Expected periodogram and smoothed spectral-density bias

Expand the finite squared modulus in $I_n(\lambda )$. By [F2] and [F3], $\mathbb{E}[(X_t-\mu )(X_s-\mu )]=\gamma (t-s)$, so $\mathbb{E}[I_n(\lambda )]=\frac{1}{2\pi n}{\sum }_{t=1}^n{\sum }_{s=1}^n\gamma (t-s)e^{-i(t-s)\lambda }\mathrm{.}$ All sums are finite and all terms integrable.

Since $K$ is continuous and $K(0)>0$, it is positive on some interval $[-r,r]$ with $r>0$. The nearest point of the Fourier grid is within $\pi \mathrm{/}n$ of ${\lambda }_0$; because $n{b}_n\to \mathrm{\infty }$, eventually $\pi \mathrm{/}(n{b}_n)<r$, so the denominator defining the weights is positive. Every positive weight is supported at a grid point within circular distance $R{b}_n$ of ${\lambda }_0$, where $[-R,R]$ contains the support of $K$; this distance tends to zero.

For each $h$ with $\mathrm{\mid }h\mathrm{\mid }<n$, exactly $n-\mathrm{\mid }h\mathrm{\mid }$ ordered pairs $(t,s)\in \{1,\dots ,n{\}}^2$ have $t-s=h$. Grouping the sum in step 1.1 by $h$ gives the asserted finite Fourier sum.

If a spectral measure $F$ is supplied, substitute $\gamma (t-s)={\int }_{\mathbb{T}}{e}^{i(t-s)\theta }F(d\theta )$ into the finite double sum and interchange the finite sum and integral. The resulting integrand is $\frac{1}{2\pi n}{\mid {\sum }_{t=1}^n{e}^{-it(\lambda -\theta )}\mid }^2={\mathrm{\Phi }}_n(\lambda -\theta ),$ which proves the Fejér convolution formula. When $F=fd\theta$, substitute that density in the same integral.

The kernel ${\mathrm{\Phi }}_n$ is nonnegative. Expanding its finite square and applying [F7] to each term gives ${\int }_{\mathbb{T}}{\mathrm{\Phi }}_n(x)dx=1$. When $F=fd\lambda$, steps 2.2 and [F4] also give $g_n(\omega ):=\mathbb{E}[I_n(\omega )]={\int }_{\mathbb{T}}{\mathrm{\Phi }}_n(\omega -\theta )f(\theta )d\theta$ as a normalized Fejér average of $f$.

For $0<\delta <\pi$, if $d_{\mathbb{T}}(x,0)\ge \delta$, choose its representative in $[-\pi ,\pi ]$, so $\delta \le \mathrm{\mid }x\mathrm{\mid }\le \pi$ by [F11]. Put $r=e^{-ix}$. Finite cancellation gives $(1-r){\sum }_{t=1}^n{r}^t=r(1-r^n)\mathrm{.}$ By [F9], $\mathrm{\mid }1-r\mathrm{\mid }=2\mathrm{\mid }\sin (x\mathrm{/}2)\mathrm{\mid }>0$ by [F10], and $\mathrm{\mid }1-r^n\mathrm{\mid }=2\mathrm{\mid }\sin (nx\mathrm{/}2)\mathrm{\mid }\le 2$. Therefore ${\mathrm{\Phi }}_n(x)=\frac{1}{2\pi n}{\mid \frac{\sin (nx\mathrm{/}2)}{\sin (x\mathrm{/}2)}\mid }^2\le \frac{1}{2\pi n{\sin }^2(\delta \mathrm{/}2)}\mathrm{.}$ Given $\epsilon >0$, continuity of $f$ at ${\lambda }_0$ gives a positive radius; shrink it so that $0<2\delta <\pi$ and $\mathrm{\mid }f(\theta )-f({\lambda }_0)\mathrm{\mid }<\epsilon$ when $d_{\mathbb{T}}(\theta ,{\lambda }_0)<2\delta$. If $d_{\mathbb{T}}(\omega ,{\lambda }_0)<\delta$, split the integral for $g_n(\omega )-f({\lambda }_0)$ into that neighborhood and its complement. The first part has modulus at most $\epsilon$ because ${\mathrm{\Phi }}_n$ has integral one; on the complement $d_{\mathbb{T}}(\omega ,\theta )\ge \delta$: choose minimizing integers $k_1,k_2$ for $d_{\mathbb{T}}(\theta ,\omega )$ and $d_{\mathbb{T}}(\omega ,{\lambda }_0)$. Then $d_{\mathbb{T}}(\theta ,{\lambda }_0)\le \mathrm{\mid }\theta -{\lambda }_0+2\pi (k_1+k_2)\mathrm{\mid }\le d_{\mathbb{T}}(\theta ,\omega )+d_{\mathbb{T}}(\omega ,{\lambda }_0),$ by [F11] and the real triangle inequality. Since the first term is at least $2\delta$ and the last term is less than $\delta$, the displayed kernel bound makes the second part at most $\frac{\mathrm{\parallel }f{\mathrm{\parallel }}_1+2\pi \mathrm{\mid }f({\lambda }_0)\mathrm{\mid }}{2\pi n{\sin }^2(\delta \mathrm{/}2)},$ which tends to zero uniformly for all such $\omega$. Hence $g_n(\omega )$ tends uniformly to $f({\lambda }_0)$ as $\omega$ ranges over any neighborhoods of ${\lambda }_0$ whose radii tend to zero.

For all sufficiently large $n$, the weights are nonnegative and sum to one, and every grid point receiving positive weight lies in a shrinking neighborhood of ${\lambda }_0$. Finite-sum linearity of expectation [F8] therefore gives $\mathbb{E}[{\tilde{f}}_n({\lambda }_0)]={\sum }_{j=0}^{n-1}{w}_{n,j}({\lambda }_0)g_n({\lambda }_{j,n})⟶f({\lambda }_0)$ by the uniform local convergence in step 4.1.