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Let $D=d\mathbb{Q}/d\mathbb{P}$ on $\mathcal{F}$. The Radon-Nikodym derivative is nonnegative and ${\mathbb{E}}_{\mathbb{P}}D=1$. For $A\in {\mathcal{F}}_t$,Uniqueness of the restricted Radon-Nikodym derivative therefore identifiesFor $s\le t$, the tower property of conditional expectation gives ${\mathbb{E}}_{\mathbb{P}}(Z_t\mid {\mathcal{F}}_s)=Z_s$, and ${\mathbb{E}}_{\mathbb{P}}{Z}_t=1$. Thus $Z$ is a nonnegative true martingale, indeed a uniformly integrable martingale, since it consists of conditional expectations of one integrable random variable. This is the Radon-Nikodym density martingale. Under the usual conditions for a filtration, it has the standard càdlàg version; continuity is the additional assumption in part (c), not a consequence of absolute continuity alone.

For $0\le s\le t$ and $E\in {\mathcal{F}}_s$, the definition of the Radon-Nikodym derivative givesThe density is nonnegative and integrable; with the probability normalization used here its expectation is one. Henceso $Z$ is a martingale. When $Z_t=h(X_t,t)$, the Itô formula under Wiener measure givesThe first term is a continuous local martingale, by localization on compact space-time sets. Since $Z$ is a martingale, uniqueness of the continuous semimartingale decomposition makes ${\int }_0^{t}g(X_s,s)ds=0$ identically. The integrand is continuous in time along every continuous path, so $g(X_t,t)=0$ for every $t$, outside one common null event.

For each fixed $t>0$, $X_t$ has a Gaussian distribution with strictly positive density everywhere on $\mathbb{R}$. Since $g(\cdot ,t)$ is continuous, $g(X_t,t)=0$ almost surely implies $g(x,t)=0$ for every real $x$. ThereforeIf the stated smoothness includes time zero, continuity extends the equation there. This is the backward heat equation. The full-support argument is needed to pass from an identity along Brownian paths to an equation at every spatial point.

At an interior statistical parameter $\theta$, local asymptotic normality means that there are random vectors ${\Delta }_{n,\theta }$ and a symmetric nonnegative Fisher information matrix $I_{\theta }$ such that, for each fixed $h\in {\mathbb{R}}^p$,A common stronger definition requires this expansion also for $h_n\to h$. The fixed-$h$ version suffices here. A regular identifiable model usually has positive definite Fisher information; nonsingularity is unnecessary for the contiguity conclusion.

Here are sufficient differentiability conditions for a proof by Taylor's theorem. Suppose the local probability density functions have common support, ${\ell }_{\vartheta }(y)=\log f(\vartheta ,y)$ is twice continuously differentiable near $\theta$, differentiation can pass through the normalization integral twice, the score function $s_{\theta }=\nabla {\ell }_{\theta }$ has finite second moment, and the local Hessian matrix is bounded in operator norm by an integrable envelope of a function class. These hypotheses implyThe first identity differentiates $\int f_{\theta }=1$ once; the second uses ${\nabla }^2{f}_{\theta }=f_{\theta }({\nabla }^2{\ell }_{\theta }+s_{\theta }{s}_{\theta }^T)$ and differentiates twice. The central limit theorem gives ${\Delta }_{n,\theta }=n^{-1/2}{\sum }_i{s}_{\theta }(Y_i)\stackrel{d}{\to }{N}_p(0,I_{\theta })$. Taylor's integral remainder givesThe weak law of large numbers and the continuity and integrable-envelope hypotheses make the bracket converge in probability to $-I_{\theta }/2$. Indeed, the expectation of the supremum of the Hessian matrix difference over a shrinking neighborhood tends to zero by dominated convergence, and Markov inequality controls its empirical average. This proves local asymptotic normality and explains the role of both derivatives.

