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The Jacobian matrix isAt $(0,0)$ its eigenvalues are $-k$ and $-1$. It is therefore a stable node for $k>0$, a saddle for $k<0$, and nonhyperbolic at $k=0$, consistent with the proposed transcritical bifurcation.

At $(0,1)$ the matrix iswith eigenvalues $1-k$ and $1$. It is an unstable node for $k<1$, a saddle for $k>1$, and nonhyperbolic at $k=1$, again consistent with a transcritical bifurcation.

At an equilibrium, each product inmust vanish. This gives the two parameter-independent equilibriathe equilibria on $y=0$and the equilibria with both coordinates nonzerowhere matching signs are used.

The $y=0$ pair is born at $k=-9/4$, suggesting a saddle-node bifurcation. At $k=0$, one of these equilibria meets $(0,0)$, suggesting a transcritical bifurcation, while the two equilibria near $(1,2)$ are simultaneously born in a saddle-node. At $k=1$, the minus branch $(1-\sqrt{k},2-\sqrt{k})$ meets $(0,1)$, suggesting another transcritical bifurcation. At $k=4$, that branch meets the $y=0$ equilibrium $(-1,0)$, again suggesting a transcritical bifurcation. Thus the bifurcation parameter values are

Since $y=x^{-1/2}u$, the two linearly independent solutions of the original equation have the asymptotic expansionsAs a check, the substitution $z=1/x$ converts the equation into the modified Bessel equation of order zero, whose solutions $I_0(z)$ and $K_0(z)$ have exactly these large-$z$ expansions.

Write the normal-form equation asFor sufficiently small positive $x$, $Q(x)>0$. Moreover,and the corresponding higher derivative condition also tends to zero, so the amplitude varies slowly relative to the exponential phase and the Liouville-Green approximation applies.

The phase and amplitude have expansionsHence two independent Liouville-Green solutions satisfyIn particular, after choosing the signs according to growth and decay,

The removal of the first derivative from a second-order differential equation usesIt givesso

After division by $x^2$, the equation becomesThe regular singular point criterion for a second-order equation would require $x(1/x)$ and $x^2(-1/x^4)$ to be analytic at zero. The second expression is $-x^{-2}$, so the criterion fails:

The stopping rule ensures that the returned function belongs toIndeed, when adding the next coefficient would make the sum exceed one, the algorithm returns the preceding sum, and at the first step it returns zero.

Since each $h_m$ takes values in $\{-1,1\}$, every $f\in \mathcal{H}$ satisfies $|f(x)|\le 1$. On $[-1,1]$, the map $z\mapsto e^{-z}$ is $\exp (1)$-Lipschitz. The expected Rademacher complexity generalization inequality followed by the Rademacher contraction lemma givesA linear functional attains the same supremum over a set and its convex hull. Since $B=-B$, adjoining zero and allowing total coefficient at most one does not increase the supremum, so the Rademacher complexity of a convex hull givesCombining the two bounds proves

The exponential loss is $\varphi (z)=e^{-z}$. For a score $f:\mathcal{X}\to \mathbb{R}$, its exponential classification risk and empirical risk areFor the set of evaluation vectorsits empirical Rademacher complexity iswhere the ${\sigma }_i$ are independent uniform random signs.

The convex hull of $S\subseteq {\mathbb{R}}^d$ isThus $v$ is a convex combination of $v_1,\dots ,v_m$ when it has precisely this form. Equivalently, $\mathrm{conv}S$ is the smallest convex set containing $S$. The Caratheodory theorem sharpens the finite-combination description in ${\mathbb{R}}^d$: every point of $\mathrm{conv}S$ is a convex combination of at most $d+1$ points of $S$.

The AdaBoost algorithm starts with weights $w_i^{(1)}=1/n$. For $m=1,\dots ,M$:

Because $h\in B$ implies $-h\in B$, the selected classifier can always have weighted error at most $1/2$, so ${\hat{\beta }}_m\ge 0$. The output score is

For the European call option, risk-neutral valuation givesDefine the stock-numeraire measure in a binomial market byIt is a probability measure because the risk-neutral martingale property gives ${\mathbb{E}}_Q{S}_N=S_0(1+r{)}^N$. Consequently

Tilting one up move by its stock factor changes its probability toSimilarly $1-\hat{q}=(1-q)(1+a)/(1+r)$. Thus under $\hat{Q}$ the number of up moves is binomial with parameter $\hat{q}$, and

The backward option pricing recursion iswith terminal value $V(N,s)=g(s)$. At a node with $S_{n-1}=s$, define the replicating stock holding byThis is a predictable process because it depends only on information available at time $n-1$.

Indeed, if $X_{n-1}=V(n-1,s)$, substituting the risk-neutral formula for $q$ into the self-financing update shows separately in the up and down states thatAn induction over $n$ therefore gives $X_N=V(N,S_N)=g(S_N)$. The required initial capital is

Under a risk-neutral measure, the discounted stock must be a martingale. If $q$ is the probability of the factor $1+b$, this condition issoThe hypothesis $a<r<b$ ensures $0<q<1$, making this equivalent to the original probability measure. Since the $N$ returns are independent under $Q$, the number of up moves has a binomial distribution, and hence

Between times $n-1$ and $n$, the investor holds ${\theta }_n$ shares worth ${\theta }_n{S}_{n-1}$ and places the remaining wealth $X_{n-1}-{\theta }_n{S}_{n-1}$ in the risk-free asset. The self-financing condition therefore gives

For an exponential proposal of rate $\lambda >0$,and the smallest valid rejection constant isbecause the exponent is maximized at $x=\lambda$. Differentiating its logarithm givesThe unique minimum occurs at $\lambda =1$. Since rejection efficiency is $1/M(\lambda )$, choosing any $\lambda \ne 1$ makes the algorithm less efficient:

The accepted $Y$ has the half-normal distribution, so $Y=|G|$ in distribution for $G\sim N(0,1)$. Multiplying it by an independent random sign that is positive and negative with equal probabilities restores the two symmetric halves of the Gaussian density. Thus

Each independent proposal is accepted with probability $1/M$. The number of proposals is therefore a geometric distribution with mean

Take the proposal density $h(x)=e^{-x}{1}_{x\ge 0}$. For the target half-normal distribution,The ratio is maximized at $x=1$, so the sharp envelope constant isGenerate $X=-\log U_1$ and an independent uniform $U_2$, and accept $X$ when

In one proposal of rejection sampling, the probability of both proposing a value in $dx$ and accepting it isIntegrating shows that the acceptance probability is $1/M$. Conditional on acceptance, the density of the proposed value is thereforeEach unsuccessful proposal restarts independently, so the eventual output $Y$ has the same conditional density:

Generate $U\sim \mathrm{Uniform}[0,1]$ and setFor $x\ge 0$,which is the distribution function of an exponential distribution with rate one. This is inverse transform sampling.