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