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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.

Assume ${\sigma }^2>0$; otherwise the scaled sum is deterministic and no nontrivial rate function is possible. The sum has normal distribution $B_N\sim N(N\mu ,N{\sigma }^2)$. For $X_N=B_N/N^{1+\epsilon }$, use the faster large-deviation speed $b_N=N^{1+2\epsilon }$. The moment-generating function of a normal distribution yieldsThe limiting cumulant-generating function is finite and differentiable on all of $\mathbb{R}$, hence satisfies the Gärtner–Ellis theorem. Its Legendre-Fenchel transform is obtained by maximizing $\theta x-{\sigma }^2{\theta }^2/2$, whose maximizer is $\theta =x/{\sigma }^2$. ThusThe drift disappears because the scaled mean is $\mu /N^{\epsilon }\to 0$. The rate function is nontrivial and good. The corrected scaled cumulant-generating function used here contains an expectation; the corresponding formula in the paper's reference material omits it.

The exit time $T$ is finite almost surely. Indeed, if the Brownian motion is still in $(-b,a)$ at an integer time, its next unit increment exceeds $a+b$ with a fixed probability $p>0$ independent of the past; on this event it must exit before the next integer time. Consequently $\mathbb{P}(T>n)\le (1-p{)}^n\to 0$. Path continuity gives $B_T\in \{a,-b\}$, and the two exit events partition the probability space up to null sets.

For each real $\lambda$, the exponential Brownian martingale $Z_t^{\lambda }=\exp (\lambda B_t-{\lambda }^{2}t/2)$ has expectation one. This follows directly from the moment-generating function of a normal distribution and independent increments. At $t\wedge T$ its value is bounded by $e^{|\lambda |\max (a,b)}$. Bounded-time optional stopping and then dominated convergence therefore giveWrite $p_{\lambda }=\mathbb{E}[e^{-{\lambda }^{2}T/2}{1}_{\{T=T_a\}}]$ and $q_{\lambda }=\mathbb{E}[e^{-{\lambda }^{2}T/2}{1}_{\{T=T_{-b}\}}]$. Applying the identity to $\lambda$ and $-\lambda$ yieldsFor $\lambda \ne 0$, multiply the first equation by $e^{\lambda b}$, the second by $e^{-\lambda b}$, and subtract. Solving the resulting system givesAdding and using the sum formula for the hyperbolic sine gives the asymmetric Brownian interval-exit transform:At $\lambda =0$, the printed sine ratio is $0/0$ and needs its removable extension. Dominated convergence as $\lambda \to 0$ givesThus the formulas hold for every real parameter with this endpoint convention. The expectation on the left of the first formula is legible in the original PDF; the converted TeX's nested exponential is an OCR error.

Use column vectors for dollar holdings ${\theta }_t$ and put $b=\mu -r1$, $h={\sigma }^{-1}b$ and $V=\sigma {\sigma }^T$. The normalized state-price density isThe vector $h$ is the market price of risk. The density $e^{rt}{\zeta }_t$ defines an equivalent risk-neutral measure, under which $W_t^{\mathbb{Q}}=W_t+ht$ is Brownian motion and every stock has drift $r$. A terminal claim $H$ with finite state-price cost is priced at ${\zeta }_t^{-1}\mathbb{E}[{\zeta }_{T}H\mid {\mathcal{F}}_t]$. Nonsingularity of $\sigma$ and the natural Brownian market information make this a complete market; in particular the optimal nonnegative payoff below can be replicated.

Let $q=n^{-1}1$ and $v={\sigma }^{T}q$. Apply the Itô formula to each log price, whose return variance is $V_{ii}$:Summing and dividing by $n$ proves the geometric stock index formulaThus $J_t=J_0\exp (v^T{W}_t+a_{J}t)$, where $a_J=q^T\mu -\mathrm{tr}(V)/(2n)$. The factor $1/n$ applies to the whole bracket. The geometric average need not itself be a self-financing portfolio; its instantaneous drift is $a_J+|v{|}^2/2$.

If ${\theta }_t^i$ is the dollar amount invested in stock $i$, the bank account holds $w_t-1^T{\theta }_t$. The self-financing portfolio equation isEquivalently, ${\pi }_t={\theta }_t/w_t$ gives $d{w}_t/w_t=(r+{\pi }_t^{T}b)dt+{\pi }_t^T\sigma d{W}_t$. For share-holding notation replace ${\theta }_t^i$ by $S_t^i$ times the number of shares.

