Variance of Sample Mean in Time Series

The following is a proposition which is important for deriving the distribution of the sample mean for a covariance stationary process.

Proposition:

Let ${x_{t}}$ be a mean zero covariance stationary time series with $sum_{k=0}^{infty}|gamma_{x}(k)|<infty$ ( where $gamma_{x}(k)$ is the autocovariance function at lag $k$).

Then $frac{1}{T} mathbb{V} left[ sum_{t=1}^{T} x_{t} right] = gamma_{x}(0) + 2 sum_{k=1}^{infty}gamma_{x}(k)$

Proof:

My proof thus far is:

begin{align*}
frac{1}{T} mathbb{V} left[ sum_{t=1}^{T} x_{t} right] &= frac{1}{T}left[ sum_{t=1}^{T} mathbb{V}[x_{t}] + mathop{sum^Tsum^T}_{ineq j} Cov(x_{i}, x_{j}) right]\
&= frac{1}{T}left[ sum_{t=1}^{T} mathbb{V}[x_{t}] + 2 mathop{sum^Tsum^T}_{i < j} Cov(x_{i}, x_{j}) right]\
&= frac{1}{T}left[ Tcdot gamma_{x}(0) + 2 (T-1) cdot gamma_{x}(1) + 2(T-2) cdot gamma_{x}(2) + ldots + 4 gamma_{x}(T-2) + 2 gamma_{x}(T-1) right]\
&= frac{1}{T}left[ Tcdot gamma_{x}(0) + 2 sum_{i=1}^{T-1} (T-i) cdot gamma_{x}(i) right]\
&= gamma_{x}(0) + 2 sum_{i=1}^{T-1} frac{(T-i)}{T} cdot gamma_{x}(i) \
end{align*}
Now all I need to show is:
begin{align*}
lim_{Ttoinfty} gamma_{x}(0) + 2 sum_{i=1}^{T-1} frac{(T-i)}{T} cdot gamma_{x}(i) = gamma_{x}(0) + 2 sum_{i=1}^{infty} gamma_{x}(i) \
end{align*}

However I am having trouble with this limit and am worried it does not converge.

Context:

This proposition was stated by my professor in a time series class and has been used to prove many other results. However it is possible I have approached the proof in the wrong way. Any input will help.

edited Feb 22 '15 at 14:41

asked Feb 21 '15 at 23:48

möbius

8611021

$begingroup$
related: stats.stackexchange.com/questions/154070/…
$endgroup$
– Taylor
Apr 29 '17 at 21:34

add a comment |

The following is a proposition which is important for deriving the distribution of the sample mean for a covariance stationary process.

Proposition:

Let ${x_{t}}$ be a mean zero covariance stationary time series with $sum_{k=0}^{infty}|gamma_{x}(k)|<infty$ ( where $gamma_{x}(k)$ is the autocovariance function at lag $k$).

Then $frac{1}{T} mathbb{V} left[ sum_{t=1}^{T} x_{t} right] = gamma_{x}(0) + 2 sum_{k=1}^{infty}gamma_{x}(k)$

Proof:

My proof thus far is:

However I am having trouble with this limit and am worried it does not converge.

Context:

edited Feb 22 '15 at 14:41

asked Feb 21 '15 at 23:48

möbius

8611021

$begingroup$
related: stats.stackexchange.com/questions/154070/…
$endgroup$
– Taylor
Apr 29 '17 at 21:34

add a comment |

The following is a proposition which is important for deriving the distribution of the sample mean for a covariance stationary process.

Proposition:

Let ${x_{t}}$ be a mean zero covariance stationary time series with $sum_{k=0}^{infty}|gamma_{x}(k)|<infty$ ( where $gamma_{x}(k)$ is the autocovariance function at lag $k$).

Then $frac{1}{T} mathbb{V} left[ sum_{t=1}^{T} x_{t} right] = gamma_{x}(0) + 2 sum_{k=1}^{infty}gamma_{x}(k)$

Proof:

My proof thus far is:

However I am having trouble with this limit and am worried it does not converge.

Context:

edited Feb 22 '15 at 14:41

asked Feb 21 '15 at 23:48

möbius

8611021

The following is a proposition which is important for deriving the distribution of the sample mean for a covariance stationary process.

Proposition:

Let ${x_{t}}$ be a mean zero covariance stationary time series with $sum_{k=0}^{infty}|gamma_{x}(k)|<infty$ ( where $gamma_{x}(k)$ is the autocovariance function at lag $k$).

Then $frac{1}{T} mathbb{V} left[ sum_{t=1}^{T} x_{t} right] = gamma_{x}(0) + 2 sum_{k=1}^{infty}gamma_{x}(k)$

Proof:

My proof thus far is:

However I am having trouble with this limit and am worried it does not converge.

