Periodogram Averaging

Question

FOR CONSISTENT ESTIMATION OF PSD

Let, $x_i(n)$ be a sequence of random variables for $i=1,2, ... , K$ uncorrelated realizations and $\widehat{P}_{(i)}^{per}(e^{jω})$ is their corresponding periodograms. Then, the average of these periodograms is -

$$\widehat{P}_x(e^{jω})=\frac{1}{K}\sum_{i=1}^K\widehat{P}_{(i)}^{per}(e^{jω})$$

Let, $x_i(n)$ be a WSS process also.

Question:

Are the following relations correct?

$E\left\{\widehat{P}_x(e^{jω})\right\}=E\left\{\widehat{P}_{(i)}^{per}(e^{jω})\right\}$ and $Var\left\{\widehat{P}_x(e^{jω})\right\}=\frac{1}{K}Var\left\{\widehat{P}_{(i)}^{per}(e^{jω})\right\}$

I found these two equations in my textbook (Statistical Digital Signal Processing and Modeling, Monson H. Hayes, Page 412).

I am failing to understand these relations (meaning I don't get how you derive them). Can someone please explain them for me?

That depends – is $x$ stationary? Is your periodogram (which is a PSD estimate) calculated using a consistent estimator? With the data you're giving, we can't say! — Marcus Müller, Nov 04 '17 at 10:03
I found these equations under Barlette's method and yes, it's a consistent estimate and the realizations of $x_i(n)$ are uncorrelated. — Sadist_Tanmoy, Nov 04 '17 at 10:12

score 1 · Accepted Answer · answered Nov 04 '17 at 18:26

Since the data records $x_i$ are uncorrelated realizations of the same random process, $\hat P^{per}_{(i)}(e^{j\omega})$ are all uncorrelated random variables with identical means and variances (given by the mean and variance of the Bartlett Method's PSD estimator). So, for any fixed $\omega$, $\mathbf E[\hat P^{per}_{(i)}(e^{j\omega})]$ are equal $\forall i$ and also $\mathrm{Var}[\hat P^{per}_{(i)}(e^{j\omega})]$ are equal $\forall i$. So, \begin{eqnarray} \mathbf E[\hat P_{x}(e^{j\omega})] &=& \mathbf E\left[\frac{1}{K} \sum_{i=1}^K\hat P^{per}_{(i)}(e^{j\omega})\right] \\ &=& \frac{1}{K}\sum_{i=1}^K \mathbf E\left[\hat P^{per}_{(i)}(e^{j\omega})\right] \\ &=& \frac{1}{K} K \;\mathbf E\left[\hat P^{per}_{(i)}(e^{j\omega})\right]\\ &=& \mathbf E\left[\hat P^{per}_{(i)}(e^{j\omega})\right] \end{eqnarray}

and similarly,

\begin{eqnarray} \mathrm{Var}[\hat P_{x}(e^{j\omega})] &=& \mathrm{Var}\left[\frac{1}{K} \sum_{i=1}^K\hat P^{per}_{(i)}(e^{j\omega})\right] \\ &=& \frac{1}{K^2}\sum_{i=1}^K \mathrm{Var}\left[\hat P^{per}_{(i)}(e^{j\omega})\right] \\ &=& \frac{1}{K^2} K \;\mathrm{Var}\left[\hat P^{per}_{(i)}(e^{j\omega})\right]\\ &=& \frac{1}{K}\mathrm{Var}\left[\hat P^{per}_{(i)}(e^{j\omega})\right]. \end{eqnarray}

Periodogram Averaging

1 Answers1