Static expected utility maximization uses the terminal budget $\mathbb{E}[{\zeta }_T{w}_T]\le w_0$. Since $J_T$ is an exogenous positive benchmark, differentiating $U(x/J_T)$ in $x$ gives $J_T^{-1}{U}^{\prime }(x/J_T)$. An interior Lagrange multiplier $\lambda >0$ for the budget therefore givesFor CRRA utility, write $p=1/R$. Since $U^{\prime }(y)=y^{-R}$, solving this marginal equation givesEvery positive lognormal moment is finite here, so the budget determines the constant uniquely. PutThenThis is also a sufficiency condition: concavity gives $U(X/J_T)-U(w_T^{\ast }/J_T)\le \lambda {\zeta }_T(X-w_T^{\ast })$, and taking expectations and using the budget inequality proves global optimality.

To find the whole optimal process, set $c=a_J-r-|h{|}^2/2$ andSince ${\zeta }_t{J}_t=J_0\exp ((v-h{)}^T{W}_t+ct)$, the moment-generating function of a normal distribution givesPricing the terminal payoff yields the explicit benchmark-relative power-utility portfolio wealthIts proportional diffusion coefficient is $\ell =ph+(1-p)v$. Since the wealth diffusion coefficient for dollar fractions is ${\sigma }^T\pi$, replication requiresUsing ${\sigma }^{-T}h=V^{-1}b$ gives the requested fixed proportions:The drift matches the self-financing portfolio equation as well. Indeed, $\ell =h+(1-p)(v-h)$, so the drift of $\log w^{\ast }$ from the explicit formula isConsequently an equivalent form of the solution isThe bank account fraction is $1-1^T{\pi }^{\ast }$, with negative values representing borrowing. The first stock term is ordinary power-utility risk exposure, while the second hedges the random benchmark. At the logarithmic limit $R=1$ the benchmark term disappears, consistently with $\log (w_T/J_T)=\log w_T-\log J_T$.

The moment-generating function of a normal distribution givesIndependence of $X_{n+1}$ from ${\mathcal{F}}_n$ impliesEvery exponential here is integrable. Thus the process is a martingale exactly when $-\lambda +{\lambda }^2/2=0$. Its two roots are $0$ and $2$, and the unique positive root isThe resulting exponential martingale of a random walk is nonnegative and has expectation $1$ at every finite time.

The natural filtration of $S$ is the natural Brownian filtration: since $\sigma >0$, the equation $W_t=(\log (S_t/S_0)-\mu t)/\sigma$ recovers the entire Brownian motion history from the stock history. For $s<t$, independent increments and the moment-generating function of a normal distribution giveAll these expectations are finite. Since $S_s>0$, the martingale identity holds for every $s<t$ exactly when the last exponential equals one. The required logarithmic drift isThis is the distinction between the drift of $\log S$ and the drift of $S$: here the latter is zero.

The Itô formula verifies the geometric Brownian motion solutionThis is valid for negative $x$ as well as positive $x$; $x=0$ gives the zero solution. Pathwise uniqueness follows from the Lipschitz continuity of the linear coefficients. The moment-generating function of a normal distribution gives $\mathbb{E}{e}^{k\sigma B_t}=e^{k^2{\sigma }^{2}t/2}$, and thereforeThese moments are finite on every finite horizon. If the convention $0\in \mathbb{N}$ is used, the zeroth moment is $1$.

Let $\tau =T-t$. Conditional on ${\mathcal{F}}_t$ under the risk-neutral measure,Using the moment-generating function of a normal distribution in the risk-neutral expectation gives the Power payoff in the Black-Scholes model valueIn particular,The delta hedge holdsunits of the risky asset.

For every $\lambda \in {\mathbb{R}}^2$, the scalar ${\lambda }^{T}X$ is normal with mean ${\lambda }^T\mu$ and variance ${\lambda }^{T}V\lambda$. The moment-generating function of a normal distribution therefore givesThus $\mu$ and $V$ determine the moment-generating function on all of ${\mathbb{R}}^2$. By the uniqueness theorem for moment-generating functions, they uniquely determine the probability distribution of $X$.

Any unbiased linear estimator has the form considered in part (b), and its error is the linear combination of independent normal random variablesThe moment-generating function of a normal distribution therefore gives, for every nonzero real $\theta$,Since the exponential function is strictly increasing and ${\theta }^2{\sigma }^2/2>0$, minimizing this exponential moment is exactly the same as minimizing ${\sum }_i{a}_i^2$. Part (b) shows that the unique minimizer is $a_i=x_i/S_{xx}$. Thus, independently of the sign or magnitude of $\theta$,

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.

Define a new probability measure by the strictly positive Radon-Nikodym derivativeThe moment-generating function of a normal distribution shows that this density has expectation one, so $\mathbb{Q}$ is a probability measure equivalent to $\mathbb{P}$. Moreover,Thus the discounted risky asset is a one-period $\mathbb{Q}$-martingale. The fundamental theorem of asset pricing implies that there is no arbitrage.