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Suppose an unknown has prior distribution $\Pi$ and its conditional observation law $P_u$ has jointly measurable Radon-Nikodym derivative $L(u,m)$ relative to a fixed observation law $P_0$. If $L>0$ for $\Pi (du)P_0(dm)$-almost every pair, thendefines the posterior distribution for almost every observation. Indeed, $\int L(u,m)P_0(dm)=1$ for almost every $u$, so Tonelli theorem gives $\int Z(m)P_0(dm)=1$. Hence $Z$ is finite almost everywhere, and positivity of $L$ and Fubini's theorem give $Z>0$ almost everywhere. The data law is $Z{P}_0$, and integration of the proposed posterior against this data law recovers the joint law. This proves the conditional distribution property and also shows that the data law is an equivalent probability measure to $P_0$.

For $\mu =\mathcal{N}(0,\Sigma )$ with injective covariance operator of a Gaussian measure, its translated law ${\mu }_h=\mathcal{N}(h,\Sigma )$ is an equivalent probability measure to $\mu$ if and only if $h\in E_{\mu }$, the Cameron-Martin space of a Gaussian measure. For such $h$,The series has mean-square convergence and converges almost surely under $\mu$, with normal distribution $\mathcal{N}(0,\Vert h{\Vert }_{E_{\mu }}^2)$. Its exponential density has expected value one by the moment-generating function of a normal distribution. Finite-dimensional projections give the formula by ratios of multivariate normal densities; convergence of these likelihood ratios gives the infinite-dimensional result. The expression $⟨h,x{⟩}_{E_{\mu }}$ is only formal when the sample is outside $E_{\mu }$. For $h\notin E_{\mu }$, translation gives mutually singular measures.

For a Radon-Nikodym derivative $L_T\ge 0$ with $\mathbb{E}{L}_T=1$, its conditional-expectation process is a nonnegative martingale. If $L_T>0$ almost surely, it defines an equivalent probability measure. A stochastic exponential satisfying the Novikov condition is a common positive density process.

For an isonormal Gaussian process $X$ on a real Hilbert space $H$ and $k\in H$, the positive density $\exp (X(k)-\Vert k{\Vert }^2/2)$ has expectation one and defines an equivalent probability measure. Under this measure, $X(h)$ has mean $⟨h,k⟩$ and the same covariance as before. The shifted family $X(h)-⟨h,k⟩$ is again an isonormal Gaussian process. This follows by evaluating the joint Gaussian exponential formula for any finite collection of arguments. It is the Hilbert-space version of exponential tilting.

Normalize the riskless financial asset to $B_0=1$, $B_1=b>0$, and write $S_0,S_1$ for the $s$ risky asset prices. Their discounted gain random vector is $Y=S_1/b-S_0$. A zero-cost portfolio holding $h\in {\mathbb{R}}^s$ in the risky assets and $-h\cdot S_0$ in the riskless asset has terminal value $bh\cdot Y$. Thus an arbitrage meansAn equivalent martingale measure is a probability measure $Q$ with the same null sets as $P$, under which $Y$ is integrable and ${\mathbb{E}}_{Q}Y=0$. The discounted risky prices are then martingales over the single trading period.

If such $Q$ exists, a nonnegative terminal gain has expectation zero under $Q$. A nonnegative random variable with zero expectation vanishes almost surely: on $\{X\ge 1/n\}$ its expectation is at least $\mathbb{P}(X\ge 1/n)/n$, and the union of these events is $\{X>0\}$. Equivalence transfers this conclusion to $P$, excluding arbitrage.

For the converse we prove the needed positive-density separation proof of the one-period asset-pricing theorem. First make the gain integrable without changing its null sets. Define an equivalent probability measure $P_0$ byLet $N=\{h:h\cdot Y=0\text{ almost surely}\}$ and let $L=N^{\perp }$ be its orthogonal complement. Taking a finite basis of $N$ shows $Y\in L$ almost surely. If $L=\{0\}$, the gain is zero and $P_0$ already gives the required measure. Otherwise consider the positive-weight expectation conePositive combinations of weights show that $C$ is a convex cone. It is open relative to $L$. Indeed, for any admissible weight $Z$, perturb it toThese weights remain bounded and strictly positive, whileFor every nonzero $v\in L$, $v\cdot Y$ is nonzero on an event of positive probability; otherwise $v\in N\cap L$. ConsequentlyThe matrix is finite because $\Vert Y{\Vert }^2/(1+\Vert Y\Vert )\le \Vert Y\Vert$. The kernel of $A_Z$ on $L$ is zero, so it defines an invertible matrix on the finite-dimensional vector space $L$. For any sufficiently small $\delta >0$, every $y$ with $\Vert y\Vert <\delta /\Vert A_Z^{-1}\Vert$ is the image of a vector in the radius-$\delta$ ball, since $\Vert A_Z^{-1}y\Vert <\delta$. Hence the perturbations give a neighborhood of zero, proving relative openness of $C$.

