Finding the Autocovariance function

[nacho-man]

let $X_t = 0.5X_{t-1} + Z_t$ where $Z_t$ ~ $ WN(0,\sigma^2)$

I want to find the ACVF of both $X_t$ and $Z_t$, but I am a little bit confused.
Say for $X_t$
$$\gamma(h) = COV(0.5X_{t-1} + Z_t, 0.5X_{t-1+h} + Z_{t+h}$$
$ = 0.5^2COV(X_{t-1},X_{t-1+h}) + 0.5COV(X_{t-1},Z_{t+h}) + 0.5COV(Z_{t},X_{t-1+h}) + COV(Z_{t},Z_{t+h}) $then for say $h =-1$ could I still say that the
$0.5^2 COV(X_{t-1},Z_{t+h}) = 0.5 \sigma$ ? Or is there a difference because of the differing $X $ and $Z$ terms? How do I actually find the Covariance between X and Z, given that I only know the variance of Z?
Is it unmathematical to say since X = Two different values of Z, it will have the same variance?

 

[Siron]

First, the only information you have is $\{X_t: t \geq 0\}$ is a stochastic process, defined as $\forall t \geq 0: X_t = 0.5 X_{t-1}+Z_t$ where $Z_t \sim WN(0, \sigma^2)$. Am I right here? Is there no further information about $\{X_t: t \geq 0\}$?

Second, your calculation of the ACVF looks fine. Now, can you show how to compute a covariance?

then for say $h =-1$ could I still say that the
$0.5^2 COV(X_{t-1},Z_{t+h}) = 0.5 \sigma$ ?

What's your argument here?

 

[nacho-man]

First, the only information you have is $\{X_t: t \geq 0\}$ is a stochastic process, defined as $\forall t \geq 0: X_t = 0.5 X_{t-1}+Z_t$ where $Z_t \sim WN(0, \sigma^2)$. Am I right here? Is there no further information about $\{X_t: t \geq 0\}$?

Second, your calculation of the ACVF looks fine. Now, can you show how to compute a covariance?


What's your argument here?

Yup, there is no further information about $X$. And this is what I was confused about.

Since $X$ is valued in terms of $Z$ but at different time steps, would it also have the same Covariance as $Z$?

Otherwise, how would I determine the Covariance.

 

[Siron]

In my opinion:

the only things we can conclude are the following:
1. Since $Z_t \sim WN(0, \sigma^2)$ we can compute $\mbox{cov}[Z_t,Z_s], \forall s,t$.
2. To compute $\mbox{cov}[X_t,Z_t]$:
$$\mbox{cov}[X_t,Z_t]=\mbox{cov}[0.5 X_{t-1}+Z_t,Z_t]= 0.5 \mbox{cov}[X_{t-1},Z_t]+\mbox{cov}[Z_t,Z_t]$$

and ofcourse $\mbox{cov}[Z_t,Z_t]$ can be computed. The problem is that $X_t$ is a stochastic process where each $X_t$ is defined in function of $X_{t-1}$ and $Z_t$. So if we have no information about any of the $X_t$ then we can't get any concrete results.

Ps: where did you get this exercice?

 

[nacho-man]

In my opinion:

the only things we can conclude are the following:
1. Since $Z_t \sim WN(0, \sigma^2)$ we can compute $\mbox{cov}[Z_t,Z_s], \forall s,t$.
2. To compute $\mbox{cov}[X_t,Z_t]$:
$$\mbox{cov}[X_t,Z_t]=\mbox{cov}[0.5 X_{t-1}+Z_t,Z_t]= 0.5 \mbox{cov}[X_{t-1},Z_t]+\mbox{cov}[Z_t,Z_t]$$

and ofcourse $\mbox{cov}[Z_t,Z_t]$ can be computed. The problem is that $X_t$ is a stochastic process where each $X_t$ is defined in function of $X_{t-1}$ and $Z_t$. So if we have no information about any of the $X_t$ then we can't get any concrete results.

Ps: where did you get this exercice?

It was from a lecture slide. Professor has a habit of chucking a few questions from lecture slides into the exam, so I thought it'd be best if I had a concrete answer to it. He's gone away for now so I can't get into contact with him!

Is there any way to conclude that since $Z_t$ is White noise, ie IID that $Cov(X_t,Z_t) = 0$ ? I think that is the likely answer here anyway.

 

[Siron]

It was from a lecture slide. Professor has a habit of chucking a few questions from lecture slides into the exam, so I thought it'd be best if I had a concrete answer to it. He's gone away for now so I can't get into contact with him!

Is there any way to conclude that since $Z_t$ is White noise, ie IID that $Cov(X_t,Z_t) = 0$ ? I think that is the likely answer here anyway.

Maybe that'll clear some things up, but I'm not familiar with the concept White noise.