Context:

probability limits stochastic-processes time-series

edited Feb 22 '15 at 14:41

asked Feb 21 '15 at 23:48

möbius

8611021

edited Feb 22 '15 at 14:41

asked Feb 21 '15 at 23:48

möbius

8611021

edited Feb 22 '15 at 14:41

asked Feb 21 '15 at 23:48

möbius

8611021

asked Feb 21 '15 at 23:48

möbius

8611021

asked Feb 21 '15 at 23:48

möbius

8611021

$begingroup$
related: stats.stackexchange.com/questions/154070/…
$endgroup$
– Taylor
Apr 29 '17 at 21:34

add a comment |

$begingroup$
related: stats.stackexchange.com/questions/154070/…
$endgroup$
– Taylor
Apr 29 '17 at 21:34

related: stats.stackexchange.com/questions/154070/…

– Taylor
Apr 29 '17 at 21:34

add a comment |

1 Answer
1

active

oldest

votes

The upper limit of your sum should be $T-1$ (since the $t=T$ term is zero). We can write

$$sum_{i=1}^{T-1}frac{T-i}Tgamma_x(i)=sum_{i=1}^{T-1}left(1-frac iTright)gamma_x(i):=a_T. $$
Clearly $a_T$ need not converge as $Ttoinfty$. However, $a_T$ converges if and only if
$$sum_{i=-infty}^infty gamma_x(i)<infty $$
(from my time series notes, a proof of this statement is escaping me at the moment). When that limit exists, your result holds.

answered Feb 22 '15 at 14:14

Math1000

19.4k31746

$begingroup$
I have incorporated this minor change to the limit of the sum. However, the issue of the convergence of $a_{T}$ is at the crux of the proof; not only must $a_{T}$ converge, but it must converge to the desired limit above. Your comment gives me hope, but the proof of your comment is needed to resolve the issue.
$endgroup$
– möbius
Feb 22 '15 at 14:45

$begingroup$
Well, my notes say that the proof is in this book onlinelibrary.wiley.com/book/10.1002/9781118186428
$endgroup$
– Math1000
Feb 22 '15 at 16:23

$begingroup$
Actually, elementary computations show that the result holds if and only if $$frac1Tsum_{i=1}^Tigamma(i)to0,$$ a condition which is implied by the convergence of the series $$sum_{i}gamma(i).$$
$endgroup$
– Did
Feb 22 '15 at 16:32

$begingroup$
@Did Can you elaborate on why the convergence of your first series to zero is implied by the convergence of the second series? Even a hint would be helpful!
$endgroup$
– möbius
Feb 22 '15 at 18:31

1

$begingroup$
Rewrite the first sums in terms of the rests $Gamma(i)=sumlimits_{kgeqslant i}gamma(k)$ of the second series, which, by hypothesis, are finite and converge to zero.
$endgroup$
– Did
Feb 22 '15 at 19:01

|
show 2 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

The upper limit of your sum should be $T-1$ (since the $t=T$ term is zero). We can write

answered Feb 22 '15 at 14:14

Math1000

19.4k31746

$begingroup$
I have incorporated this minor change to the limit of the sum. However, the issue of the convergence of $a_{T}$ is at the crux of the proof; not only must $a_{T}$ converge, but it must converge to the desired limit above. Your comment gives me hope, but the proof of your comment is needed to resolve the issue.
$endgroup$
– möbius
Feb 22 '15 at 14:45

$begingroup$
Well, my notes say that the proof is in this book onlinelibrary.wiley.com/book/10.1002/9781118186428
$endgroup$
– Math1000
Feb 22 '15 at 16:23

$begingroup$
Actually, elementary computations show that the result holds if and only if $$frac1Tsum_{i=1}^Tigamma(i)to0,$$ a condition which is implied by the convergence of the series $$sum_{i}gamma(i).$$
$endgroup$
– Did
Feb 22 '15 at 16:32

$begingroup$
@Did Can you elaborate on why the convergence of your first series to zero is implied by the convergence of the second series? Even a hint would be helpful!
$endgroup$
– möbius
Feb 22 '15 at 18:31

1

$begingroup$
Rewrite the first sums in terms of the rests $Gamma(i)=sumlimits_{kgeqslant i}gamma(k)$ of the second series, which, by hypothesis, are finite and converge to zero.
$endgroup$
– Did
Feb 22 '15 at 19:01

|
show 2 more comments

The upper limit of your sum should be $T-1$ (since the $t=T$ term is zero). We can write

answered Feb 22 '15 at 14:14

Math1000

19.4k31746

$begingroup$
I have incorporated this minor change to the limit of the sum. However, the issue of the convergence of $a_{T}$ is at the crux of the proof; not only must $a_{T}$ converge, but it must converge to the desired limit above. Your comment gives me hope, but the proof of your comment is needed to resolve the issue.
$endgroup$
– möbius
Feb 22 '15 at 14:45

$begingroup$
Well, my notes say that the proof is in this book onlinelibrary.wiley.com/book/10.1002/9781118186428
$endgroup$
– Math1000
Feb 22 '15 at 16:23