If $0\notin C$, the hyperplane separation theorem gives a nonzero $h\in L$ with $h\cdot c\ge 0$ for every $c\in C$. The separating constant can be taken to be zero because $C$ is a cone. Put $A=\{h\cdot Y<0\}$ and use the strictly positive weights $Z=1_A+\epsilon$. Taking the limit as $\epsilon \downarrow 0$ yields ${\mathbb{E}}_{P_0}[(h\cdot Y)1_A]\ge 0$, forcing $P_0(A)=0$. But $h\notin N$, so $P_0(h\cdot Y>0)>0$. This is an arbitrage, a contradiction. Hence $0\in C$: some bounded strictly positive $Z$ satisfies ${\mathbb{E}}_{P_0}[ZY]=0$. Normalize it by $dQ/d{P}_0=Z/{\mathbb{E}}_{P_0}Z$. This gives an equivalent martingale measure with integrable gains, establishing the fundamental theorem of asset pricing here without any further existence theorem.

Now retain only the $m$ positive-probability states of the finite probability space. Let $V\subseteq {\mathbb{R}}^m$ be the linear span of the discounted terminal asset payoffs; it contains the constant vector $1$. A complete market means $V={\mathbb{R}}^m$. An equivalent martingale measure is a vector $q$ of strictly positive state probabilities whose pairing with each asset payoff gives its initial price. By linearity it prices every attainable payoff. If $V={\mathbb{R}}^m$, the indicator functions of individual states are attainable, so these prices determine every $q_j$ uniquely.

If $V$ is proper, choose $0\ne v\in V^{\perp }$. Since $1\in V$, ${\sum }_j{v}_j=0$. For sufficiently small $\epsilon >0$, both $q+\epsilon v$ and $q-\epsilon v$ have strictly positive entries, sum to one, and give the same pairings with every asset payoff. They are distinct equivalent martingale measures. Therefore market completeness is equivalent to uniqueness of the equivalent martingale measure.

A useful finite-horizon form of the Girsanov theorem is as follows. Let $W$ be a Brownian motion and $\eta$ an adapted, progressively measurable process with ${\int }_0^T{\eta }_s^{2}ds<\infty$ almost surely. Suppose its stochastic exponentialis a true martingale with mean one. The Novikov condition $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{\eta }_s^{2}ds)<\infty$ is a sufficient condition. Under the equivalent probability measure $dQ=Z_{T}dP$, the process $W_t^Q=W_t+{\int }_0^t{\eta }_{s}ds$ is a Brownian motion.

For the proof sketch, the Itô formula gives $dZ=-Z\eta dW$. Integration by parts cancels the drift in $d(Z{W}^Q)$, since $Z\eta dt$ from $d{W}^Q$ cancels $d[Z,W^Q]=-Z\eta dt$. After localization, the Bayes formula for conditional expectation makes $W^Q$ a continuous local martingale under $Q$. Its quadratic variation is still $t$, because adding a finite-variation process does not change quadratic variation. One can see the Brownian conclusion directly: for real $z$, the Itô formula makes $\exp (iz{W}_t^Q+z^{2}t/2)$ a local martingale. Its modulus is bounded on the fixed horizon, so it is a true martingale. ThusThis is the conditional characteristic function of a centered normal distribution with variance $t-s$, independent of the past. Continuity and these independent normal increments give the defining properties of Brownian motion.

