.

First fix K=1 in (1). Then w₀=1 and ${{\bf{v}}}^{(0)} = (1/\sqrt{2 \pi T}, ... , 1/\sqrt{2 \pi T})$ corresponds to the familiar periodogram. The periodogram is inconsistent and generally a biased estimate of the true spectral density. It is inconsistent because its variance does not approach zero as the number of observations increases. The remedy is to ``smooth'' it, which is equivalent to multiplying the data by a taper before computing the periodogram. Because of the side-lobes of the Fej$\rm{\acute{e}}$r kernel power is transferred among different regions of the spectrum (leakage) and cause the periodogram to be biased. The formula below illustrates the point:

$$\begin{eqnarray} E \{ \hat {S(f)} \} = {(2 \pi T)}^{-1}\int_{-1/2}^{1/2} F(f-\mu) S(\mu) d\mu, \end{eqnarray}$$

where $F(\lambda) = {(2 \pi T)}^{-1} \frac{\sin^2 (N \lambda/2)}{ \sin^2(\lambda/2 )}$ is the Fej$\rm{\acute{e}}$r kernel.