Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 209 / 2

The empirical characteristic function replaces the expected value in a characteristic function by the sample average:Its centered fluctuation at the central limit theorem scale is the empirical characteristic process:In particular, ${\phi }_n(0)=1$ and ${\mathcal{C}}_n(0)=0$.

For real $v$, $\overline{\phi (v)}=\phi (-v)$. Expand the complex covariance of the sample averages. Terms indexed by distinct observations vanish because the observations are independent; each diagonal term equals $\mathbb{E}{e}^{i(u-v)X_1}-\phi (u)\overline{\phi (v)}$. HenceTaking $v=u$ and using $\phi (0)=1$ gives the exact variance:The upper bound follows from $|\phi (u)|\le \mathbb{E}|e^{iu{X}_1}|=1$, and holds without any moment assumptions on $X_1$.

Fix $u$. Put $m=\phi (u)$ and form the centered real vectorsThey are bounded independent and identically distributed random variables, so the multivariate central limit theorem applies to $n^{-1/2}{\sum }_j{Y}_j=(\mathrm{Re}{\mathcal{C}}_n(u),\mathrm{Im}{\mathcal{C}}_n(u){)}^T$.

To identify its limiting covariance matrix, put $Z=e^{iu{X}_1}-m$. Its complex covariance and pseudo-covariance are respectivelyFor $Z=A+iB$, $|Z{|}^2=A^2+B^2$ and $Z^2=A^2-B^2+2iAB$. ThusThe two conditions on $\Gamma (u)$ identify exactly this real covariance matrix: the second is ${\mathrm{Cov}}_{\mathbb{C}}(\Gamma (u),\overline{\Gamma (u)})=\mathbb{E}\Gamma (u{)}^2$, with a conjugate in its second argument. A centered multivariate normal distribution is determined by its covariance matrix, including when that matrix is singular. ThereforeThis proves convergence in distribution at each fixed $u$; it does not assert convergence of the entire empirical characteristic process.

Write $J_{n,h}$ for the squared integral norm in question. The scaling property of the Fourier transform gives $\mathcal{F}{K}_h(u)=\mathcal{F}K(hu)$, which vanishes outside $[-1/h,1/h]$. Also $|\mathcal{F}K(v)|\le \Vert K{\Vert }_1$. The condition $\int K=1$ does not imply $\Vert K{\Vert }_1=1$, because $K$ need not be nonnegative.

Part (b), now for observations of $Y$, gives $\mathbb{E}|{\phi }_n^Y(u)-{\phi }^Y(u){|}^2\le 1/n$. Apply the Tonelli theorem to this nonnegative integrand, use the bound on $1/{\phi }_{ϵ}$, and change variables $v=hu$:The characteristic function ${\phi }_{ϵ}$ is continuous and nonzero on the compact interval of integration, so ${\tilde{M}}_h<\infty$ for every $h>0$. The Markov inequality gives, for every $z>0$,Consequently $J_{n,h}^{1/2}=O_{\mathbb{P}}({\tilde{M}}_h/\sqrt{nh})$, including for any deterministic bandwidth sequence $h=h_n>0$. The stochastic order follows directly from this uniform probability bound. Independence of $X$ and $ϵ$ gives ${\phi }^Y={\phi }^X{\phi }_{ϵ}$, explaining why division by ${\phi }_{ϵ}$ is the Fourier deconvolution operation.