There are some restrictons of the parametric estimation. Obviously, when the series does not follow the assumed model, the estimation would be off.

Nonparametric estimation

Our goal is to estimate spectram $s(\omega)$, which is the Fourier transform of $\gamma_j$.

Idea: use smoothness of $s(\omega)$: continuity: $s(\lambda) \rightarrow s(\omega)$ as $\lambda \rightarrow \omega$.

Local average with weights: (here the weights depend on the distance between $\omega$ and $\lambda$. )

Let $\omega_j = 2\pi j/T$. Kernel estimation for $s(\omega_j)$ : \(\tilde{S}(\omega_j) = \sum_{m=-h}^h K(\omega_{j+m}, \omega_j ) \hat{s}(\omega_{j+m})\) here the $h$ is the bandwidth. $2\hat{s}(\omega)/s(\omega) \rightarrow \chi^2_2$ and $\hat s(\omega)$ and $\hat s(\lambda)$ are asymptoticaly independent. Here we would let $h\rightarrow \infty$ as $T\rightarrow \infty$ and thus, we have a vanishing variance. Thus, we trade a small bias with a large reduce of variance. If $s(\omega)$ is less smoother, the average introduces a lot of bias. For rough $s(\omega)$, we need smaller $h$. We need some normalizatoin restriction on $k$ as \(\sum_{m=-h}^h K(\omega_{j+m}, \omega_j) = 1\) A simple example is \(K(\omega_{j+m}m \omega_j) = \frac{h+1 - |m| }{c(h)}\) The normalization $c(h) = (h+1)^2$. (linear decay with distance $m$)

If $h\rightarrow\infty$, the variance of $\tilde S$ vanishes. If $h=o(T)$, the bias is controlled.

Another look at the kernel method. We do this by using the continuous spectrum although this is not the case in reall application. \(\tilde{s}(\omega) = \frac{1}{2\nu}\int_{\omega-\nu}^{\omega+\nu} \hat{s}(\lambda)d\lambda\) where we used a uniform kernel with band width $\nu$. Recall that \(\begin{aligned} \hat{s}(\lambda) &= \frac{1}{2\pi} \sum_{j=-T+1}^{T-1} \hat{\gamma}_j e^{-i\lambda j} \\ &=\frac{1}{2\pi} \left[\hat\gamma_0+2\sum_{j=1}^{T-1} \hat\gamma_j \cos(\lambda j)\right] \end{aligned}\) thus, \(\begin{aligned} \tilde S(\omega) &= \frac{1}{4\pi \nu} \int_{\omega-\nu}^{\omega+\nu} \left(\hat\gamma_0 ++2\sum_{j=1}^{T-1} \hat\gamma_j \cos(\lambda j) \right) d\lambda\\ &=\frac{1}{4\pi\nu} \left[\hat\gamma_0(2\nu) + 2\sum_{j=1}^{T-1} \hat\gamma_j \int_{\omega-\nu}^{\omega+\nu}\cos(\lambda j) d\lambda \right] \\ &= \frac{1}{2\pi}\hat\gamma_0 + \frac{1}{2\pi \nu} \sum_{j=1}^{T-1} \frac{\hat\gamma_j}{j}\left[\sin((\omega+\nu) j) - \sin((\omega-\nu) j)\right]\\ &= \frac{1}{2\pi}\left[\hat\gamma_0 + 2\sum_{j=1}^{T-1} \frac{\sin(\nu j)}{\nu j} \hat\gamma_j\cos(\omega j)\right] \end{aligned}\) The periodogram is \(\hat S(\omega) = \frac{1}{2\pi}\left[\hat\gamma_0 + 2\sum_{j=1}^{T-1} \hat\gamma_j\cos(\omega j)\right]\) Thus, the kernel method is a weighted average with weights $\sin(\nu j)/\nu j$ for $\hat\gamma_j$ for lag $j$.

In the limiting case that $\nu\rightarrow 0$, then $\sin(\nu j)/\nu j \rightarrow 1$ and this reduces to the simple periodogram.


kernel $K(\omega_{j+m},\omega_j)$ becomes weights for $\hat\gamma_j$ and we call this weight $\kappa_j^*$.

The Bartlett kernel is \(\kappa_j^* = \left\{ \begin{aligned} &1 - \frac{j}{q+1} , \quad j= 1,\dots,q\\ &0, \quad j>q \end{aligned}\right.\) This shows \(\tilde{S}(\omega) = \frac{1}{2\pi}\left[\hat\gamma_0 + 2 \sum_{j=1}^q \left(1-\frac{j}{1+q} \right)\gamma_j \cos(\omega j)\right]\) Summary: bias increases as $q$ become larger and variance decreasess. The bias depends on the smoothness of $S(\omega)$.

i.e. more smoothness $\Rightarrow$ can have larger h or q​. less smoothness $\Rightarrow$ need to have small $h$ or $q$.

More on Kernel regression :

If we have sample $x_1,\dots x_T$ and $f(x)$. Then we want to esimate $\hat{f}(x)$ by \(\hat f (x) = \arg\min_{c} \sum_{t=-h}^h K(x,x_t) (c-f(x))^2\) Take derivative and get \(\hat f(x) = \frac{1}{\sum_{t=-h}^h K(x,x_t)} \sum_{t=-h}^h K(x,x_t)f(x_t)\) we can replace $c$ by some function.


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