Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 210 (Topics in Statistical Theory)

For a centered random variable $X$, being sub-Poisson in the right tail with variance parameter ${\sigma }^2$ meansIt is sub-Gamma in the right tail with variance parameter $v$ and scale parameter $c$ when

For $q\ge 2$, $q!\ge 23^{q-2}$. Hence, for $0\le \lambda <3$,Substitution into the defining moment-generating function bound shows that $X$ is sub-Gamma in the right tail with variance parameter ${\sigma }^2$ and scale parameter $1/3$.

Fix $0\le \lambda <c^{-1}$. Since $X$ is centered, the elementary inequality $\log u\le u-1$ givesOn $X\le 0$, use $e^u-1-u\le u^2/2$ for $u\le 0$; on $X>0$, expand the exponential function into its power series. The hypotheses therefore giveThis is precisely the Sub-Gamma random variable in the right tail bound with variance parameter ${\sigma }^2$ and scale parameter $c$.

A centered random variable $Y$ is sub-Gaussian with variance parameter ${\sigma }^2$ whenThe Chernoff bound, applied to $Y$ and $-Y$, yieldsThe tail integral formula for moments and the substitution $u=t^2/(2{\sigma }^2)$ now give

Write $s={\sigma }_Y{\sigma }_Z$ and $W=YZ-\mathbb{E}(YZ)$. The variables $Y$ and $Z$ need not be independent random variables. Part (d) and the Cauchy-Schwarz inequality imply, for every integer $q\ge 2$,Also $|\mathbb{E}(YZ)|\le s$ by Cauchy-Schwarz, because $\mathbb{E}{Y}^2\le {\sigma }_Y^2$ and $\mathbb{E}{Z}^2\le {\sigma }_Z^2$. Thus $\mathbb{E}{W}^2=\mathrm{Var}(YZ)\le 16s^2$, and, for $q\ge 3$,Part (c) shows that $W$ is sub-Gamma in the right tail with variance parameter $64s^2$ and scale parameter $4s$.

If $\mathbb{E}(YZ)\ge 0$, then $W_{+}\le (YZ{)}_{+}\le |YZ|$. Consequentlywhile $\mathbb{E}{W}^2\le 16s^2$. Another application of part (c) gives the improved variance parameter $16{\sigma }_Y^2{\sigma }_Z^2$ and scale parameter $2{\sigma }_Y{\sigma }_Z$.

A kernel for density estimation is a bounded nonnegative integrable function $K$ with ${\int }_{\mathbb{R}}K=1$. Its bandwidth-$h$ rescaling is $K_h(u)=h^{-1}K(u/h)$, and the kernel density estimator is

Fix $x$ and put $V_i=K_h(x-X_i)$. Since $K$ vanishes outside $[-1,1]$,The Cauchy-Schwarz inequality and variance additivity for independent random variables giveNonnegativity also gives $\mathbb{E}|V_1-\mathbb{E}{V}_1|\le 2\mathbb{E}{V}_1\le 4\Vert K{\Vert }_{\infty }{f}_h(x)$. Taking the better estimate at each $x$ and applying Tonelli theorem proves

For nonnegative $a,b$ and $0\le \theta \le 1$, $\min a,b\le a^{\theta }{b}^{1-\theta }$. Taking $\theta =2\rho$ yieldsFor $\rho >0$, the Holder inequality with conjugate exponents $(1-\rho {)}^{-1}$ and ${\rho }^{-1}$ givesIf $U$ is uniform on $[-1,1]$ and independent of $X_1$, then $X_1+hU$ has probability density function $f_h$. Sincethe last integral is at most $(1+h{)}^{\delta }\int (1+|x|{)}^{\delta }f(x)dx$. Substitution proves the second displayed bound. The case $\rho =0$ is the first bound integrated using $\int f_h=1$ and follows directly.

A linear estimator in nonparametric regression at $x_0$ has the form ${\hat{m}}_n(x_0)={\sum }_i{W}_i(x_0)Y_i$, where the weights may depend on the design, $x_0$, $p$, $h$, and $K$, but not on the responses. Putand define the local polynomial Gram matrixWhen ${\lambda }_0>0$, the weighted least squares normal equations have the unique solutionTherefore ${\hat{m}}_n(x_0)={\sum }_i{W}_i{Y}_i$, where the effective kernel weight is

The identityshows that these weights exactly reproduce at $x_0$ every multivariate polynomial of total degree at most $p$. This is the required polynomial reproduction property of local polynomial regression.

Only grid points with $\Vert x_i-x_0{\Vert }_{\infty }\le h$ have nonzero weight. There are at most $(2n_{1}h+1{)}^d\le (4n_{1}h{)}^d=4^{d}n{h}^d$ such points because $n_{1}h\ge 1/2$. On this support,so the operator norm bound $\Vert B_n^{-1}{\Vert }_{\mathrm{op}}={\lambda }_0^{-1}$ givesBecause the errors are independent random variables with the stated variance bounds,Thus ${\gamma }_1=d$.

Let $T_{x_0}$ be the Multivariate Taylor polynomial of $m$ at $x_0$ through total degree ${\beta }_0$. The assumed Hölder continuity of the derivatives andgive the Taylor remainder boundPolynomial reproduction cancels $T_{x_0}$ in the bias. The same support count and weight bound giveConsequentlyso ${\gamma }_2=\beta$.

The pushforward measure of $\mu$ under $g$ is $g_{\ast }\mu$, defined for $B\in \mathcal{B}$ byThe Lebesgue decomposition theorem says that if $\nu$ and $\mu$ are sigma-finite measures on the same measurable space, then uniquely

Let $f:[0,\infty )\to \mathbb{R}\cup +\infty$ be convex with $f(1)=0$. If $\lambda$ dominates the probability distributions $P,Q$, with densities $p,q$, their f-divergence isusing the lower-semicontinuous perspective convention where $q=0$. This definition is independent of the dominating measure.

The data processing inequality for f-divergences statesTo prove it, take $\lambda =P+Q$ and let $\mathcal{G}=g^{-1}(\mathcal{B})$. If $p=dP/d\lambda$ and $q=dQ/d\lambda$, then the pullbacks of the densities of $g_{\ast }P$ and $g_{\ast }Q$ with respect to $g_{\ast }\lambda$ are respectively ${\mathbb{E}}_{\lambda }[p\mid \mathcal{G}]$ and ${\mathbb{E}}_{\lambda }[q\mid \mathcal{G}]$. The perspective $F(a,b)=bf(a/b)$ of a convex function is jointly convex. Conditional Jensen inequality therefore givesIntegration proves the claim.

The Squared Hellinger distance isIf $P,Q$ have densities $p,q$ with respect to a sigma-finite measure $\mu$, this becomes

Fix any probability distribution $Q$ and set $p_j=P_j(A_j)$, $q_j=Q(A_j)$, and $t_j=H^2(P_j,Q)$. Applying the data processing inequality to the indicator function of $A_j$ gives the Bernoulli Hellinger boundThe hinted inequality impliesThe function $t\mapsto \sqrt{t(1-t/4)}$ is concave on $[0,2]$. Since the $A_j$ form a set partition, ${\sum }_j{q}_j=1$, and Jensen inequality givesTaking the infimum over $Q$ proves the stated inequality.