Write $Q_n◃P_n$ for contiguity of probability measures: for every measurable sequence $A_n$, $P_n(A_n)\to 0$ implies $Q_n(A_n)\to 0$. Mutual contiguity requires this in both directions. A useful form of Le Cam's first lemma is the following: if the Radon-Nikodym derivative $L_n$ of the absolutely continuous part of $Q_n$ relative to $P_n$ converges in distribution under $P_n$ to $L$, with $L>0$ almost surely and $\mathbb{E}L=1$, then the two sequences are mutually contiguous. This includes the usual case $L_n=d{Q}_n/d{P}_n$. The mean-one condition ensures uniform integrability of $L_n$ and vanishing mass of any singular part; positivity ensures the reverse implication.

Take $P_n=P_{\theta }^n$ and $Q_n=P_{\theta +h/\sqrt{n}}^n$. Their likelihood ratio on the support of $P_n$ satisfies, by local asymptotic normality and the continuous mapping theorem,This limit is positive and $\mathbb{E}L=e^{-v/2}{e}^{v/2}=1$, including $v=0$. The conditions of Le Cam's first lemma therefore hold, giving mutual contiguity of the product sampling laws. The products, rather than the one-observation laws, are the relevant sequences.

For any $\epsilon >0$, let $A_n=\{\Vert {\hat{\theta }}_n-\theta \Vert >\epsilon \}$. Consistency under $P_{\theta }^n$ makes $P_{\theta }^n(A_n)\to 0$, and contiguity gives $Q_n(A_n)\to 0$. Thus the estimator remains consistent for $\theta$ under each fixed local alternative. Since $\Vert \theta +h/\sqrt{n}-\theta \Vert \to 0$, it is also consistent for the moving parameter in the sense that $\Vert {\hat{\theta }}_n-(\theta +h/\sqrt{n})\Vert \to 0$ in probability under $Q_n$.

For $0\le t\le T$, set $Z_t=\cosh x{Y}_t$. Part (b) gives $Z_0=1$, $\mathbb{E}{Z}_T=1$, and $Z_t>0$. Its stochastic differential isThus $Z$ is the stochastic exponential of $\int \eta dW$. The Novikov condition also holds, since $\exp (\frac{1}{2}{\int }_0^T{\eta }_t^{2}dt)\le e^{T/2}$.

The Girsanov theorem says that under the measure with Radon-Nikodym derivative $Z_T$, the process $W_t-{\int }_0^t{\eta }_{s}ds$ is a Brownian motion up to $T$. Substituting the sign of $\eta$ and the original stochastic differential equation givesThe density is strictly positive, so $\mathbb{Q}$ and $\mathbb{P}$ are equivalent probability measures.

Under $\mathbb{Q}$, $X_T=x+W_T^{\mathbb{Q}}$ has density ${\phi }_T(z-x)=(2\pi T{)}^{-1/2}\exp (-(z-x{)}^2/(2T))$. The reciprocal Radon-Nikodym derivative depends only on $X_T$:Consequently the probability density function under the original measure isIt integrates to one because ${\mathbb{E}}_{\mathbb{Q}}\cosh X_T=e^{T/2}\cosh x$. Completing the square also gives the useful mixture distribution formThus the terminal law is a mixture of $N(x+T,T)$ and $N(x-T,T)$ with the displayed positive weights. The diffusion with hyperbolic tangent drift density also exhibits the Doob h-transform with $h(z)=\cosh z$.

For $1\le p<\infty$, the real Lp space is the vector space of real measurable functions with ${\int }_{\Omega }|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. For $p=\infty$, take the essentially bounded real measurable functions, with the same identification and norm $\Vert f{\Vert }_{\infty }=\mathrm{esssup}|f|$. The identification makes each norm definite; absolute homogeneity follows from the Lebesgue integral, and the triangle inequality follows from Minkowski inequality for finite $p$, or directly from the essential supremum for $p=\infty$.

