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An equivalent martingale measure is a probability measure $\mathbb{Q}$ on the terminal sigma-algebra such that $\mathbb{Q}$ and the physical measure $\mathbb{P}$ are mutually absolutely continuous and every asset price discounted by the chosen numéraire is a $\mathbb{Q}$-martingale with respect to the market filtration.

Under the null model, $\pi =1$, which is a boundary point of the parameter space. Standard interior-point asymptotic normality of the maximum-likelihood estimator therefore need not apply.

Indeed, put $a=1-e^{-\lambda }$. Under $\pi =1$, $N_{+}\sim \mathrm{Binomial}(n,a)$. For the unconstrained estimator $\tilde{\pi }=N_{+}/(na)$, the central limit theorem givesSince $\hat{\pi }=\min \{1,\tilde{\pi }\}$, the continuous mapping theorem yieldsThis distribution has probability $1/2$ at zero and a continuous normal half-density on the negative half-line, so it is not normal.

Let $N_{+}=\#\{i:Y_i>0\}$, $N_0=n-N_{+}$, and $a=1-e^{-\lambda }$. Ignoring factors independent of $\pi$, the likelihood isThe unconstrained critical point satisfiesso, after enforcing $0\le \pi \le 1$,

The mass function is the mixtureso it is nonnegative and sums to one. Its first two moments areThereforeFor $0<\pi <1$ this is strictly larger than the mean, so the Zero-inflated Poisson distribution models overdispersion. At $\pi =1$ it reduces to an ordinary Poisson distribution, and at $\pi =0$ it is degenerate at zero.

For $X_1\sim \mathrm{Poisson}(\theta )$, its probability generating function isThus $\mathbb{E}[X_1]=G^{\prime }(1)=\theta$ andThe log-likelihood is, up to an additive constant,so the maximum-likelihood estimator is ${\hat{\theta }}_{\mathrm{MLE}}=\overline{X}$. The central limit theorem gives

Let $m$ be a nonzero invariant measure. Balance at state $i\ge 1$ givesand hence, after choosing $m_0>0$,Balance at zero requiresThe sum telescopes:Consequently a nonzero invariant measure exists exactly whenIt can be normalized to an invariant probability distribution exactly when, in addition,where the empty product for $i=0$ is one and an overall constant has been suppressed.

Under the transience assumption, the number of returns to zero is geometrically distributed and hence finite almost surely. The final excursion is a pure ascent through all sufficiently large states. Its holding times are independent exponential random variables with meansThe nonexplosion criterion for a pure birth process says that their sum is finite almost surely exactly when the sum of their means is finite. Thus

After leaving zero, the jump chain avoids every future reset with probabilityThis probability is positive exactly when ${\prod }_{i\ge 1}(1+{\rho }_i)<\infty$. If $q>0$, the probability of ever returning to zero is less than one, so the irreducible chain is transient. If $q=0$, a reset to zero occurs almost surely after each departure; by the Strong Markov property this happens repeatedly, so the chain is recurrent. Therefore

A reversible Markov chain must satisfy detailed balance. For every $i\ge 1$, the rate from $i$ to $i+1$ is positive, while the reverse rate from $i+1$ to $i$ is zero. No positive measure can satisfy ${\pi }_i{g}_{i,i+1}={\pi }_{i+1}{g}_{i+1,i}$, so the irreducible chain is not reversible.

The total jump rate is ${\lambda }_0$ at zero and ${\lambda }_i(1+{\rho }_i)$ at $i>0$. Hence the jump chain has transition probabilitiesand, for $i>0$,All other entries vanish.

Take independent random variables $U_1,\dots ,U_d$, each uniform on $[0,1/2]$, and independent Rademacher random variables ${\epsilon }_1,\dots ,{\epsilon }_{d-1}$. Defineand let $\eta$ be the joint law of $(X_1,\dots ,X_d)$.

Each $X_i$ is uniform on $[-1/2,1/2]$. If the $d$th coordinate is omitted, the remaining signs are the independent ${\epsilon }_i$. If coordinate $j<d$ is omitted, the remaining signs arewhich are again uniformly distributed over all $2^{d-1}$ sign vectors because this map from $({\epsilon }_1,\dots ,{\epsilon }_{d-1})$ is a bijection. The independent magnitudes then show that every $(d-1)$-coordinate marginal is Lebesgue measure.

The full law is not $d$-dimensional Lebesgue measure, becausealmost surely, whereas this product is equally likely to have either sign under independent Lebesgue coordinates.

On the probability space $(X^2,\eta )$, put $A=I\times X$ and $B=X\times J$. The marginal conditions give $\eta (A)=x$ and $\eta (B)=y$, while $A\cap B=I\times J$. The elementary Fréchet inequalities yieldHence

Let $U$ be uniformly distributed on $X$ and let $\eta$ be the law of $(U,U)$. Both marginals are $\mu$, becauseHowever, $\eta$ is concentrated on the diagonal $\{(x,x):x\in X\}$, which has two-dimensional Lebesgue measure zero. Thus $\eta$ is not Lebesgue measure on $X^2$.

Saying that $U$ has law $\mu$ means that $\mu$ is the pushforward measureFor a real random variable, $\mu$ is a probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$.

The pair $U,V$ consists of independent random variables whenfor every pair of Borel sets $A,B$; equivalently, their joint law is the product of their marginal laws.

The product sigma-algebra isThe product measure ${\mu }_1\otimes {\mu }_2$ is the probability measure on this sigma-algebra satisfyingThe measurable rectangles form a pi-system generating ${\mathcal{A}}_1\otimes {\mathcal{A}}_2$. Any two probability measures agreeing on them therefore agree on the generated sigma-algebra by the pi-lambda theorem, proving uniqueness.

The fundamental theorem of regular space curves states that smooth functions $k>0$ and $\tau$ on an interval determine a smooth arc-length-parametrized curve in ${\mathbb{R}}^3$, unique up to a proper Euclidean motion. Consequently two such curves $\alpha$ and $\tilde{\alpha }$ are related parameterwise by a proper Euclidean motion if and only iffor every $s$.

If $\tau =0$ everywhere, the third of the Frenet-Serret formulas gives $B^{\prime }=0$, so $B$ is a constant vector. MoreoverThus $\alpha (s)\cdot B$ is constant, and the image of $\alpha$ lies in the plane perpendicular to $B$.