Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 94 exercise 4.1

Author: Panagiotis

1 Feb

In problem [1, p. 94 exercise 4.1] we will show that the periodogram is an inconsistent estimator by examining the estimator at $f=0$ , or

$\hat{P}_{PER}(0)=\frac{1}{N}\left(\sum\limits_{n=0}^{N-1}x[n]\right)^{2}.$

(1)

If $x[n]$ is real white Gaussian noise process with PSD

$P_{xx}(f)=\sigma_{x}^{2}$

(2)

we are asked to find the mean an variance of $hat{P}_{PER}(0)$ . We are asked if the variance converge to zero as $N \rightarrow \infty$ . The hint provided within the exercise is to note that

$\hat{P}_{PER}(0)= \sigma_{x}^{2} \left(\sum\limits_{n=0}^{N-1}\frac{x[n]}{\sigma_{x}\sqrt{N}}\right)^{2}$

(3)

where the quantiiy inside the parenthesis is $\sim N(0,1)$ Solution: Because $Y = \sum_{n=0}^{N-1}\frac{x[n]}{\sigma_{x}\sqrt{N}}$ is distributed according to a normal distribution $\sim N(0,1)$ the squared variable $Y^{2}=\frac{\hat{P}_{PER}(0)}{\sigma_{x}^{2}}$ is distributed according to a $\chi^{2}(1)$ with mean $E\left\{Y\right\}=1$ and variance $Var\left\{Y\right\}= 2$ . From the previous relations we obtain the mean and the variance of the periodogram $\hat{P}_{PER}(0)$ as

$E\left\{\hat{P}_{PER}(0)\right\} = \sigma_{x}^{2} E\left\{Y\right\} = \sigma_{x}^{2}$
$Var\left\{\hat{P}_{PER}(0)\right\} = \sigma_{x}^{2} Var\left\{Y\right\} = 2 \sigma_{x}^{2}.$