# Lecture 12 (Feb 18, 2022)

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.

Connection:

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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