Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 60 exercise 3.4
Author: Panagiotis
28
Dez
In
[1, p. 60 exercise 3.4] we are asked to prove that the estimate
is an unbiased estimator, given
![\left\{x[0],x[1],...,x[N]\right\}](https://lysario.de/wp-content/cache/tex_1cc48dbb4a4db05229880d4eafd1756b.png)
are independent and identically distributed according to a

distribution. Furthermore we are asked to also find the variance of the estimator.
Solution:
The mean of the estimator

is given by
Thus the estimator

is unbiased. The variance of the estimator about the mean

is given by:
Considering the fact that the samples are independent we can set
![E\{x[n]x[k]\}=E\{x[n]\}E\{x[k]\}=\mu^2_x](https://lysario.de/wp-content/cache/tex_051237e59c44de24ff7741d385b52df1.png)
for

and obtain
Thus it remains to obain the value of
![E\{x^2[n]\}](https://lysario.de/wp-content/cache/tex_146af3daf0adb7dfe89991c6250647f2.png)
in order to determine the variance of the estimator. We can determine
![E\{x^2[n]\}](https://lysario.de/wp-content/cache/tex_146af3daf0adb7dfe89991c6250647f2.png)
by the variance

of
![x[n]](https://lysario.de/wp-content/cache/tex_d3baaa3204e2a03ef9528a7d631a4806.png)
:
And thus the variance of the estimator (
2), considering (
3) and (
4), is given by:
We have proved thus that the estimator is unbiased and have obtained the variance of the estimator

. QED.
[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.
2 Responses for "Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 60 exercise 3.4"
[...] Rao bound of the variance of the mean estimation is bounded by : (3) We had computed in [2, (5)] that the variance of the estimator of the mean is given by . Thus the sample mean is an [...]
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