- #1
thedean515
- 11
- 0
We have two definitions for the autocovariance of finite samples $y\left(t\right)$and
it is given as
[tex]
\begin{equation}
\hat{r}\left(k\right)=\frac{1}{N-k}\sum_{t=K+1}^{N}y\left(t\right)y^{*}\left(t-k\right),\qquad0\le k\le N-1\end{equation}[/tex]
and
[tex]
\begin{equation}
\tilde{r}\left(k\right)=\frac{1}{N}\sum_{t=K+1}^{N}y\left(t\right)y^{*}\left(t-k\right),\qquad0\le k\le N-1\end{equation}
[/tex]
In addition we know that the autocovariance sequence for infinite
samples is
[tex]
\begin{equation}
r\left(k\right)=E\left\{ y\left(t\right)y^{*}\left(t-k\right)\right\} \end{equation}
[/tex]
where [tex]E\left\{ \cdot\right\}[/tex]is the expectation operator which
averages over the ensemble of realizations. Now I have been told that
[tex]\begin{equation}
E\left\{ \tilde{r}\left(k\right)\right\} =r\left(k\right)\end{equation}
[/tex]
and
[tex]
\begin{equation}
E\left\{ \hat{r}\left(k\right)\right\} =\frac{N-\left|k\right|}{N}r\left(k\right)\end{equation}
[/tex]
but they would be the other way round, can we proof it?
it is given as
[tex]
\begin{equation}
\hat{r}\left(k\right)=\frac{1}{N-k}\sum_{t=K+1}^{N}y\left(t\right)y^{*}\left(t-k\right),\qquad0\le k\le N-1\end{equation}[/tex]
and
[tex]
\begin{equation}
\tilde{r}\left(k\right)=\frac{1}{N}\sum_{t=K+1}^{N}y\left(t\right)y^{*}\left(t-k\right),\qquad0\le k\le N-1\end{equation}
[/tex]
In addition we know that the autocovariance sequence for infinite
samples is
[tex]
\begin{equation}
r\left(k\right)=E\left\{ y\left(t\right)y^{*}\left(t-k\right)\right\} \end{equation}
[/tex]
where [tex]E\left\{ \cdot\right\}[/tex]is the expectation operator which
averages over the ensemble of realizations. Now I have been told that
[tex]\begin{equation}
E\left\{ \tilde{r}\left(k\right)\right\} =r\left(k\right)\end{equation}
[/tex]
and
[tex]
\begin{equation}
E\left\{ \hat{r}\left(k\right)\right\} =\frac{N-\left|k\right|}{N}r\left(k\right)\end{equation}
[/tex]
but they would be the other way round, can we proof it?