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Could someone explain why I didn't get the same answer?

This is an AR(2),
$$X_t={\phi_0}+{\phi_1}{X_{t-1}}+{\phi_2}{X_{t-2}}+{Z_t} \hspace{6 mm} \text{where } Z_t \sim WN(0, \sigma^2)$$

$\begin{align*} \gamma(0) &= \rm{cov}(X_t, X_t)\\ &= \rm{cov}({\phi_0}+{\phi_1}{X_{t-1}}+{\phi_2}{X_{t-2}}+{Z_t},\hspace{3 mm} {\phi_0}+{\phi_1}{X_{t-1}}+{\phi_2}{X_{t-2}}+{Z_t}) \hspace{6 mm} \text{(by definition)}\\ &= \rm{cov}({\phi_1}{X_{t-1}}+{\phi_2}{X_{t-2}}+{Z_t}, \hspace{3 mm} {\phi_1}{X_{t-1}}+{\phi_2}{X_{t-2}}+{Z_t})\\ &= \rm{cov}({\phi_1}{X_{t-1}}, {\phi_1}{X_{t-1}})+ \rm{cov}({\phi_1}{X_{t-1}}, {\phi_2}{X_{t-2}}) + \rm{cov}({\phi_2}{X_{t-2}}, {\phi_1}{X_{t-1}}) + \rm{cov}({\phi_2}{X_{t-2}}, {\phi_2}{X_{t-2}}) + \sigma^2\\ &= {\phi_1^2\gamma(0)} + {\phi_1}{\phi_2}\gamma(1) + {\phi_1}{\phi_2}\gamma(1) + \phi_2^2\gamma(0) + \sigma^2\\ &= (\phi_1^2+\phi_2^2)\gamma(0) + 2{\phi_1}{\phi_2}\gamma(1) + \sigma^2\\ \text{So, we get}\\ \gamma(0) &= (\phi_1^2+\phi_2^2)\gamma(0) + 2{\phi_1}{\phi_2}\gamma(1) + \sigma^2\\ (1-\phi_1-\phi_2)\gamma(0) &= 2{\phi_1}{\phi_2}\gamma(1) + \sigma^2\\ \gamma(0) &= \frac{2{\phi_1}{\phi_2}\gamma(1) + \sigma^2}{(1-\phi_1-\phi_2)}\\ \end{align*}$

This is the final answer, but it is not correct. It is derived as the following, $\begin{align*} X_t &={\phi_0}+{\phi_1}{X_{t-1}}+{\phi_2}{X_{t-2}}+{Z_t}\\ X_tX_t &={\phi_0}+{\phi_1}{X_tX_{t-1}}+{\phi_2}{X_tX_{t-2}}+{X_tZ_t}\\ E[X_t^2] &={\phi_0}+{\phi_1}E[{X_tX_{t-1}}]+{\phi_2}E[{X_tX_{t-2}}]+E[{X_tZ_t}] \hspace{3 mm}\\ \text{(note: }\rm{cov}(x,y)=E[xy]- E[x]E[y])\\ \text{(Where did the 2nd term } E[x]E[y] \text{ go}???)\\ \gamma(0)&= {\phi_0}+{\phi_1}\gamma(1)+{\phi_2}\gamma(2)+\sigma^2 \\ \end{align*}$

Why are two answers different? Is one of them wrong?

user13985
  • 1,215

1 Answers1

3

The first computation is correct, just be carefull just after the "so we get": $$ \gamma(0) = \frac{2\phi_1\phi_2\gamma(1) - \sigma^2}{1-\phi_1^2-\phi_2^2} $$

The second one is ok: $$EX_t^2 = \phi_0 EX_t+ \phi_1 EX_t X_{t-1}+ \phi_2 EX_t X_{t-2} + E[X_t Z_t] $$ Now substract $$ [EX_t]^2 = \phi_0 [EX_t]+ \phi_1 EX_t EX_{t-1}+ \phi_2 EX_t EX_{t-2} + EX_t E Z_t $$ and using $\rm{cov}(X_t,Z_t) = \sigma^2$ you get:

$$ \gamma(0) = \phi_1 \gamma(1)+ \phi_2 \gamma(2) + \sigma^2 $$

This is a particular case of the Yule-Walker equations.

mookid
  • 28,236