$begingroup$
Actually, elementary computations show that the result holds if and only if $$frac1Tsum_{i=1}^Tigamma(i)to0,$$ a condition which is implied by the convergence of the series $$sum_{i}gamma(i).$$
$endgroup$
– Did
Feb 22 '15 at 16:32

$begingroup$
@Did Can you elaborate on why the convergence of your first series to zero is implied by the convergence of the second series? Even a hint would be helpful!
$endgroup$
– möbius
Feb 22 '15 at 18:31

1

$begingroup$
Rewrite the first sums in terms of the rests $Gamma(i)=sumlimits_{kgeqslant i}gamma(k)$ of the second series, which, by hypothesis, are finite and converge to zero.
$endgroup$
– Did
Feb 22 '15 at 19:01

|
show 2 more comments

The upper limit of your sum should be $T-1$ (since the $t=T$ term is zero). We can write

answered Feb 22 '15 at 14:14

Math1000

19.4k31746

The upper limit of your sum should be $T-1$ (since the $t=T$ term is zero). We can write

answered Feb 22 '15 at 14:14

Math1000

19.4k31746

answered Feb 22 '15 at 14:14

Math1000

19.4k31746

answered Feb 22 '15 at 14:14

Math1000

19.4k31746

answered Feb 22 '15 at 14:14

Math1000

19.4k31746

$begingroup$
I have incorporated this minor change to the limit of the sum. However, the issue of the convergence of $a_{T}$ is at the crux of the proof; not only must $a_{T}$ converge, but it must converge to the desired limit above. Your comment gives me hope, but the proof of your comment is needed to resolve the issue.
$endgroup$
– möbius
Feb 22 '15 at 14:45

$begingroup$
Well, my notes say that the proof is in this book onlinelibrary.wiley.com/book/10.1002/9781118186428
$endgroup$
– Math1000
Feb 22 '15 at 16:23

$begingroup$
Actually, elementary computations show that the result holds if and only if $$frac1Tsum_{i=1}^Tigamma(i)to0,$$ a condition which is implied by the convergence of the series $$sum_{i}gamma(i).$$
$endgroup$
– Did
Feb 22 '15 at 16:32

$begingroup$
@Did Can you elaborate on why the convergence of your first series to zero is implied by the convergence of the second series? Even a hint would be helpful!
$endgroup$
– möbius
Feb 22 '15 at 18:31

1

$begingroup$
Rewrite the first sums in terms of the rests $Gamma(i)=sumlimits_{kgeqslant i}gamma(k)$ of the second series, which, by hypothesis, are finite and converge to zero.
$endgroup$
– Did
Feb 22 '15 at 19:01

|
show 2 more comments

$begingroup$
I have incorporated this minor change to the limit of the sum. However, the issue of the convergence of $a_{T}$ is at the crux of the proof; not only must $a_{T}$ converge, but it must converge to the desired limit above. Your comment gives me hope, but the proof of your comment is needed to resolve the issue.
$endgroup$
– möbius
Feb 22 '15 at 14:45

$begingroup$
Well, my notes say that the proof is in this book onlinelibrary.wiley.com/book/10.1002/9781118186428
$endgroup$
– Math1000
Feb 22 '15 at 16:23

$begingroup$
Actually, elementary computations show that the result holds if and only if $$frac1Tsum_{i=1}^Tigamma(i)to0,$$ a condition which is implied by the convergence of the series $$sum_{i}gamma(i).$$
$endgroup$
– Did
Feb 22 '15 at 16:32

$begingroup$
@Did Can you elaborate on why the convergence of your first series to zero is implied by the convergence of the second series? Even a hint would be helpful!
$endgroup$
– möbius
Feb 22 '15 at 18:31

1

$begingroup$
Rewrite the first sums in terms of the rests $Gamma(i)=sumlimits_{kgeqslant i}gamma(k)$ of the second series, which, by hypothesis, are finite and converge to zero.
$endgroup$
– Did
Feb 22 '15 at 19:01

I have incorporated this minor change to the limit of the sum. However, the issue of the convergence of $a_{T}$ is at the crux of the proof; not only must $a_{T}$ converge, but it must converge to the desired limit above. Your comment gives me hope, but the proof of your comment is needed to resolve the issue.

– möbius
Feb 22 '15 at 14:45

Well, my notes say that the proof is in this book onlinelibrary.wiley.com/book/10.1002/9781118186428

– Math1000
Feb 22 '15 at 16:23

Actually, elementary computations show that the result holds if and only if $$frac1Tsum_{i=1}^Tigamma(i)to0,$$ a condition which is implied by the convergence of the series $$sum_{i}gamma(i).$$

– Did
Feb 22 '15 at 16:32

@Did Can you elaborate on why the convergence of your first series to zero is implied by the convergence of the second series? Even a hint would be helpful!

– möbius
Feb 22 '15 at 18:31

Rewrite the first sums in terms of the rests $Gamma(i)=sumlimits_{kgeqslant i}gamma(k)$ of the second series, which, by hypothesis, are finite and converge to zero.

– Did
Feb 22 '15 at 19:01

|
show 2 more comments

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