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For the bounded rate diffusion of linear bond pricing in a bounded short-rate diffusion, $D_t{e}^{-(T-t)}{r}_t$ is a bounded martingale with terminal value $D_T{r}_T$. Dividing its conditional expectation by $D_{t}P(t,T)$ through the Bayes formula for conditional expectation gives the displayed forward-measure expectation. It lies between zero and one.

For bounded predictable $\sigma$ on $[0,T]$, the stochastic exponential $Z_s=\exp (\frac{1}{2}{\int }_0^s{\sigma }_{u}d{W}_u-\frac{1}{8}{\int }_0^s{\sigma }_u^{2}du)$ is a positive martingale by the Novikov condition. Under $dQ/dP=Z_T$, the Bayes formula for conditional expectation givesIndeed, the square-root stock ratio is $(Z_T/Z_t)\exp (-\frac{1}{8}{\int }_t^T{\sigma }_u^{2}du)$. This exact identity gives the bounds on square-root stock implied volatility by bounding integrated variance pathwise. The Girsanov theorem gives $W^Q=W-\frac{1}{2}\int \sigma du$. The measure is an auxiliary pricing identity for this payoff, not a claim that it is an equivalent martingale measure for the original stock.

Denote the density process by $L_t=\exp (M_t-[M{]}_t/2)$. The given identities for all $A\in {\mathcal{F}}_s$ implyThus $L$ is a strictly positive true martingale, not merely a local martingale. The Itô formula gives $dL=LdM$, and henceFirst work with the zero-starting corrected increments $U=X-X_0-[X,M]$. Integration by parts givesThe finite-variation bracket has zero quadratic covariation with $L$, so no other term appears. Thus $LU$ is a P-local martingale.

To justify the change of measure, stop $U$ at ${\tau }_n=\inf \{t:|U_t|\ge n\}$, with optional further localization if needed. The product $L_t{U}_{t\wedge {\tau }_n}$ remains a P-local martingale: before the stopping time the cancellation above applies, and afterwards only its bounded constant factor multiplies $dL$. Its absolute value is bounded by $n{L}_t$. On a finite horizon the stopped values of $L$ are uniformly integrable, so Q1(c) upgrades this product to a true martingale. The Bayes formula for conditional expectation now givesFinite-horizon equivalence of the measures transfers the localizing sequence to Q. Therefore $U$ is a continuous Q-local martingale. If $X_0\in L^1(\mathbb{Q})$, adding its constant-in-time process provesThis proves the Girsanov theorem assertion under the usual zero or deterministic initial-value convention. More generally the normalization $M_0=0$ makes $L_0=1$, so P and Q agree on ${\mathcal{F}}_0$ and the required initial integrability is automatic.

There is a literal initial-integrability qualification if neither convention is imposed. On positive integers let $P(n)=2^{-n}$, $Q(n)=c/n^2$, with $c=({\sum }_{n\ge 1}{n}^{-2}{)}^{-1}$, and let every filtration equal the full sigma-algebra. Constant processes $M_t=\log (c{2}^n/n^2)$ and $X_t=n$ are P-martingales, and $e^{M_t}$ is precisely the prescribed density. But ${\mathbb{E}}_Q|X_0|=\infty$ and $[X,M]=0$, so the asserted Q-local martingale cannot have an integrable initial value. This is why initial integrability under a density change is part of the precise formulation.

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 $S_0^2$.

Fix the horizon $T$ and define the half-volatility measure for a square-root stock claim byThe Novikov condition makes $Z$ a true martingale with $\mathbb{E}{Z}_T=1$, and $Z_T>0$ makes $Q$ equivalent to $P$. The Girsanov theorem identifies $W_s^Q=W_s-\frac{1}{2}{\int }_0^s{\sigma }_{u}du$ as a Brownian motion under $Q$.

The exact factorization from the previous part isUsing the Bayes formula for conditional expectation therefore givesThe random integrated variance need not have any independence property under $Q$. Its pathwise bounds still imply $e^{-b^2\tau /8}\le {\mathbb{E}}_Q[e^{-I_{t,T}/8}\mid {\mathcal{F}}_t]\le e^{-a^2\tau /8}$. Inverting the same price function proves $a\le \Sigma (t,T)\le b$ even for volatility adapted to the Brownian motion, on the positive-price event. The zero-price convention is as above. The auxiliary measure here is not asserted to be an equivalent martingale measure for the stock.

Construct the strictly positive rolling one-period bond account using successive one-period zero-coupon bonds:At time $t-1$, invest the entire account value in the bond maturing at $t$. This gives a positive self-financing portfolio usable as a numéraire.