Now put ${\varphi }_t(x)=(2\pi t{)}^{-1/2}{e}^{-x^2/(2t)}$ and $\Phi (z)={\int }_{-\infty }^z{\varphi }_1(x)dx$, the standard normal distribution function. At zero drift the endpoint density of paths staying below $a$ isIndeed, reflect the portion of a path after its first hitting time of $a$. By continuity and the independent symmetric Brownian motion increments after this stopping time, reflection preserves its probability law and maps a crossing path ending at $x<a$ to a path ending at $2a-x>a$. This subtracts exactly ${\varphi }_t(2a-x)$ from the unrestricted endpoint density; it is the Brownian reflection principle.

Taking the constant drift tilt $dQ/dP=e^{\mu W_t-{\mu }^{2}t/2}$ in the Girsanov theorem makes $W$ under $Q$ have the same law as $W+\mu s$ under $P$. Therefore the desired finite-horizon maximum of Brownian motion with drift has probabilityCompleting squares gives $e^{\mu x-{\mu }^{2}t/2}{\varphi }_t(x)={\varphi }_t(x-\mu t)$ and $e^{\mu x-{\mu }^{2}t/2}{\varphi }_t(2a-x)=e^{2\mu a}{\varphi }_t(x-2a-\mu t)$. HenceThe maximum has no atom at a positive level, so strict and weak barrier inequalities agree. This also follows from continuity of the displayed distribution function. At $\mu =0$ the formula reduces to $2\Phi (a/\sqrt{t})-1$.

For the up-and-out power claim, write $T=t_0>0$ and use the risk-neutral measure for the Black-Scholes model, under whichwith $\sigma >0$. The barrier survives exactly when ${\sup }_{t\le T}(W_t+m_{0}t)<A$. Its risk-neutral pricing value isThe factor inside the expectation is a mean-one Exponential martingale for Brownian motion. Under the probability tilted by it, $W_t-2\sigma t$ is a Brownian motion, so the barrier process has drift $m=m_0+2\sigma =(\rho +3{\sigma }^2/2)/\sigma$. Applying the preceding survival formula yieldsAs $c\to \infty$, the bracket tends to one and the price becomes the discounted second moment $S_0^2{e}^{(\rho +{\sigma }^2)T}$. As $c\downarrow S_0$, it tends to zero. If $\sigma =0$, use the deterministic path instead: the price is $S_0^2{e}^{\rho T}$ when $S_0\max (1,e^{\rho T})<c$, and zero otherwise. At $T=0$, with $c>S_0$, the price is simply $S_0^2$.

The density $Z=\exp (X(k)-\Vert k{\Vert }^2/2)$ is strictly positive. The Gaussian moment-generating function gives $\mathbb{E}Z=1$, so it defines an equivalent probability measure.

The joint normal distribution of $(X(h),X(k))$, with covariance $⟨h,k⟩$, gives the mixed exponential formulaMultiplying by the normalizing and centring factors therefore yieldsThe characteristic function identifies the answer:This is exponential tilting of an isonormal Gaussian process: the mean shifts by the inner product while its covariance remains unchanged.

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.

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

The positive density process is the stochastic exponentialSince $a^{\prime }$ is bounded, the Novikov condition holds; $Z$ is a true martingale with $\mathbb{E}{Z}_T=1$ and $Z_T>0$. Thus it defines the stated equivalent probability measure. The positive sign in the exponent means that the Girsanov theorem givesUnder this change of measure, applying the Itô formula to ${\pi }_t=U(t,S_t)$ produces precisely the drift in its backward equation, which vanishes:Bounded $a$ and $U_S$ make $\pi$ a true martingale under $\hat{\mathbb{P}}$. Its terminal value is $g^{\prime }(S_T)$. ConsequentlyThe sign can also be checked before changing measure: under the original measure $d\pi =-a{a}^{\prime }{U}_{S}dt+a{U}_{S}dW$, while the quadratic covariation term in $d(Z\pi )$ is $Za{a}^{\prime }{U}_{S}dt$, exactly cancelling its drift.

The Cameron-Martin space of a Gaussian random variable in a Banach space is most conveniently defined using the first Gaussian chaos. Work over the real separable Banach space $B$, with continuous dual space $B^{\ast }$. For a centered Gaussian random variable in a Banach space, defineThe set inside the closure is already a linear subspace. Every $g\in \mathcal{G}$ is a centered Gaussian random variable. The Bochner integral defining $S$ exists: Fernique's theorem states that for some $a>0$, $\mathbb{E}{e}^{a\Vert X{\Vert }_B^2}<\infty$, and then Cauchy-Schwarz inequality gives $\mathbb{E}\Vert Xg{\Vert }_B<\infty$.

