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Hi Parseval: Let me put an answer here and hopefully it's clear but I NEED $\mu = 0 $ or it doesn't work out. So, I believe that the question should have said that $\mu = 0$.

(1) $Y_t = \sum_{i=0}^{q} a_{i} X_{t-i}$

(2) $Y_{t+h} = \sum_{j=0}^{q} a_j X_{t+h-j}$

$ h > q$.

Note that the last (earliest ) term in (2) is $a_{q} X_{t+h-q}$ and first (latest ) term in (1) is $a_{0} X_{t}$. Therefore, since $h > q$, none of the noise terms in $Y_t$ overlap with any of the the noise terms in $Y_{t+h}$.

Now, the usual definition of covariance, gives:

$Cov(Y_{t}, Y_{t+h}) = E(\sum_{i=0}^{q} a_{i} X_{t-i} \sum_{j=0}^{q} a_{j} X_{t+h-j}) - E(\sum_{i=0}^{q} a_{i} X_{t-i}) E(\sum_{j=0}^{q} a_j X_{t+h-j}) $

So, for the first term we have the two sums multiplying each other and then an expectation is taken. Then, for the second term, we have two expectations multiplying each other.

Each of the $X$s have expectation $\mu = 0$.

FIRST SIMPLIFY THE SECOND TERM

Taking the expectations of the terms, we get $\sum_{i=0}^{q} a_{i} \mu \sum_{j=0}^{q} a_j \mu = \mu^2 \sum_{i=0}^{q} a_{i} \sum_{j=0}^{q} a_j$.

Then, since $\mu = 0$, we get:

$ \mu^2 \sum_{i=0}^{q} \sum_{j=0}^{q} a_i a_j = 0 $

NOW SIMPLIFY THE FIRST TERM:

$ E(\sum_{i=0}^{q} a_i X_{t-i} \sum_{j=0}^{q} a_j X_{t+h-j}) = $

$( \sum_{i=0}^{q} \sum_{j=0}^{q} a_{i} a_j ) E(X_{t-i} X_{t+h-j})$

But we showed earlier than that none of the terms in the very last expectation overlap, and, since they are independent Gaussians, we can re-write the last expression as $ E(X_{t-i} X_{(t+h-j)}) = E(X_{t-i}) E(X_{(t+h-j)}) = 0$.

So, we showed that the first term equals zero and the second term equals zero which means that $Cov(Y_{t}, Y_{t+h}) = 0 $.

Note though that I needed two assumptions to obtain the result.

A) the X_{t} are independent Gaussian RV's. This is the usual assumption in the MA(q) model.

B) The expectation of the $X_t$, $E(X_{t}), = \mu = 0$. This is also the usual assumption in the MA(q) model.

Hi Parseval: Let me put an answer here and hopefully it's clear. I fixed it so that $\mu = 0 $ is not needed but of course, the independent Gaussian assumption is still needed.

(1) $Y_t = \sum_{i=0}^{q} a_{i} X_{t-i}$

(2) $Y_{t+h} = \sum_{j=0}^{q} a_j X_{t+h-j}$

$ h > q$.

Note that the last (earliest ) term in (2) is $a_{q} X_{t+h-q}$ and first (latest ) term in (1) is $a_{0} X_{t}$. Therefore, since $h > q$, none of the noise terms in $Y_t$ overlap with any of the the noise terms in $Y_{t+h}$.

Now, the usual definition of covariance, gives:

$Cov(Y_{t}, Y_{t+h}) = E(\sum_{i=0}^{q} a_{i} X_{t-i} \sum_{j=0}^{q} a_{j} X_{t+h-j}) - E(\sum_{i=0}^{q} a_{i} X_{t-i}) E(\sum_{j=0}^{q} a_j X_{t+h-j}) $

So, for the first term we have the two sums multiplying each other and then an expectation is taken. Then, for the second term, we have two expectations multiplying each other.

FIRST SIMPLIFY THE SECOND TERM

Taking the expectations of the terms, we get $\sum_{i=0}^{q} a_{i} \mu \sum_{j=0}^{q} a_j \mu = \mu^2 \sum_{i=0}^{q} a_{i} \sum_{j=0}^{q} a_j$.

Simplifying this, results in:

$ \mu^2 \sum_{i=0}^{q} \sum_{j=0}^{q} a_i a_j $

NOW SIMPLIFY THE FIRST TERM:

$ E(\sum_{i=0}^{q} a_i X_{t-i} \sum_{j=0}^{q} a_j X_{t+h-j}) = $

$( \sum_{i=0}^{q} \sum_{j=0}^{q} a_{i} a_j ) E(X_{t-i} X_{t+h-j})$

But we showed earlier than that none of the terms in the very last expectation overlap, and, since they are independent Gaussians, we can re-write the last expression as $ E(X_{t-i} X_{(t+h-j)}) = E(X_{t-i}) E(X_{(t+h-j)}) = \mu^2$.

So, we showed that the first term equals the same thing as the second term which means that $Cov(Y_{t}, Y_{t+h}) = 0 $.

