# Covariance of a Stationary Process

Let $$Y_t$$ be a stationary process such that $$Y_1 = a_1$$ and $$Y_2 = \theta a_1 + a_2$$, where $$\theta$$ is a parameter and $$a_t$$ is the white noise process with mean 2 and variance $$\sigma^2_a = 0.5$$. Find cov$$(Y_1, Y_2)$$.

According to the textbook the answer is $$\theta\sigma_a^2$$. However, I've tried using the definition $${\rm cov}(Y_1, Y_2) = E[Y_1Y_2] - \mu_{Y_1}\mu_{Y_2}$$

however I keep getting a value of 0.

I think there is a property of covariance I must be missing here. Any suggestions?

• Could you show how you determined the covariance is zero? Obviously the problem occurs somewhere in those steps.
– whuber
Commented Apr 20, 2021 at 20:16
• Try using the following properties: $Cov(aZ + Y, X) = Cov(aZ,X) + Cov(Y,X)$ and $Cov(X,X) = Var(X)$ Commented Apr 20, 2021 at 20:17

$$\begin{eqnarray*} {\rm Cov}(Y_1, Y_2) & = & {\rm Cov}(a_1, \theta a_1 + a_2) \\ & = & {\rm Cov}(a_1, \theta a_1) + {\rm Cov}(a_1, a_2) \\ & = & \theta{\rm Cov}(a_1, a_1) + 0 \\ & = & \theta {\rm Var}(a_1) \\ & = & \theta \sigma_a^2. \end{eqnarray*}$$