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 $y_0$ ?

  • $\begingroup$ What is $a_t$? Do you have more information about that process? $\endgroup$ – Abdoul Haki Jun 7 '19 at 19:19
  • $\begingroup$ Sorry typo edit it $\endgroup$ – 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 $\epsilon_t$.

I don't know if you can call it a random walk but it has a similar dynamic.

  • $\begingroup$ 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 $y_3 = y_1 + e_3$. $\endgroup$ – Felix Ha Jun 9 '19 at 8:41
  • $\begingroup$ 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 $t$. $\endgroup$ – Abdoul Haki Jun 10 '19 at 13:36

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