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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?

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

1 Answer 1

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\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*}

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