Here is a completeness proof valid on any measure space. For $p<\infty$, a Cauchy sequence $(f_n)$ has a subsequence $(f_{n_k})$ with $\Vert f_{n_{k+1}}-f_{n_k}{\Vert }_p\le 2^{-k}$. Choose measurable function representatives and put $h_k=f_{n_{k+1}}-f_{n_k}$. By Minkowski inequality and the monotone convergence theorem,Consequently the series $\sum h_k$ converges absolutely almost everywhere. Define $f=f_{n_1}+\sum h_k$ there, and define it to be zero on the measurable exceptional null set. Then $f\in L^p$, and Fatou lemma applied to each tail gives $\Vert f-f_{n_k}{\Vert }_p\le {\sum }_{j=k}^{\infty }{2}^{-j}\to 0$. The original Cauchy sequence also converges in Lp norm, by the triangle inequality. For $p=\infty$, choose the same subsequence using the essential supremum norm. Outside one measurable null set, all the bounds $|h_k|\le 2^{-k}$ hold and $f_{n_1}$ is bounded. The series then converges uniformly there, with an essentially bounded measurable function limit and the same tail estimate in essential supremum norm. Thus all these spaces are Banach spaces.

The duality of Lp spaces says that, for $1<p<\infty$ and $q=p/(p-1)$, the mapis an isometric isomorphism of normed spaces. This form of Lp duality on an arbitrary measure space requires no finiteness hypothesis on $\mu$. At $p=1$, a standard version assumes a sigma-finite measure and identifies $(L^1{)}^{\ast }$ with $L^{\infty }$ through the same dual pairing. That endpoint assertion must not be made without a suitable measure-space hypothesis.

We first prove the required duality of Lp spaces for a finite measure. Let $F\in (L^p{)}^{\ast }$ and set $\nu (E)=F(1_E)$. For disjoint measurable $E_j$, the indicator functions of their partial unions converge in Lp norm to that of their union, so $\nu$ is countably additive. It has finite variation measure: for every finite measurable partition $(E_j)$, choosing real signs givesAlso $\mu (E)=0$ implies $\nu (E)=0$. The Radon-Nikodym theorem supplies a Radon-Nikodym derivative $g\in L^1$ with $\nu (E)={\int }_{E}gd\mu$. Linearity gives $F(f)=\int fgd\mu$ for simple functions. Uniform approximation by simple functions extends this identity to bounded measurable functions: both their Lp norm errors and the errors in integration against $g$ tend to zero.

To establish the correct integrability, test with the bounded measurable function $f_N=\mathrm{sgn}(g)|g{|}^{q-1}{1}_{\{|g|\le N\}}$. Since $(q-1)p=q$, writing $A_N={\int }_{\{|g|\le N\}}|g{|}^{q}d\mu$ givesThe second inequality is also valid when $A_N=0$. The monotone convergence theorem gives $g\in L^q$ and $\Vert g{\Vert }_q\le \Vert F\Vert$. Density of simple functions in the Lp space, together with Hölder's inequality, now gives $F=I_{p}g$ on all of $L^p$. Conversely Hölder's inequality gives $\Vert I_{p}g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, testing againstgives $\Vert f{\Vert }_p=1$ and $(I_{p}g)(f)=\Vert g{\Vert }_q$, so $\boxed{\Vert I_{p}g\Vert =\Vert g{\Vert }_q}$. This also proves uniqueness of the representing Radon-Nikodym derivative.

For completeness, the passage to an arbitrary measure space can be made without losing a hypothesis in the reflexive Banach space argument below. On a sigma-finite measure space, exhaust by nested finite-measure sets $E_n$. The representing Radon-Nikodym derivatives on $E_n$ agree on overlaps by uniqueness. Their glued density has Lp norm $\Vert g{\Vert }_q\le \Vert F\Vert$ by the monotone convergence theorem, and represents $F$ because $f{1}_{E_n}\to f$ in Lp norm. Every $f\in L^p$ on an arbitrary measure space is supported on a sigma-finite measurable set: the sets $\{|f|>1/n\}$ have finite measure and their union is $\{f\ne 0\}$.

Use support localization of an Lp functional as follows. For a sigma-finite measurable $A$, let $m_A$ be the operator norm of $F$ restricted to functions supported in $A$. The support observation gives ${\sup }_A{m}_A=\Vert F\Vert$. Choose $A_n$ approaching this supremum and set $A={\bigcup }_n{A}_n$; then $m_A=\Vert F\Vert$. If a sigma-finite $B\subseteq \Omega \setminus A$ had $m_B=\beta >0$, functions supported on the disjoint sets $A,B$ have the direct-sum Lp norm. Optimizing their two scalar coefficients by Hölder's inequality, and using functions approaching the two restriction operator norms, would givea contradiction. Thus $F$ vanishes on functions supported outside $A$. The density on $A$, extended by zero, represents $F$ globally. The case $F=0$ simply uses $g=0$. This proves the stated Lp duality on an arbitrary measure space.

