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 -


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


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?

  • $\begingroup$ 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! $\endgroup$ – Marcus Müller Nov 4 '17 at 10:03
  • $\begingroup$ I found these equations under Barlette's method and yes, it's a consistent estimate and the realizations of $x_i(n)$ are uncorrelated. $\endgroup$ – Sadist_Tanmoy Nov 4 '17 at 10:12
  • $\begingroup$ I think, x(n) is WSS. $\endgroup$ – Sadist_Tanmoy Nov 4 '17 at 10:19

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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.