Note though that I needed the assumption that the X_{t} are independent Gaussian RV's but this is the usual assumption in the MA(q) model.

Hi Parseval: Let me put an answer here and hopefully it's clear but I NEED $\mu = 0 $ or it doesn't work out. So, I believe that the question should have said that $\mu = 0$.

(1) $Y_t = \sum_{i=0}^{q} a_{i} X_{t-i}$

(2) $Y_{t+h} = \sum_{j=0}^{q} a_j X_{t+h-j}$

$ h > q$.

Note that the last (earliest ) term in (2) is $a_{q} X_{t+h-q}$ and first (latest ) term in (1) is $a_{0} X_{t}$. Therefore, since $h > q$, none of the noise terms in $Y_t$ overlap with any of the the noise terms in $Y_{t+h}$.

Now, the usual definition of covariance, gives:

$Cov(Y_{t}, Y_{t+h}) = E(\sum_{i=0}^{q} a_{i} X_{t-i} \sum_{j=0}^{q} a_{j} X_{t+h-j}) - E(\sum_{i=0}^{q} a_{i} X_{t-i}) E(\sum_{j=0}^{q} a_j X_{t+h-j}) $

So, for the first term we have the two sums multiplying each other and then an expectation is taken. Then, for the second term, we have two expectations multiplying each other.

Each of the $X$s have expectation $\mu = 0$.

FIRST SIMPLIFY THE SECOND TERM

Taking the expectations of the terms, we get $\sum_{i=0}^{q} a_{i} \mu \sum_{j=0}^{q} a_j \mu = \mu^2 \sum_{i=0}^{q} a_{i} \sum_{j=0}^{q} a_j$.

Then, since $\mu = 0$, we get:

$ \mu^2 \sum_{i=0}^{q} \sum_{j=0}^{q} a_i a_j = 0 $

NOW SIMPLIFY THE FIRST TERM:

$ E(\sum_{i=0}^{q} a_i X_{t-i} \sum_{j=0}^{q} a_j X_{t+h-j}) = $

$( \sum_{i=0}^{q} a_i \sum_{j=0}^{q} a_j ) \sum_{i=0}^{q}\sum_{j=0}^{q} E(X_{t-i} X_{t+h-j})$

Note that none of the terms in the very last expectation overlap, so they are independent which means that the  $ E(X_{t-i} X_{(t+h-j)}) = E(X_{t-i}) E(X_{(t+h-j)}) = 0$.

So, we showed that the first term equals zero and the second term equals zero which means that $Cov(Y_{t}, Y_{t+h}) = 0 $.

Hi Parseval: Let me put an answer here and hopefully it's clear but I NEED $\mu = 0 $ or it doesn't work out. So, I believe that the question should have said that $\mu = 0$.

(1) $Y_t = \sum_{i=0}^{q} a_{i} X_{t-i}$

(2) $Y_{t+h} = \sum_{j=0}^{q} a_j X_{t+h-j}$

$ h > q$.

Note that the last (earliest ) term in (2) is $a_{q} X_{t+h-q}$ and first (latest ) term in (1) is $a_{0} X_{t}$. Therefore, since $h > q$, none of the noise terms in $Y_t$ overlap with any of the the noise terms in $Y_{t+h}$.

Now, the usual definition of covariance, gives:

$Cov(Y_{t}, Y_{t+h}) = E(\sum_{i=0}^{q} a_{i} X_{t-i} \sum_{j=0}^{q} a_{j} X_{t+h-j}) - E(\sum_{i=0}^{q} a_{i} X_{t-i}) E(\sum_{j=0}^{q} a_j X_{t+h-j}) $

So, for the first term we have the two sums multiplying each other and then an expectation is taken. Then, for the second term, we have two expectations multiplying each other.

Each of the $X$s have expectation $\mu = 0$.

FIRST SIMPLIFY THE SECOND TERM

Taking the expectations of the terms, we get $\sum_{i=0}^{q} a_{i} \mu \sum_{j=0}^{q} a_j \mu = \mu^2 \sum_{i=0}^{q} a_{i} \sum_{j=0}^{q} a_j$.

Then, since $\mu = 0$, we get:

$ \mu^2 \sum_{i=0}^{q} \sum_{j=0}^{q} a_i a_j = 0 $

NOW SIMPLIFY THE FIRST TERM:

$ E(\sum_{i=0}^{q} a_i X_{t-i} \sum_{j=0}^{q} a_j X_{t+h-j}) = $

$( \sum_{i=0}^{q} \sum_{j=0}^{q} a_{i} a_j ) E(X_{t-i} X_{t+h-j})$

But we showed earlier than that none of the terms in the very last expectation overlap, and, since they are independent Gaussians, we can re-write the last expression as $ E(X_{t-i} X_{(t+h-j)}) = E(X_{t-i}) E(X_{(t+h-j)}) = 0$.

So, we showed that the first term equals zero and the second term equals zero which means that $Cov(Y_{t}, Y_{t+h}) = 0 $.

Note though that I needed two assumptions to obtain the result.

