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Let $h:A\to B$ be a homeomorphism, and let $f:B\to B$ be continuous. The conjugate mapis continuous. By hypothesis it has a fixed point $a\in A$. Thereforeand applying $h$ gives $f(h(a))=h(a)$. Hence every continuous self-map of $B$ has a fixed point. This is the homeomorphism invariance of the fixed-point property.

LetThe function $h$ is continuous, withThe intermediate value theorem gives $c\in [0,1]$ such that $h(c)=0$, and hence $f(c)=c$. This proves the fixed-point property of a closed interval.

Now suppose $0,1\in f([0,1])$. For any $y\in [0,1]$, choose $a,b\in [0,1]$ with $f(a)=0$ and $f(b)=1$. Applying the intermediate value theorem to $f$ on the interval with endpoints $a,b$ gives a point $x$ with $f(x)=y$. Thus $f$ is a surjective function.

For the American put option, compare exercise with continuation at time one. At $S_1=6$, both values are zero. At $S_1=4$, immediate exercise pays $1$, whereas the continuation value is the European value $4/9$, so exercise is optimal. At time zero, immediate exercise pays zero and continuation isThe optimal policy is therefore to continue at time zero, exercise at time one if $S_1=4$, and otherwise continue to maturity. This nodewise maximum is the Snell envelope value.

The European call option with strike $5$ pays $2,0,0$ at terminal stock prices $7,5,3$. Its lower-node value is zero, while its upper-node value isA second step of backward option pricing givesBy the fundamental theorem of asset pricing, this discounted expectation under the unique risk-neutral probabilities is the unique no-arbitrage price.

Now $R=1+r=9/8$. The local risk-neutral up probabilities solve $RS=q{S}_u+(1-q)S_d$. They areThe terminal payoff of the European put option with strike $5$ is $0,0,2$ at stock prices $7,5,3$. Backward option pricing gives time-one valuesand henceThe first-period stock holding in the replicating portfolio in a binomial market isThus the hedge initially shorts $2/9$ of a share.

The stock tree has $S_0=5$, first-period values $6,4$, and terminal successors $7,5$ from $6$ and $5,3$ from $4$. When $r=1/6$, the risk-free growth factor is $R=7/6$. At the upper time-one node, investing $6$ in the risk-free asset produces $7$ at time two, while one share bought for $6$ produces only $7$ or $5$.

Use the following predictable self-financing portfolio. Hold nothing initially. If $S_1=6$, short one share and invest the proceeds $6$ in the risk-free asset; if $S_1=4$, continue to hold nothing. Its terminal payoff ison the upper branch, and zero on the lower branch. It costs zero, is never negative, and is positive with positive probability, so it is an arbitrage.

Apply part (b) simultaneously at every rational $t$; the intersection of these probability-one events still has probability one. On that event, monotonicity and continuity extend the convergence from the rationals to every real $t$.

More directly, fix $\epsilon >0$. Since $F$ is a continuous cumulative distribution function, choose rational pointsso that the two tails have $F(t_0)<\epsilon$ and $1-F(t_k)<\epsilon$, while every increment $F(t_i)-F(t_{i-1})<\epsilon$. Part (b) makes $|F_m^{\ast }(t_i)-F(t_i)|<\epsilon$ at all these finitely many points for all sufficiently large $m$. If $t_{i-1}\le t\le t_i$, monotonicity givesso $|F_m^{\ast }(t)-F(t)|<2\epsilon$; the same estimate follows in the tails. This proves the uniform convergence of distribution functions to a continuous limit:almost surely.

Conditionally on $Y_1,\dots ,Y_m$, the cumulative distribution function of $X^{\ast }$ is the weighted empirical distributionFor fixed $t$, the strong law of large numbers gives, almost surely,whileThe ratio therefore converges almost surely to $F(t)$. This is precisely the claimed convergence in distribution of the self-normalized importance sampling resample.

Assume that $f(x)=0$ wherever $h(x)=0$ on the part of the space relevant to $g$. The importance sampling algorithm draws independent and identically distributed random variables $Y_1,\dots ,Y_m$ from $h$, forms the weightsand returnsA change of measure givesHence the estimator is unbiased whenever the expectation exists, and, if ${\mathbb{E}}_h|g(Y)w(Y)|<\infty$, the strong law of large numbers gives ${\hat{\mu }}_m\to {\mathbb{E}}_f[g(X)]$ almost surely.

For one visitor, letand defineA visitor arriving at time $s\le t$ is in room $j$ at time $t$ exactly when its independent service-time mark satisfies $T_{j-1}\le t-s<T_j$. Thus it receives mark $j$ with probability $p_j(t-s)$; a further mark records that it has already left the gallery.

The Independent marking theorem for Poisson point processes makes the point processes carrying the different room marks independent. Consequently $V_1(t),\dots ,V_{10}(t)$ are independent, and Poisson thinning givesEquivalently,

Let $N$ be the homogeneous Poisson point process in ${\mathbb{R}}^3$ with intensity measure $\lambda dx$, and map each point $x$ to its radius $g(x)=|x|$. For an interval $(a,b]\subseteq {\mathbb{R}}_{+}$, the pushforward measure of the intensity isThe Mapping theorem for Poisson point processes therefore shows that the radii form a Non-homogeneous Poisson point process with intensity function

If $R$ is the distance to the closest star, the event $\{R>r\}$ says that the ball of radius $r$ contains no points. The zero-count probability of a Poisson distribution givesDifferentiating its cumulative distribution function yields the probability density function

A Non-homogeneous Poisson point process with intensity function $\lambda$ is a random counting measure $N$ such that, for every finite family of pairwise disjoint bounded Borel sets $A_1,\dots ,A_k$, the counts $N(A_1),\dots ,N(A_k)$ are independent random variables andThe assumed local integrability makes each Poisson parameter finite. This count law and independence characterize the process.

