The periodogram $\widehat{P_x}(e^{jω})$ is an estimate of PSD $P_x(e^{jω})$ and for mean-square convergence -

$\lim_{N \rightarrow \infty}E\left\{\widehat{P_x}(e^{jω})\right\}=P_x(e^{jω})$

My question:

If I calculate the expected value, which is according to definition an average, this way

$E\left\{\widehat{P_x}(e^{jω})\right\} = \frac{1}{2π}\int_{<2π>} \widehat{P_x}(e^{jω})dω$

Then, $E\left\{\widehat{P_x}(e^{jω})\right\}$ should have a single value. If this was true, then it would mean that $P_x(e^{jω})$ has the same value for any $ω$ and this clearly is not true. So, where am I mistaking?



The mistake is here:

If I calculate the expected value, which is according to definition an average, this way

$E\left\{\widehat{P_x}(e^{jω})\right\} = \frac{1}{2π}\int_{<2π>} > \widehat{P_x}(e^{jω})dω$

The mistake is in your interpretation of expectation. It is not an average over all frequencies.

For each fixed $\omega$, the expectation is taken over many realizations of the random signal $x$.

Gory details

Suppose you are given a complex valued random signal $x(n)$ for $0\leq n\leq N-1$. I would emphasize that this is a random signal. If you repeat your experiment you'll obtain a different set of $N$ values each time (called "realizations" of $x$).

Recall how the periodogram $\hat P_x(e^{j\omega})$ is computed. You first compute the sample autocorrelation at each lag $m$: $$ \hat r_x(m) = \frac{1}{N-m}\sum_{n=0}^{N-m-1}x(n)x^*(n+m) \;\;\;\;\;\;\;\;\;\;\;\;(1) $$ and then get the periodogram estimate by taking the DTFT: $$ \hat P_x(e^{j\omega}) = \sum_{m=-(N-1)}^{N-1} \hat r_x(m) e^{j \omega m}. $$ Since we started with a random $x$ we would like to know how this estimate performs on average i.e. average over the randomness in $x$:. So we take the expectation: \begin{eqnarray} \hat P_x(e^{j\omega}) &=& \mathbf E\left[\sum_{m=-(N-1)}^{N-1} \hat r_x(m) e^{j \omega m}\right]\\ &=& \sum_{m=-(N-1)}^{N-1} \mathbf E[ \hat r_x(m)] e^{j \omega m}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)\\ &=& \sum_{m=-(N-1)}^{N-1} \mathbf E\left[ \frac{1}{N-m}\sum_{n=0}^{N-m-1}x(n)x^*(n+m) \right ] e^{j \omega m} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(3)\\ &=& \sum_{m=-(N-1)}^{N-1} \frac{1}{N-m}\sum_{n=0}^{N-m-1}\underbrace{\mathbf E[x(n)x^*(n+m)]}_{\mbox{autocorrelation function of }x} \; e^{j \omega m}\\ &=& \sum_{m=-(N-1)}^{N-1} r_x(m) e^{j\omega m} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(4) \\ &=& P_x(e^{j\omega}) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(5) \end{eqnarray}

where (3) follows by substituting (1) in (2), (4) is by definition of autocorrelation $r_x$ of the random sequence $x$ and (5) is by definition of $P_x$.

Once again, note that the expectation was taken to "average out" the randomness introduced by $x$. It is not an average over frequencies because the frequencies are not random here.


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