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Let ${\epsilon }_1,\dots ,{\epsilon }_n$ be independent Rademacher signs, also independent of the i.i.d. sample $Z_{1:n}$. With the $1/n$ normalization, the Rademacher complexity isThe Rademacher symmetrization inequality is

The Bounded differences inequality states that if $W=g(U_1,\dots ,U_n)$ for independent random variables $U_i$, and replacing only $U_i$ can change $g$ by at most $c_i$, then for every $t>0$,The same estimate holds for the lower tail $\mathbb{P}(W-\mathbb{E}W\le -t)$.

Because $S_1$ is Gaussian, the terminal wealth $W_1$ is normal. For constant absolute risk aversion utilitythe moment-generating function of a normal variable givesMaximizing this expected utility is equivalent to maximizingafter dropping a constant. The unique first-order condition isand strict concavity givesIt agrees with part (a) exactly whenFor the usual risk-averse convention $\gamma >0$, the asserted correspondence assumes $w_1>(1+r)w_0$; under that natural target-return condition the choice is unique.

The time-zero budget constraint isso ${\theta }^0=w_0-{\theta }^T{S}_0$. The terminal wealth is thereforePutThenand the target mean is the linear constraint ${\theta }^{T}b=c$.

Since the invertible covariance matrix $V$ is positive definite, the Lagrange multiplier equation for minimizing ${\theta }^{T}V\theta$ isThusThe constraint determinesThis is the one-period Gaussian minimum-variance portfolio; the riskless holding is then ${\theta }^0=w_0-{\theta }^T{S}_0$.

Suppose for contradiction that $f$ has no zero in the closed disc. Thenis a homotopy through loops in $\mathbb{C}\setminus \{0\}$. At $s=1$ it is $\alpha$, while at $s=0$ it is the constant loop with value $f(0)$. A constant loop has winding number zero, so homotopy invariance givescontrary to the hypothesis. Hence

DefineThe strict inequality $|s\sigma (t)|<|\gamma (t)|$ ensures that $H(s,t)\ne 0$ for every $(s,t)$, while $H(s,0)=H(s,1)$. Thus $H$ is a homotopy through closed loops in $\mathbb{C}\setminus \{0\}$ from $\gamma$ to $\gamma +\sigma$. By homotopy invariance of winding number,

Normalize the loop byChoose a continuous lift $\vartheta :[0,1]\to \mathbb{R}$ satisfyingBecause $\gamma (0)=\gamma (1)$, the difference $\vartheta (1)-\vartheta (0)$ is an integer independent of the chosen lift. The winding number of a continuous closed path is

If $\eta$ is a lift of $\delta /|\delta |$, then $\vartheta +\eta$ is a lift ofTherefore

For ${\lambda }_n=\alpha n+\beta$,so part (b) ensures nonexplosion. The Markov jump-process generator acts on the identity function $f(n)=n$ asConsequently $m(t)=\mathbb{E}N(t)$ satisfiesSolving this linear ordinary differential equation gives the mean of the linear birth process with immigration:

IfthenA nonnegative random variable with finite expectation is finite almost surely, so the process explodes.

Conversely, suppose the series diverges. For $s>0$, independence and the Laplace transform of an exponential variable giveThe sum in the exponent diverges. If infinitely many ${\lambda }_n\le s$, infinitely many terms are at least $\log 2$; otherwise, eventually $s/{\lambda }_n\le 1$ andThus $\mathbb{E}{e}^{-s{T}_{\infty }}=0$. Since $e^{-s{T}_{\infty }}$ is strictly positive on $\{T_{\infty }<\infty \}$, this forces $T_{\infty }=\infty$ almost surely. Hence the nonexplosion criterion for a pure birth process is

A pure birth process $N=(N(t):t\ge 0)$ is a continuous-time Markov chain on ${\mathbb{N}}_0$ whose only possible transition from state $n$ isEquivalently, after entering state $n$ it waits an independent exponential time $X_n\sim \mathrm{Exp}({\lambda }_n)$ and then moves to $n+1$.

The process is nonexplosive when it makes only finitely many jumps during every bounded time interval. If it starts from zero, this is equivalent to its explosion timebeing infinite almost surely.

Fix an integer $k>n$ and partition $[0,1]$, up to endpoints of measure zero, intoFor $p\ne q$, defineThese functions have integral zero and $L^2$ norm one, so they are admissible.

Letbe the interval occupancy counts. Since each $Z_i$ is uniform, $\mathbb{E}{N}_p=n/k$, and henceFor every outcome,because the nonnegative integer counts satisfy $N_p^2\ge N_p$ and ${\sum }_p{N}_p=n$. No independence assumption was used.

Averaging over all ordered pairs givesThe diagonal terms vanish, so some admissible pair $p\ne q$ satisfies the same lower bound. ThereforeLetting $k\to \infty$ proves the dependent uniform sampling variance lower bound

Put ${\mu }_f={\int }_0^{1}fd\lambda$. Subtracting this constant does not affect variance, and pairwise independent random variables have zero pairwise covariance. ThereforeOnly pairwise independence is needed for the off-diagonal covariance terms to vanish.

Let $U$ have the uniform distribution on $[0,1]$, and setThe variables have the same law. They are not independent random variables, since

The space $L^2(X,\mathcal{F},\nu )$ consists of equivalence classes, modulo equality almost everywhere, of measurable real functions satisfyingIt has inner product and induced normThe Cauchy-Schwarz inequality makes the inner product finite and verifies the norm axioms.

To prove completeness, let $(f_n)$ be Cauchy and choose a subsequence such thatBy the triangle inequality, the partial sumshave uniformly bounded $L^2$ norm. The monotone convergence theorem shows that their pointwise limit $g$ belongs to $L^2$ and is finite almost everywhere. The telescoping series therefore converges almost everywhere to a measurable function $f$, andThe original Cauchy sequence then converges to $f$ in $L^2$. Thus the result that an L2 space is a Hilbert space gives

LetAt $x\in F^{-1}(-1)$,This differential is nonzero because $x\ne 0$, so $-1$ is a regular value. The regular level set theorem makes $F^{-1}(-1)$ a three-dimensional embedded submanifold; intersecting it with the open set $\{x_0>0\}$ preserves that property. Therefore $X$ is the three-dimensional hyperboloid model, and

For an embedded submanifold $X\subseteq {\mathbb{R}}^N$ and $p\in X$, the tangent space isEquivalently, in a local parametrization $\phi :U\subseteq {\mathbb{R}}^d\to X$ with $\phi (u)=p$, one has $T_{p}X=\mathrm{im}D{\phi }_u$.