We use the finite-discrete-time fundamental theorem of asset pricing: in a frictionless market with finitely many adapted assets and trading dates, no arbitrage is equivalent to existence of an equivalent martingale measure for asset prices, including dividends, expressed in a positive traded numéraire. Maturing bond payoffs are reinvested in that numéraire. Let $\mathbb{Q}$ be such a measure and $L_t=\mathbb{E}[d\mathbb{Q}/d\mathbb{P}\mid {\mathcal{F}}_t]$ its positive density process. The martingale pricing relation for a unit zero-coupon bond isDefine $Z_t=L_t/{\beta }_t$. The Bayes formula for conditional expectation then givesThe positive expectations are finite because these are the traded finite bond prices; in particular $\mathbb{E}{Z}_T=Z_0{P}_0(T)$ when initial information is trivial. Normalize $Z_0=1$ without changing any ratio. With nontrivial initial information the same identities are conditional on that information. The state-price density need not be unique when the bond market is incomplete; existence is sufficient.

Since the density martingale is uniformly integrable, $Z_t={\mathbb{E}}_{\mathbb{P}}[Z_{\infty }\mid {\mathcal{F}}_t]$. Equivalence of measures, as stipulated, means $Z_{\infty }>0$ almost surely. Write $L=\int Z^{-1}dZ$ for the stochastic logarithm. The Itô formula gives $\log Z=L-⟨L⟩/2$, and hence the quadratic covariation in the drift correction isIt involves the local-martingale part of $\log Z$; its continuous finite-variation part has zero covariation.

With $Y=X-C$, the Itô product rule now gives an exact cancellation:Thus $ZY$ is a P-local martingale. To transfer this conclusion rigorously, set ${\tau }_n=\inf \{t:|Y_t|\ge n\}$. The same computation for the stopped $Y$ givesThis is again a P-local martingale. Moreover its absolute value is at most $n{Z}_t$. Uniform integrability of $Z$ makes the family over bounded stopping times uniformly integrable, so the product is a true martingale. This is the bounded-process density-product criterion.

The Bayes formula for conditional expectation therefore gives, for $s\le t$,Continuity gives ${\tau }_n\uparrow \infty$, proving $X-⟨\log Z,X⟩$ is a Q-local martingale. This proves the needed Girsanov theorem rather than invoking it.

For the Brownian-filtration conclusion, use this precise Brownian martingale representation theorem: every continuous local martingale in the usual augmentation of the natural Brownian filtration is an Itô integral with a predictable integrand locally square integrable in time. In particularDefineOn each finite horizon a strictly positive continuous $Z$ has a positive pathwise minimum. Thus ${\int }_0^t{\alpha }_s^{2}ds<\infty$ almost surely, andThe already proved measure-change result makes $\hat{W}=W-\int \alpha ds$ a continuous Q-local martingale. Its quadratic variation is $t$, unchanged by its finite-variation correction. The Lévy characterization of Brownian motion provesNo additional exponential-integrability condition is needed, since the equivalent uniformly integrable density process is already given.

The terminal density is strictly positive and has expectation one under the usual deterministic initial bond-price convention. Its density process isFor $0\le s\le t\le T$, the Bayes formula for conditional expectation under a change of measure givesThe same calculation at $s=0$ gives the finite expectation $1/P(0,T)$, so this is a true martingale, not just a formal conditional identity. Thus the continuous-time bank account measured in units of the maturity-$T$ bond is a ${\mathbb{Q}}_T$-martingale. This is the forward measure change of numéraire. If the initial bond price were random rather than given, integrability of its reciprocal would need to be included for this true-martingale assertion.

Use the same affine drift cancellation with terminal conditions $a(0)=0$, $b(0)=1$. It gives $a(\tau )=0$, $b(\tau )=e^{-\tau }$, soThe process is bounded, which justifies the conditional expectation identity. The Bayes formula for conditional expectation for the forward measure now givesThe denominator is positive; the ratio lies in $[0,1]$ and equals $r_T$ at maturity. This is the forward-measure terminal rate in a linear bond model.

Since $|Z_t{Y}_t|\le C{Z}_t$ and $Z$ is uniformly integrable, the continuous local martingale $ZY$ is locally in Doob's class and hence is a true martingale. For $A\in {\mathcal{F}}_s$ and $s\le t$, Bayes formula for conditional expectation givesThe bounded process $Y$ is integrable under $\tilde{\mathbb{P}}$, so this is exactly the martingale property.