Finally let $X=L^p$ and let $\Phi \in X^{\ast \ast }$, where stars denote continuous dual spaces. Compose with $I_p$ to obtain the bounded linear functional $g\mapsto \Phi (I_{p}g)$ on $L^q$. Applying duality of Lp spaces with the exponents reversed gives $f\in L^p$ with $\Phi (I_{p}g)=\int fgd\mu$. For the canonical embedding into the bidual $J_{X}f$, its value on $I_{p}g$ is also $\int fgd\mu$. Since $I_p$ is onto, $\Phi =J_{X}f$. Its norm is $\Vert f{\Vert }_p$ by the same dual pairing norm identity. Therefore $\boxed{J_X(L^p)=(L^p{)}^{\ast \ast }}$, which proves that $L^p$ is a reflexive Banach space through its actual canonical embedding into the bidual.

The Radon-Nikodym theorem for positive measures says: if $\mu$ and $\nu$ are sigma-finite measures on the same measurable space, and $\nu$ is absolutely continuous with respect to $\mu$, then there is a nonnegative measurable function $g$, unique $\mu$-almost everywhere, such thatHere sigma-finiteness means that the space is a countable union of measurable sets of finite measure; absolute continuity of measures, written $\nu \ll \mu$, means that $\mu (E)=0$ implies $\nu (E)=0$. The function $g=d\nu /d\mu$ is the Radon-Nikodym derivative. For a finite signed or complex measure $\nu$ of finite total variation norm of a measure, absolutely continuous with respect to a sigma-finite $\mu$, the corresponding density belongs to $L^1(\mu )$. This follows by applying the positive theorem to the positive and negative parts of the real and imaginary parts of $\nu$.

For every measure space and $1<p<\infty$, $(L^p{)}^{\ast }$ is isometrically $L^q$, where $1/p+1/q=1$. We use the complex-linear pairingWith the convention $\int h\overline{g}$, the same identification is conjugate-linear in $g$. By Hölder's inequality, ${\Phi }_g$ is a bounded linear functional with $\Vert {\Phi }_g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, setinterpreting the numerator as zero where $g=0$. The identity $(q-1)p=q$ gives $\Vert h{\Vert }_p=1$ and ${\Phi }_g(h)=\Vert g{\Vert }_q$. ThusIt remains to represent an arbitrary $\Phi \in (L^p{)}^{\ast }$, rather than merely produce functionals from $L^q$.

We prove Lp duality on an arbitrary measure space without imposing sigma-finiteness on $\mu$. Write $C=\Vert \Phi \Vert$. For every measurable set $E$ with $\mu (E)<\infty$, define a complex measure on $E$ byIt is countably additive: for disjoint $F_j\subseteq E$, the partial sums of their indicator functions tend in $L^p$ to $1_{\bigcup F_j}$, because the measure of the omitted tail tends to zero. It is absolutely continuous with respect to $\mu {|}_E$, since indicator functions of null sets represent zero in $L^p$.

Its total variation norm of a measure is finite. For any finite measurable partition $E={\bigcup }_j{E}_j$, choose scalars $c_j$ of modulus $1$ with $c_j{\nu }_E(E_j)=|{\nu }_E(E_j)|$. ThenTaking the supremum over partitions gives the variation bound. Since $\mu {|}_E$ is finite, the Radon-Nikodym theorem supplies $g_E\in L^1(E)$ with ${\nu }_E(F)={\int }_F{g}_{E}d\mu$. By uniform approximation with simple functions,for every bounded measurable $h$ supported in $E$.