A) the X_{t} are independent Gaussian RV's. This is the usual assumption in the MA(q) model.

B) The expectation of the $X_t$, $E(X_{t}), = \mu = 0$. This is also the usual assumption in the MA(q) model.

Hi Parseval: Let me put an answer here and hopefully it's clear but I NEED $\mu = 0 $ or it doesn't work out. So, I believe that the question should have said that $\mu = 0$.

(1) $Y_t = \sum_{i=0}^{q} a_{i} X_{t-i}$

(2) $Y_{t+h} = \sum_{j=0}^{q} a_j X_{t+h-j}$

$ h > q$.

Note that the last (earliest ) term in (2) is $a_{q} X_{t+h-q}$ and first (latest ) term in (1) is $a_{0} X_{t}$. Therefore, since $h > q$, none of the noise terms in $Y_t$ overlap with any of the the noise terms in $Y_{t+h}$.

Now, the usual definition of covariance, gives:

$Cov(Y_{t}, Y_{t+h}) = E(\sum_{i=0}^{q} a_{i} X_{t-i} \sum_{j=0}^{q} a_{j} X_{t+h-j}) - E(\sum_{i=0}^{q} a_{i} X_{t-i}) E(\sum_{j=0}^{q} a_j X_{t+h-j}) $

So, for the first term we have the two sums multiplying each other and then an expectation is taken. Then, for the second term, we have two expectations multiplying each other.

Each of the $X$s have expectation $\mu = 0$.

FIRST SIMPLIFY THE SECOND TERM

Taking the expectations of the terms, we get $\sum_{i=0}^{q} a_{i} \mu \sum_{j=0}^{q} a_j \mu = \mu^2 \sum_{i=0}^{q} a_{i} \sum_{j=0}^{q} a_j$.

Then, since $\mu = 0$, we get:

$ \mu^2 \sum_{i=0}^{q} \sum_{j=0}^{q} a_i a_j = 0 $

NOW SIMPLIFY THE FIRST TERM:

$ E(\sum_{i=0}^{q} a_i X_{t-i} \sum_{j=0}^{q} a_j X_{t+h-j)}) = $

$( \sum_{i=0}^{q} a_i \sum_{j=0}^{q} a_j ) \sum_{i=0}^{q}\sum_{j=0}^{q} E(X_{t-i} X_{(t+h-j)})$

Note that none of the terms in the very last expectation overlap, so they are independent which means that the $ E(X_{t-i} X_{(t+h-j)}) = E(X_{t-i}) E(X_{(t+h-j)}) = 0$.

So, we showed that the first term equals zero and the second term equals zero which means that $Cov(Y_{t}, Y_{t+h}) = 0 $.

Hi Parseval: Let me put an answer here and hopefully it's clear but I NEED $\mu = 0 $ or it doesn't work out. So, I believe that the question should have said that $\mu = 0$.

(1) $Y_t = \sum_{i=0}^{q} a_{i} X_{t-i}$

(2) $Y_{t+h} = \sum_{j=0}^{q} a_j X_{t+h-j}$

$ h > q$.

Note that the last (earliest ) term in (2) is $a_{q} X_{t+h-q}$ and first (latest ) term in (1) is $a_{0} X_{t}$. Therefore, since $h > q$, none of the noise terms in $Y_t$ overlap with any of the the noise terms in $Y_{t+h}$.

Now, the usual definition of covariance, gives:

$Cov(Y_{t}, Y_{t+h}) = E(\sum_{i=0}^{q} a_{i} X_{t-i} \sum_{j=0}^{q} a_{j} X_{t+h-j}) - E(\sum_{i=0}^{q} a_{i} X_{t-i}) E(\sum_{j=0}^{q} a_j X_{t+h-j}) $

So, for the first term we have the two sums multiplying each other and then an expectation is taken. Then, for the second term, we have two expectations multiplying each other.

Each of the $X$s have expectation $\mu = 0$.

FIRST SIMPLIFY THE SECOND TERM

Taking the expectations of the terms, we get $\sum_{i=0}^{q} a_{i} \mu \sum_{j=0}^{q} a_j \mu = \mu^2 \sum_{i=0}^{q} a_{i} \sum_{j=0}^{q} a_j$.

Then, since $\mu = 0$, we get:

$ \mu^2 \sum_{i=0}^{q} \sum_{j=0}^{q} a_i a_j = 0 $

NOW SIMPLIFY THE FIRST TERM:

$ E(\sum_{i=0}^{q} a_i X_{t-i} \sum_{j=0}^{q} a_j X_{t+h-j}) = $

$( \sum_{i=0}^{q} a_i \sum_{j=0}^{q} a_j ) \sum_{i=0}^{q}\sum_{j=0}^{q} E(X_{t-i} X_{t+h-j})$

Note that none of the terms in the very last expectation overlap, so they are independent which means that the $ E(X_{t-i} X_{(t+h-j)}) = E(X_{t-i}) E(X_{(t+h-j)}) = 0$.

So, we showed that the first term equals zero and the second term equals zero which means that $Cov(Y_{t}, Y_{t+h}) = 0 $.