The map $S$ is injective. Indeed, $Sg=0$ implies $\mathbb{E}[g\ell (X)]=\ell (Sg)=0$ for all $\ell \in B^{\ast }$. Density of these observations in $\mathcal{G}$ then implies $\mathbb{E}{g}^2=0$. Define the Reproducing kernel Hilbert space byThe injectivity makes the inner product unambiguous, and the completeness of $\mathcal{G}$ makes $H$ a Hilbert space. The embedding into $B$ is continuous sinceFor $\ell \in B^{\ast }$ put $Q\ell =S(\ell (X))$. The reproducing property isThis defines the Banach-space Gaussian RKHS even for a degenerate Gaussian measure.

For $h=Sg\in H$, write $\hat{h}(X)=g$. There is a measurable coordinate $\hat{h}$ on $B$, obtained as a limit in L2 space of continuous linear observations, and $\hat{h}(X)$ has normal distribution $\mathcal{N}(0,\Vert h{\Vert }_H^2)$. The Cameron-Martin theorem for a Gaussian measure, in its real separable Banach space form, states that if $\mu$ is the probability law of $X$, then the probability law ${\mu }_h$ of $X+h$ is an equivalent probability measure to $\mu$ precisely when $h\in H$. For such a shift,For $h\notin H$ the two laws are mutually singular measures. The coordinate $\hat{h}$ need not be a continuous functional on $B$, and the sample $X$ need not belong to $H$.

Apply this theorem to the shift $-h$. With $a=\Vert h{\Vert }_H^2/2$, it givesChoose the measurable coordinate so that $\hat{h}(-x)=-\hat{h}(x)$ on a symmetric set of full $\mu$-measure. This is possible by taking an almost-surely convergent subsequence of the approximating linear observations and intersecting its convergence set with its negative. Alternatively the joint law of $(X,\hat{h}(X))$ is invariant under simultaneous negation, directly from those observations and their mean-square convergence. Since $C=-C$ and the centered Gaussian measure is symmetric, the two exponential integrals with signs $+$ and $-$ agree. Consequentlybecause $\cosh v\ge 1$. The symmetric Gaussian translation lower bound is thereforeOnly symmetry and Borel measurability of $C$ are required; convexity is not needed.

There is one degenerate flaw in the last printed claim. The separable Banach space $B=\{0\}$ with $X=0$ has $H=\{0\}$ dense in $B$, and all positive-radius balls have probability one. Nevertheless $\Phi (t)=1$ for every $t>0$. Thus the statement needs the additional hypothesis $B\ne \{0\}$. If the course convention excludes the zero space, this hypothesis is already implicit.

Under this necessary hypothesis, density of $H$ implies that $H$ contains a nonzero vector. Given $0<a<b$, scale that vector to obtain $h\in H$ with $\Vert h{\Vert }_B=(a+b)/2$, and choose $0<\delta <(b-a)/2$. The triangle inequality givesThe centered closed ball $C_{\delta }=\{x:\Vert x{\Vert }_B\le \delta \}$ is a symmetric Borel set. Apply the symmetric Gaussian translation lower bound to getHence the Gaussian norm distribution is strictly increasing when $B$ is nonzero. This proves the strict increase of a Gaussian norm distribution without assuming that its distribution has a density or that spheres have zero probability.

The Cameron-Martin space of a Gaussian measure $\mu =\mathcal{N}(0,\Sigma )$ isThe inverse is defined on the range of ${\Sigma }^{1/2}$. In this infinite-dimensional setting “positive definite” must mean $⟨\Sigma h,h⟩>0$ for every nonzero $h$, rather than a uniform lower bound: a trace-class operator cannot be uniformly positive on an infinite-dimensional Hilbert space.

By the spectral theorem for compact Hermitian operators, choose an orthonormal basis $(e_j)$ with $\Sigma e_j={\lambda }_j{e}_j$, ${\lambda }_j>0$, and ${\sum }_j{\lambda }_j<\infty$. (A strictly positive trace-class operator also forces the ambient Hilbert space to be separable.) In coordinates $h_j=⟨h,e_j⟩$,This is a Hilbert space with its indicated norm, even though its range need not be closed in the ambient norm.