To improve $g_E$ from $L^1$ to $L^q$, test with the bounded functionIf $I_N={\int }_{E\cap \{|g_E|\le N\}}|g_E{|}^{q}d\mu$, thenThus $I_N^{1/q}\le C$ when $I_N>0$, and the same bound is trivial when it is zero. The monotone convergence theorem givesIf $E,F$ both have finite measure, the densities $g_E,g_F$ agree almost everywhere on $E\cap F$: their integrals over every measurable subset of the intersection equal the same functional value. This is uniqueness in the Radon-Nikodym theorem.

We now perform support localization of an Lp functional. SetChoose finite-measure sets $E_n$ whose displayed integrals tend to $M$, and let $D_n=E_1\cup \cdots \cup E_n$, $S={\bigcup }_n{D}_n$. If $M=0$, take $E_n=\varnothing$. Compatibility allows us to define a measurable function $g$ on $S$ by taking $g_{D_n}$ on the disjoint measurable sets $D_n\setminus D_{n-1}$, and put $g=0$ off $S$. It agrees almost everywhere with $g_{D_n}$ on every $D_n$. Moreover,Indeed each integral is at most $M$, and it is at least the integral over $E_n$, which tends to $M$.

For any finite-measure set $F\subseteq \Omega \setminus S$, compatibility on the disjoint union $D_n\cup F$ givesLetting $n\to \infty$ forces $g_F=0$ almost everywhere. For an arbitrary finite-measure set $E$, compatibility on $E\cap D_n$ and the preceding conclusion on $E\setminus S$ show that $g_E=g$ almost everywhere on $E$. Consequently $\Phi (h)=\int hgd\mu$ for every simple function supported on a finite-measure set.

Those simple functions are dense in $L^p$ even for this arbitrary measure space. To see the needed finite-support property, for $h\in L^p$ the sets $\{|h|>1/n\}$ have finite measure, bounded by $n^p\Vert h{\Vert }_p^p$. First truncate $h$ to such sets and to bounded values, then approximate by simple functions; the discarded $L^p$ integral tends to zero. Continuity of $\Phi$ and Hölder's inequality therefore extend the representation to every $h\in L^p$.

We have constructed $g\in L^q$ with $\Phi ={\Phi }_g$, and the previously proved norm identity gives $\Vert g{\Vert }_q=\Vert \Phi \Vert$. It also proves uniqueness: if ${\Phi }_g={\Phi }_{\tilde{g}}$, then $\Vert g-\tilde{g}{\Vert }_q=0$. This completes the isometric duality of Lp spaces.

The dominated sequence actually converges to zero in norm, and hence weakly. The assumptions give $|f_n{|}^p\le |f{|}^p\in L^1$ and $|f_n{|}^p\to 0$ almost everywhere. The dominated convergence theorem yieldsFor every $\Phi \in (L^p{)}^{\ast }$, $|\Phi (f_n)|\le \Vert \Phi \Vert \Vert f_n{\Vert }_p\to 0$. Therefore

On the same measurable space, absolute continuity of measures meansThe Radon-Nikodym theorem then supplies a nonnegative measurable density $Z$ with ${\mathbb{E}}_{\mathbb{P}}Z=1$ and $\tilde{\mathbb{P}}(A)={\mathbb{E}}_{\mathbb{P}}[Z{1}_A]$. The converse follows directly from this formula. Strict positivity of the Radon-Nikodym derivative gives equivalence of the two probability measures; absolute continuity of measures alone need not give equivalence.

The reference Brownian motion must start at $x$, and both laws must use the same stopped-path rule. Under the reference probability measure let $R_s=x+W_s$, let $\tau$ be its exit time from $(a,b)$, and put $T=t\wedge \tau$ and $c=(\delta -1)/2$. The Girsanov density for a stopped Bessel process, as a functional of the stopped reference path, isUse the predictable process ${\theta }_s=c{R}_s^{-1}{1}_{\{s\le \tau \}}$, with any harmless choice at the endpoint. On $[0,t]$, $\int {\theta }_s^{2}ds\le c^{2}t/a^2$. The Novikov condition holds; under the density $D$, the Girsanov theorem makes ${\tilde{W}}_s=W_s-{\int }_0^s{\theta }_{u}du$ a Brownian motion. Before exit, $d{R}_s=c{R}_s^{-1}ds+d{\tilde{W}}_s$, which is the required stochastic differential equation. The drift is a locally Lipschitz function on $(0,\infty )$, so pathwise uniqueness and uniqueness in law hold up to this exit time.

