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Expand the payoff as $(S_T-2K+K^2/S_T)1_{\{S_T>K\}}$. The same risk-neutral measure applies. For any real $p$, completing the square gives the truncated lognormal momentUsing $p=1,0,-1$ and discounting by $e^{-rT}$ givesThe inverse-price moment explains both the extra volatility factor and the shifted normal argument in the last term.

In the Black-Scholes model, absence of arbitrage and completeness give the price as the discounted expectation under the risk-neutral measure. Under that measure,The physical drift $\mu$ therefore does not enter the option price. PutFor the standard normal distribution function $\Phi$, $\mathbb{P}(S_T>K)=\Phi (d_2)$. Completing the square in the Gaussian density also gives $\mathbb{E}[S_T{1}_{\{S_T>K\}}]=S_0{e}^{rT}\Phi (d_1)$. Expanding the call payoff into these two truncated moments yields

A standard Brownian motion has $B_0=0$, continuous paths almost surely, and independent increments with $B_t-B_s\sim N(0,t-s)$ for $0\le s<t$. With a specified filtration, the increment is also independent of the information available at time $s$.

Put $s=\sqrt{n}/2$. The posterior is $\mathrm{Beta}(X+s,n-X+s)$, andThe denominator is positive for $n\ge 1$. For $n>4$, both parameters exceed one and the mode simplifies to $(X+s-1)/(n+\sqrt{n}-2)$. For $n=4$ this is the uniform-prior mode, exactly $X/n$; for $n=1,2$ the endpoint or symmetric cases also make the mode rule coincide with $X/n$. Otherwise it is generally a different rule. The posterior mean agrees with $X/n$ only at $X=n/2$.

To find the minimax estimator under quadratic loss, consider the posterior mean ${\delta }_s=(X+s)/(n+2s)$. Its risk function isFor $s^2=n/4$, the numerator is identically $n/4$, soThis constant-risk binomial proportion estimator is the unique Bayes estimator for its proper beta prior under quadratic loss. A competing rule's maximum risk is at least its risk averaged over that prior, which is at least this Bayes risk. The constant-risk rule attains the bound, so the posterior mean in (iii) is the unique minimax rule. Consequently any coincident rule is minimax too: the mean in (i) is the same rule when $n=4$, and the mean in (ii) is the same rule when $n=1$. No posterior mode listed here equals that mean rule for every observation, since at $X=0$ its value is positive whereas their endpoint modes differ from it. This accounts for the small-$n$ coincidences as well as the general answer.

The Jeffreys prior is proportional to $[\theta (1-\theta ){]}^{-1/2}$, hence is $\mathrm{Beta}(1/2,1/2)$. Its posterior is $\mathrm{Beta}(X+1/2,n-X+1/2)$, givingThe interior case is used only when $n\ge 2$. Generally neither rule equals $X/n$: the mean agrees only at $X=n/2$; the mode agrees at the endpoints and, in the interior, at $X=n/2$. For the small experiments $n=1,2$, all possible observations happen to be among these cases, so the whole mode rule then coincides with the maximum-likelihood estimator.

The uniform prior has $s=1$, so the posterior is $\mathrm{Beta}(X+1,n-X+1)$. Its mean and mode areThus the uniform-prior posterior mode is the maximum-likelihood estimator, including $X=0,n$. The mean shrinks toward $1/2$ and equals the maximum-likelihood estimator only when $X=n/2$.

The random clock is understood to be independent of the discrete-time Markov chain; without this hypothesis its law need not be determined by $K$ alone. Conditional on $N_t=n$, the transition matrix is $K^n$, soHence $q_{xy}=\lambda K_{xy}$ for $x\ne y$ and $q_{xx}=-\lambda (1-K_{xx})$. If $\pi K=\pi$, then $\pi P_t=\pi$ by the series, proving that the invariant distribution is unchanged.

Let ${\xi }_i$ be the independent exponential interarrival times of the Poisson process. The absorption time is $T_a={\sum }_{i=1}^{{\tau }_a}{\xi }_i$. By independence of the chain and clock and Tonelli theorem,This includes infinite expectations and absorption at time zero. The same argument proves the hinted random-sum identity for an independent nonnegative integer-valued count $M$.

A rate-$\lambda$ Poisson process is a counting process starting at zero, with right-continuous nondecreasing integer-valued paths, independent stationary increments, and $N_t-N_s\sim \mathrm{Pois}(\lambda (t-s))$ for $0\le s<t$. Its jumps have size one almost surely.

For independent Poisson processes $N,M$, increments of $N+M$ on disjoint intervals are independent, because all the corresponding increments of the two processes are jointly independent. On an interval of length $t$, multiplying their probability generating functions givesThus the increment is Poisson with mean $(\lambda +\mu )t$. The path requirements hold, and independent processes almost surely have no simultaneous jumps. Therefore $N+M$ is a Poisson process of rate $\lambda +\mu$.

