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Let $N=T-t$ and use the discounted multipliers $u,d$. Under $Q$, the number $K$ of up moves in the remaining $N$ periods has the binomial distribution $\mathrm{Binomial}(N,q)$. Sincethe conditional expectation in part (b) becomesFor an undiscounted stock, replace $u,d$ by $U,D$ and discount the resulting payoff by the appropriate power of $R$.

The arbitrage-free value of the discounted claim is the conditional expectationConditional on the observed path $(S_0^1,\dots ,S_t^1)=(x_0,\dots ,x_t)$, the future multiplicative increments are independent of ${\mathcal{F}}_t$ and have the same law as a fresh stock path divided by its initial value. ThereforewhereThis is Risk-neutral valuation in the Cox--Ross--Rubinstein model for a path-dependent claim.

In the one-period building block of the Cox--Ross--Rubinstein model, the money-market account grows by a deterministic factor $R>0$, while the stock grows independently at each time byThus $S_t^0=R^t{S}_0^0$ and $S_t^1=S_0^1{\prod }_{j=1}^t{Y}_j$. Equivalently, the discounted stock has multipliers $u=U/R$ and $d=D/R$.

The model has no arbitrage precisely when the risk-free return lies strictly between the stock returns:Indeed, failure of either strict inequality gives a one-period arbitrage, while under the strict inequalities the probabilitylies in $(0,1)$. Taking the moves independent under $Q$ with up probability $q$ makes the discounted stock a martingale, since $qu+(1-q)d=1$. This is the unique Risk-neutral probability in the Cox--Ross--Rubinstein model, hence the unique equivalent martingale measure.

For observations $x=(x_1,\dots ,x_n)$, the likelihood function isA maximum-likelihood estimator is any measurable choiceFor one observation in a regular scalar model, the score function isand the Fisher information iswhere the second identity requires the usual differentiation-under-the-integral regularity. For $n$ independent observations, Tensorization of Fisher information gives $I_n(\theta )=nI(\theta )$.

Here $\Lambda$ is planar Lebesgue measure. For a Borel set $B\subseteq [0,\infty )$, polar coordinates giveThe pushforward is therefore locally finite and atomless. By the mapping theorem for Poisson point processes, $f(\Pi )$ is a homogeneous Poisson process on $[0,\infty )$ with constant intensity $\pi$. This is the two-dimensional radial volume transform of a homogeneous Poisson point process.

Put $S_k=R_k^2$. The variable $S_k$ is the $k$th arrival time of that rate-$\pi$ Poisson process, hence has the gamma distribution with shape $k$ and rate $\pi$:Since $s=r^2$ and $ds/dr=2r$, the change-of-variables formula for a probability density gives the Kth-nearest-neighbour distance in a homogeneous Poisson point process

The mapping theorem for Poisson point processes applies provided $f$ is a measurable function and its pushforward measureis locally finite, meaningfor every bounded Borel set $B\subseteq {\mathbb{R}}^s$. If “Poisson process” means a simple point process, one also requires $f_{\ast }\Lambda$ to be atomless, equivalentlyUnder these conditions, $f(\Pi )$ is a Poisson point process with mean measure

For every Borel set $A\subseteq {\mathbb{R}}^d$, letbe its random point count. A Non-homogeneous Poisson point process with intensity function of a point process $\lambda$ is characterized by:

Because $\Lambda$ has a density with respect to Lebesgue measure, it is atomless, so the process is simple: almost surely no location occurs more than once.

No. Fix $0<A<1$ and, near $s=0$, letThen $f^{\prime 2}+g^{\prime 2}=1$, so this is a regular unit-speed profile with $f>0$ on a sufficiently small interval. Rotate it around the $z$-axis. For an arc-length surface of revolution, $K=-f^{\prime \prime }/f$, and thereforeAt $s=0$, the formulas from part (a) give the principal curvaturesThey are unequal because $A\ne 1$, so the point is not an umbilical point. Every point of a sphere is umbilical. Consequently no neighbourhood of this point can lie on a sphere, giving a non-spherical surface of constant Gaussian curvature one.

A regular surface is a minimal surface when its mean curvature vanishes identically, equivalently when its two principal curvatures sum to zero everywhere.

The catenoidis an example. In part (a), take $f(v)=\cosh v$ and $g(v)=v$, so $q=\cosh v$. The formula there givesA plane and a circular cylinder have zero Gaussian curvature, as does a flat cone away from its vertex. Because a local isometry preserves Gaussian curvature, the catenoid is minimal and is not locally isometric to any of those surfaces; this is the catenoid is a minimal surface.

WriteThe regularity of the profile makes $q>0$. The coordinate tangent vectors areso the first fundamental form of a surface of revolution has coefficientsChoose the unit normalTaking scalar products with the second derivatives of $\varphi$ gives the second fundamental form coefficientsThus the coordinate directions are principal directions, with principal curvaturesThe curvatures of a parametrized surface of revolution are thereforeReversing the orientation of a surface reverses the sign of $H$ but leaves $K$ unchanged.