For a version involving just the endpoint and an ordinary integral, the Itô formula for $\log R$ givesThusThe stopped density is measurable from the stopped path, so it is also the Radon-Nikodym derivative of the stopped-path laws. The stopping times in this formula are evaluated on the reference path, rather than taken from a different realization of $X$. A zero-start reference Brownian motion would have a different deterministic initial value and could not dominate this law.

Since $P(T,T)=1$, part (b) shows that the density process of the T-forward measure isIts terminal expectation is one and it is strictly positive, so the stated Radon-Nikodym derivative defines an equivalent probability measure. By the Girsanov theorem,is a Brownian motion under ${\mathbb{Q}}_T$. Substitution of $d{W}_t=d{W}_t^T-b_t^{T}dt$ cancels the entire instantaneous forward rate drift:Because the integrand is deterministic and bounded, its Itô integral is square-integrable on $[0,T]$. With the usual fixed initial instantaneous forward rate curve, the requested true martingale is

Use the Radon-Nikodym derivative at $T_1$ and the discounted bond price martingale from part (b):For the variance, put $b_t^i={\int }_t^{T_i}\sigma (t,u)du$ and $d_t=b_t^2-b_t^1={\int }_{T_1}^{T_2}\sigma (t,u)du$, for $t\le T_1$. The ratio $R_t=P(t,T_2)/P(t,T_1)$ has, by the Itô formula,Under the T-forward measure ${\mathbb{Q}}_{T_1}$, $d{W}_t=d{W}_t^{T_1}-b_t^{1}dt$, so its drift becomes $-d_t^2/2$. Since $R_{T_1}=P(T_1,T_2)$,The first two terms are deterministic; the Itô isometry supplies the variance of the last. The two requested answers areIn fact the terminal zero-coupon bond price has a log-normal distribution under this forward measure, with the deterministic negative half-variance correction ensuring the displayed mean.

Let $P_0,P_1$ be simple hypotheses with densities $p_0,p_1$ (their Radon-Nikodym derivatives) relative to a common dominating measure. A possibly randomised statistical test is a measurable function $\phi \in [0,1]$; its size of a statistical test is $E_0\phi$ and its statistical power is $E_1\phi$. The Neyman-Pearson lemma says that a likelihood ratio threshold testchosen to have size $\alpha$, is a most powerful test among tests of size at most $\alpha$. A constant randomisation on the equality set suffices to reach the required size; sets where $p_0=0<p_1$ are always rejected. For $0<\alpha <1$, a quantile of the likelihood ratio under $P_0$ gives such a threshold. If the threshold is zero, randomisation where $p_1=0$ may be needed to use the remaining size. At $\alpha =0$, one can reject only a $P_0$-null set and optimally reject the part supporting $P_1$; at $\alpha =1$, always reject.

For the proof, every competing $\phi$ satisfiespointwise: the signs agree off the equality set, where the product is zero. Integrating givesThus the likelihood-ratio threshold test has maximum power at the stated level. If $c>0$, a competing most-powerful test must have size $\alpha$ and agree with ${\phi }_{\ast }$ off the equality set, apart from sets null for $P_0+P_1$; both inequalities must then be equalities. For $c=0$, equal size is not necessary for optimality, because changing decisions where $p_1=0$ does not affect power. These qualifications account for ties and singular supports rather than assuming all densities are strictly positive.

For probability measures on $(\Omega ,\mathcal{F})$, absolute continuity of measures means preservation of null sets in one direction:Equivalently, by the Radon-Nikodym theorem, there is an $\mathcal{F}$-measurable Radon-Nikodym derivative $Z\ge 0$ with ${\mathbb{E}}_{\mathbb{P}}Z=1$ and $\tilde{\mathbb{P}}(A)={\mathbb{E}}_{\mathbb{P}}[Z{1}_A]$. The derivative may vanish, so absolute continuity of measures does not require equivalence of the two probability measures. In particular, the reverse null-set implication is not part of the definition.