Let $(Y_n)$ be independent and identically distributed random variables with the standard Cauchy distribution, whose probability density function isIts first absolute moment diverges sinceOn the other hand, its characteristic function is $\phi (t)=e^{-|t|}$. The characteristic function of a sum of independent variables therefore givesThus the stability of the Cauchy distribution implies that $S_n/n$ has the standard Cauchy distribution for every $n$. In particular,where $Z$ is standard Cauchy and $\stackrel{d}{\to }$ denotes convergence in distribution.

Write $S_n={\sum }_{j=1}^n{Y}_j$ for the partial sum. Fix an integer $M\ge 1$. The tail-sum formula for expectation, applied to the nonnegative random variable $|Y_1|$, givesbecause $\mathbb{E}|Y_1|=\infty$. The events $\{|Y_n|>Mn\}$ are independent, so the second Borel--Cantelli lemma shows that $|Y_n|/n>M$ infinitely often with probability one. Taking the countable intersection over $M\in \mathbb{N}$ yieldsalmost surely.

Now $Y_n=S_n-S_{n-1}$, and the triangle inequality givesIf the limit superior of $|S_n|/n$ were finite, the right-hand side would have finite limit superior, contradicting the preceding conclusion. Thereforealmost surely.

Again, no. On the unit round sphere and for a sufficiently small nonzero $\epsilon$, considerThis is a smooth embedded closed curve and may be reparametrized by arc length. It obeysso the antipodal isometry preserves the curve and exchanges the two complementary discs. Its restriction therefore gives an isometry $S_1\to S_2$.

The curve is not contained in any plane through the origin: its third-harmonic height cannot satisfy a nontrivial linear relation with $\cos \theta$ and $\sin \theta$. It is therefore not a great circle. Since the closed geodesics of the round sphere are precisely the great circles, $\gamma$ is not geodesic. This antipodally symmetric nongeodesic spherical curve shows that the point-transitivity assumption cannot be omitted.

No. On the round sphere, take a latitude circle other than the equator. Rotations about the polar axis preserve the circle and act transitively on its points, so all the isometries ${\varphi }_{x,y}$ exist. The latitude separates the sphere into two caps of unequal area, so those caps are not isometric. It is not a great circle and therefore is not a geodesic; equivalently, its geodesic curvature is nonzero. This is a homogeneous nongeodesic latitude.

Because $S_1$ and $S_2$ are isometric,The two induced orientations on their common boundary $\gamma$ are opposite. Ifis computed with the boundary orientation from $S_1$, then its value for $S_2$ is $-B$. Applying the Gauss-Bonnet theorem to the two halves givesThe equalities above imply $B=0$. This is the isometric halves give zero total boundary geodesic curvature argument.

Every ambient isometry ${\varphi }_{x,y}$ preserves the magnitude of geodesic curvature and sends $x$ to $y$. Hence $|k_g|$ is constant along the connected curve $\gamma$ by the transitive curve symmetry makes geodesic-curvature magnitude constant. If that constant were positive, continuity would force the signed $k_g$ to have one fixed sign, making its integral nonzero. Therefore $k_g\equiv 0$. A curve has zero geodesic curvature exactly when it is a geodesic, so $\gamma$ is geodesic.

For a positively oriented local parametrization $X(u,v)$ with first fundamental formthe area element of a surface isUnder an orientation-preserving coordinate change, the Jacobian from $du\wedge dv$ cancels the inverse Jacobian in the square root of the metric determinant. Hence these local expressions agree and define a global two-form.

The Euler characteristic may be defined from any finite triangulation byor equivalently by the alternating sum of the dimensions of the rational homology groups. Subdivision leaves $V-E+F$ unchanged, and the homological formula shows that it is a topological invariant, so the definition does not depend on the triangulation.

Give the boundary its induced orientation and parametrize it by arc length. If $T$ is its unit tangent and $N$ is the chosen unit normal to the surface, its signed geodesic curvature iswhere $D_s$ is the surface covariant derivative. The Gauss-Bonnet theorem for a compact oriented surface with smooth boundary and no corners iswhere $K$ is the Gaussian curvature.

Use the centered Hardy-Littlewood maximal functionIf $x>0$, choosing $0<r<x$ makes the whole interval lie in the positive half-line, so $M\theta (x)=1$. If $x=0$, every centered interval has exactly half its length in the positive half-line, so $M\theta (0)=1/2$.

If $x<0$, the average is zero for $r\le -x$. For $r>-x$, it isand these values tend to $1/2$ as $r\to \infty$. HenceFor the uncentered maximal-function convention, intervals extending arbitrarily far to the right make the supremum equal to $1$ at every $x$.

We prove the estimate first for $u$ in the Schwartz space. By the Fourier inversion theorem and the Cauchy-Schwarz inequality,The integral is finite because $n>n/2$.

For $0<\alpha \le 1$, the elementary boundgivesThe last integral is finite near zero for every $\alpha >0$ and at infinity exactly when $2\alpha <n$. Choose, for example, $\alpha =1/4$, which works for every $n\ge 1$. We obtainDensity of the Schwartz space in $H^n$ extends the estimate and supplies a unique Hölder-continuous representative. Thereforecontinuously. This is the Fourier proof of Hölder regularity from a Sobolev norm.