The Cameron-Martin theorem for a Gaussian measure states that the translated law ${\mu }_h(B)=\mu (B-h)=\mathcal{N}(h,\Sigma )(B)$ is an equivalent probability measure to $\mu$ exactly when $h\in E$. For such $h$, its Radon-Nikodym derivative isThe series has mean-square convergence and converges almost surely under $\mu$, since its independent summands have total variance ${\sum }_j{h}_j^2/{\lambda }_j$. Its law is $\mathcal{N}(0,\Vert h{\Vert }_E^2)$, so the density has expected value one. It is sometimes formally written ${\ell }_h(x)=⟨h,x{⟩}_E$, but a typical infinite-dimensional Gaussian sample is not in $E$; the series interpretation is essential. If $h\notin E$, the two laws are mutually singular measures.

The notation $P_0=\mathcal{N}(0,I)$ denotes generalized Gaussian white noise over $H=L^2({\mathbb{R}}^2)$, not a Gaussian measure supported on $H$: the identity is not a trace-class operator there. For example, realize the observation on a suitable space of tempered distributions. Its Cameron-Martin space of a Gaussian measure is $H$. For deterministic $h\in H$, the stochastic pairing $W_m(h)$ has normal distribution $\mathcal{N}(0,\Vert h{\Vert }_H^2)$ under $P_0$ and is the isonormal Gaussian process indexed by $H$.

Interpret the stated maps between Sobolev spaces as bounded linear operators, as usual. In particular $K:H\to H^2({\mathbb{R}}^2)\subset H$ is bounded, so its adjoint operator $K^{\ast }$ is bounded on $H$. Hence $A=K^{\ast }K$ maps $H$ into $H$, which is the property needed here. The stronger smoothing follows by Sobolev duality: boundedness of $K:H^{-4}\to H^{-2}$ gives $K^{\ast }:H^2\to H^4$, and therefore $A:H\to H^4\subset H$. As $\Pi (H)=1$, the shift $Au$ belongs to the noise Cameron-Martin space of a Gaussian measure for $\Pi$-almost every $u$.

The white-noise likelihood for a square-integrable shift, obtained from the Cameron-Martin theorem for a Gaussian measure in its white-noise form, isHere $P_u$ is the law of $Au+\eta$. In an orthonormal basis $(e_j)$ of $H$, with white-noise coordinates $m_j$, the pairing under $P_0$ isFor fixed $u$ this has mean-square convergence and converges almost surely, since ${\sum }_j|⟨Au,e_j⟩{|}^2=\Vert Au{\Vert }_H^2$. The stipulated joint measurability allows the likelihood to be used under ${\nu }_0(du,dm)=\Pi (du)P_0(dm)$. One must not replace this expression by a finite $\Vert m-Au{\Vert }_H^2$, because white noise is not $H$-valued. On the unbounded domain ${\mathbb{R}}^2$, membership of white noise in a global unweighted negative Sobolev space must not be assumed either; the stochastic pairing avoids that issue.

Let $\nu (du,dm)=\Pi (du)P_u(dm)$ be the actual joint law. The shift formula gives $\nu \ll {\nu }_0$ with Radon-Nikodym derivative $L$. DefineFor each fixed admissible $u$, the moment-generating function of a normal distribution givesBy Tonelli theorem, $\int Z(m)P_0(dm)=1$, so $Z(m)<\infty$ for $P_0$-almost every $m$. The likelihood is finite and strictly positive for ${\nu }_0$-almost every $(u,m)$; Fubini's theorem then gives $Z(m)>0$ for $P_0$-almost every $m$. The marginal observation law is $Z(m)P_0(dm)$ and is an equivalent probability measure to $P_0$.

The Bayes formula for a dominated observation model therefore definesTo verify that it is the conditional distribution, for measurable sets $B$ of unknowns and $C$ of data,Thus the posterior distribution is well defined for almost every observation under the actual data law. This is the appropriate almost-everywhere assertion supplied by Bayes' theorem; the stated assumptions do not prescribe a canonical posterior at every exceptional generalized datum.

Put ${\theta }_t=a{t}^{a-1}$. When $a>1/2$,The integrand is deterministic, so the Novikov condition holds. Define an equivalent probability measure $Q$ byThe Cameron-Martin-Girsanov theorem makesa $Q$-Brownian motion on $[0,T]$.