Define a probability measure on the same full measurable space byThe Radon-Nikodym derivative is positive and has expectation one, so this is a probability measure and even equivalent to $\mathbb{P}$, in particular ${\mathbb{P}}^T\ll \mathbb{P}$. For $t\le T$ its density on ${\mathcal{F}}_t$ is $\mathbb{E}[Z_T\mid {\mathcal{F}}_t]=Z_t$.

The finite-horizon Girsanov theorem says that if $W$ is a Brownian motion and $\mathcal{E}(\int hdW)$ is a true martingale through $T$ defining the density, then $W_t-{\int }_0^t{h}_{s}ds$ is a Brownian motion under that density for $t\le T$. Here $h_s=\mu$, and the stochastic exponential is exactly $Z$. The previous part verifies the necessary martingale property; alternatively the Novikov condition is $e^{{\mu }^{2}T/2}<\infty$. Hence $B_t-\mu t$ is a standard Brownian motion on $[0,T]$ relative to the same filtration under ${\mathbb{P}}^T$.

Let $p=dP/d\nu$ and $q=dQ/d\nu$ be Radon-Nikodym derivatives. Use the following normalizations of the Kullback-Leibler divergence, total variation distance and unnormalized Hellinger distance:We use $0\log (0/q)=0$ and assign infinite Kullback-Leibler divergence if $Pot\ll Q$, where $\ll$ denotes absolute continuity of measures. Common domination alone also suffices for the inequalities with the extended-value convention. The negative part of the logarithmic integral is integrable: on $p<q$, $p\log (q/p)\le q-p\le q$. Thus the extended-value integral defining the Kullback-Leibler divergence is well-defined. These expressions do not depend on which common dominating measure is chosen. The Kullback-Leibler divergence is generally asymmetric and therefore fails a defining property of a metric.

For completeness, the integral formula for total variation distance follows by taking $A=\{p\ge q\}$: since $\int (p-q)d\nu =0$, the positive and negative parts of $p-q$ have the same integral, and this set attains the supremum.

Put $a=\int \sqrt{pq}d\nu \le 1$, by the Cauchy-Schwarz inequality. A second application givesThe logarithmic inequality in the hint implies $\log t\le t-1$ for $t>0$. Apply it to $t=\sqrt{q/p}$ on $\{p>0\}$:Multiplying by $p$ and integrating provesIf $q=0<p$ on a set of positive $P$-measure, the right side is finite but the Kullback-Leibler divergence is infinite, so the inequality is immediate. ThusThis total variation–Hellinger–relative entropy inequality requires the stated Hellinger distance normalization for its first constant. With the alternative convention ${\bar{h}}^2=\frac{1}{2}\int (\sqrt{p}-\sqrt{q}{)}^2$, the correct comparison is $\mathrm{TV}\le \sqrt{2}\bar{h}$ and $\bar{h}\le \sqrt{\mathrm{KL}/2}$. In particular, $P=\mathrm{Bernoulli}(0.6)$ and $Q=\mathrm{Bernoulli}(0.4)$ have $\mathrm{TV}=0.2>\bar{h}\approx 0.1421$.

Here is a testing form of the Le Cam two-point lemma. Suppose two parameter values ${\theta }_0,{\theta }_1$ in a metric space have separation $\Delta =d({\theta }_0,{\theta }_1)>0$, and let $P_0,P_1$ be the corresponding observation laws. Every estimator $T$ satisfiesTo prove it, classify $T$ by the closer of the two parameters, breaking a tie in favour of one. Let $A=\{d(T,{\theta }_1)\le d(T,{\theta }_0)\}$. On $A$, the triangle inequality implies $d(T,{\theta }_0)\ge \Delta /2$, while on $A^c$ it implies $d(T,{\theta }_1)>\Delta /2$. Hence the sum of the two probabilities in the display is at leastAt least one is at least half the sum, proving the claim. It also gives a lower bound $\Delta (1-\mathrm{TV})/4$ for the maximum expected metric loss.