# Derive of special AR process the autocovariance function

Given the $$y_t = y_{t-2} + \epsilon_t$$ process, which starts at $$y_1$$ and $$\epsilon_t$$ is White Noise :

a) what kind of special process is $$y_t$$ ?

I would say random walk

b) Derive the autocovariance function of $$y_t$$

My problem is the even case, for example $$t = 2$$, then i would get $$y_2 = y_0 + e_2$$, but the process starts at 1. How can I handle $$y_0$$ ?

• What is $a_t$? Do you have more information about that process? – Abdoul Haki Jun 7 '19 at 19:19
• Sorry typo edit it – Felix Ha Jun 7 '19 at 19:27

$$\forall t\geq 2, y_t = \left\{\begin{array}{lcl} y_1 + \sum_{i=1}^{\tfrac{t-1}{2}} \epsilon_{1+2i} & \text{if} & t \text{ odd }\\ y_2 + \sum_{i=1}^{\tfrac{t}{2}} \epsilon_{2i} & \text{if} & t \text{ even } \end{array}\right.$$ then $$\mathbb{C}ov(y_t,y_{t+2k+1}) = 0$$ and $$\mathbb{C}ov(y_t,y_{t+2k}) = \mathbb{C}ov\left(y_t,y_t+\sum_{i=1}^k \epsilon_{t+2i}\right)=\mathbb{C}ov\left(y_t,y_t\right) = \mathbb{V}ar\left(y_t,y_t\right) = t\sigma^2$$ where $$\sigma^2$$ is the variance of $$\epsilon_t$$.
• Could you please explain your solution a little more(How to derive the equation). With your solution $y_3 = y_2 + e_2 + e_4 + e_6$. But with original formula we get for $y_3 = y_1 + e_3$. – Felix Ha Jun 9 '19 at 8:41
• you're right. I correct it. If fact, I meant if $t$ is odd, $y_t$ equal to $y_1$ plus all the innovations with odd index from $3$ to $t$. And almost the same thing with the even $t$. – Abdoul Haki Jun 10 '19 at 13:36