Hölder's inequality states that for conjugate exponents $1/p+1/q=1$, $f\in L^p$ and $g\in L^q$ imply $fg\in L^1$ andIt includes $p=1,q=\infty$ (and the reversed case), with the essential-supremum norm for $L^{\infty }$.

For $1\le p<\infty$, $L^p(E,\mathcal{E},\mu )$ consists of measurable functions with ${\int }_E|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^p{)}^{1/p}$. Thus membership is a condition on an almost-everywhere equivalence class, not on individual point values.

The equality is false. In $A=k[x,y]$, take $I=(y)$ and $J=(y-x^2)$. Both are prime ideals, since their quotient rings are isomorphic to $k[x]$, so both are radical. ButThe element $x$ belongs to the latter ideal and not the former. This also shows why taking the radical of an ideal can enlarge a sum of already radical ideals.

If a power of $a$ lies in $I\cap J$, then $a\in \sqrt{I}\cap \sqrt{J}$. Conversely, if $a^m\in I$ and $a^n\in J$, the power $a^{\max (m,n)}$ lies in both ideals. Therefore $\sqrt{I\cap J}=\sqrt{I}\cap \sqrt{J}$ for any commutative ring.

If $f\in \sqrt{I}$, a power $f^m$ vanishes at every point of $Z(I)$, so $f$ itself vanishes there. This gives $\sqrt{I}\subseteq I(Z(I))$.

For the reverse inclusion, take $f$ vanishing on $Z(I)$ and use the Rabinowitsch trick. In $k[x_1,\dots ,x_n,t]$ form $J=(I,1-tf)$. A common zero would have $x\in Z(I)$, whence $f(x)=0$ and $1-tf(x)=1$, a contradiction. The assumed Weak Hilbert Nullstellensatz gives $J=(1)$. Thus there is a finite expressionIf $f=0$ the conclusion is immediate. Otherwise substitute $t=f^{-1}$ in the localized polynomial ring. The last term vanishes, and clearing a sufficiently high power of $f$ gives $f^N\in I$. Hence $f\in \sqrt{I}$, proving the Strong Hilbert Nullstellensatz:

The radical of an ideal $I$ in a commutative ring $A$ is $\sqrt{I}=\{a\in A:a^m\in I\text{ for some }m\ge 1\}$. It is an ideal: if $a^m,b^n\in I$, every term of $(a+b{)}^{m+n}$ has either at least $m$ factors $a$ or at least $n$ factors $b$, so the sum belongs to $I$. Also $(ca{)}^m\in I$ for every $c\in A$, and additive inverses preserve the defining condition.

A surjective inverse-polynomial germ projection without even covering comes from the inverse of $P(w)=w^3-3w$ over $G=\mathbb{C}$. Consider $R=\mathbb{C}\setminus \{-1,1\}$, with $\pi (w)=P(w)$. Since $P^{\prime }(w)=3(w^2-1)\ne 0$ on $R$, each $w$ defines a holomorphic inverse germ of $P$ at $P(w)$. This identifies $R$ with a component of the space of germs of holomorphic functions: inverse germs continue along paths in $R$, and the holomorphic identity $P(f(z))=z$ persists under continuation. A germ with value $\pm 1$ cannot exist, because differentiating that identity would give $P^{\prime }(\pm 1)f^{\prime }(z)=1$. Since $R$ is connected, all its inverse germs belong to one complete analytic function.

The projection is surjective. A generic value has three distinct regular preimages; the critical values $2,-2$ each retain a simple regular preimage, respectively $2,-2$, even after their double critical preimages $-1,1$ are removed. Thus fibre size drops from $3$ to $1$ at these values, making an evenly covered neighbourhood impossible. This is surjective but not a regular topological covering.

If “regular” is specifically used for deck transitivity among genuine coverings, there is also a genuine nonnormal example. Remove the critical values from the base and all their preimages from the source:This is a connected three-sheeted analytic covering. A small loop about each critical value swaps the two roots meeting there. The two swaps are distinct and generate $S_3$ (connectedness also forces transitive monodromy). The stabilizer of one sheet is an order-$2$ subgroup, which is not normal in $S_3$. The classification of connected covering spaces therefore shows this covering is not regular in the deck-transitive sense. Either convention has a fully explicit example.

Take $G={\mathbb{C}}^{\ast }$ and the complete analytic function generated by a logarithm germ. Its continuation surface identifies with $\mathbb{C}$ via $w\mapsto [\log {]}_z$ with $z=e^w$ and logarithm value $w$. Thus $\pi (w)=e^w$. It is a universal cover; the deck transformations $w\mapsto w+2\pi in$ act transitively on each fibre, so it is also a regular covering map in the deck-transitive sense. It is not bijective, since every nonzero $z$ has infinitely many logarithm values. It also meets the alternative convention in which “regular” merely requires an evenly covered neighbourhood at each base point.