MHB Finding the Autocovariance function

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The discussion revolves around finding the autocovariance function (ACVF) for the stochastic process defined by $X_t = 0.5X_{t-1} + Z_t$, where $Z_t$ is white noise with variance $\sigma^2$. Participants express confusion about calculating the covariance between $X_t$ and $Z_t$, particularly whether it can be assumed that $Cov(X_t, Z_t) = 0$ due to the independence of white noise. It is clarified that while the ACVF calculation appears correct, the lack of information about the values of $X_t$ limits the ability to derive concrete results. The conversation highlights the importance of understanding the relationship between the stochastic process and its components, especially regarding their covariance properties. Overall, the key takeaway is that without additional information about $X_t$, definitive conclusions about its covariance with $Z_t$ cannot be made.
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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?
 
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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?

nacho said:
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?
 
Siron said:
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.
 
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?
 
Siron said:
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.
 
nacho said:
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.
 
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