the process to be weakly stationary, the absolute value of this root must be greater than 1, which implies that |φ1| must be less than 1.

The parameter θ1 does not affect the stationarity of the ARMA(1,1) model, so it does not need to be less than 1 for the process to be weakly stationary.