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The deterministic clock is $a(t)=e^{2\lambda t}$, so $a(0)=1$. For $s\le t$, the martingale property of Brownian motion givesIntegrability follows from the Gaussian second moment. Thus $X$ is a martingale in the time-changed filtration.

Its increments have variance $a(t)-a(s)$ and are independent of ${\mathcal{G}}_s$. Equivalently, $X_t^2-a(t)$ is a martingale. Since a quadratic variation starts at zero, its compensator must be normalized at $t=0$:The initial random variable $X_0=B_1$ does not contribute to quadratic variation. The expression $e^{2\lambda t}$ alone would violate $[X{]}_0=0$.

If $X$ is a P-Brownian motion, its starting value is deterministic, so Q5(a) applies. The correction $[X,M]$ has finite variation and does not alter quadratic variation; Q1(b) gives $[\tilde{X}{]}_t=t$ under Q. The Lévy characterization of Brownian motion therefore identifies $\tilde{X}$ as a Q-Brownian motion with the same starting value.

For the likelihood, use Wiener measure ${\mathbb{Q}}^x$ as the reference measure and write its coordinate as ${\omega }_t=x+W_t$. Setand let $L=\mathcal{E}(M)$ be its stochastic exponential. We must prove it is a true martingale; unverified large-horizon Novikov estimates would not suffice.

Fix $T$ and stop at ${\tau }_n=\inf \{s:|{\omega }_s|\ge n\}$, with $n>|x|$. The bounded integrand satisfies the Novikov condition, so $L_{T\wedge {\tau }_n}$ defines a probability $P_n$ on ${\mathcal{F}}_T$. By the result just established,is Brownian under $P_n$ on this horizon. Therefore $d{\omega }_s=d{W}_s^{(n)}-\mu 1_{\{s\le {\tau }_n\}}{\omega }_{s}ds$. The Gronwall inequality givesThe Doob L2 maximal inequality then bounds its squared expectation by $C_T=2e^{2|\mu |T}(x^2+4T)$, uniformly in $n$. In particular $P_n({\tau }_n\le T)\le C_T/n^2$. On $\{{\tau }_n>T\}$ the stopped and unstopped densities agree, soContinuous Brownian paths are bounded on $[0,T]$, and the indicators increase to one. The monotone convergence theorem proves $\mathbb{E}{L}_T=1$. Since a nonnegative local martingale is a supermartingale, constant expectation at every time makes $L$ a true martingale.

Under its tilted measure, ${\hat{W}}_t=W_t+\mu {\int }_0^t{\omega }_{s}ds$ is Brownian and $d{\omega }_t=d{\hat{W}}_t-\mu {\omega }_{t}dt$. The explicit linear solution gives uniqueness of this law, identifying it with ${\mathbb{P}}^x$. Consequently the Ornstein-Uhlenbeck likelihood relative to Wiener measure isThis is a finite-horizon density statement; it does not assert absolute continuity of the full infinite-horizon path laws.

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 $S_0^2$.

A one-factor short-rate model uses a single scalar state, normally the short rate itself. Under the physical probability measure writeThe second process is the continuous-time bank account. Specifying the physical diffusion alone does not determine bond prices: it must be accompanied by the market price of risk $\lambda (t,r)$, or equivalently by pricing-measure dynamics. If the Girsanov theorem applies to the likelihood $\mathcal{E}(-\int \lambda dW)$, then under the risk-neutral measureChanging measure alters drift, not quadratic variation. The coefficients need conditions for a nonexplosive solution and for the relevant discounted prices and measure-change density to be true martingales.

A unit zero-coupon bond maturing at $T$ has priceThe Markov property makes current time and rate sufficient state variables. Apply Itô formula to $P(t,r_t;T)/B_t$. Its drift must vanish, giving the short-rate bond pricing equationConversely, under the appropriate boundary and integrability conditions the Feynman-Kac formula identifies its solution with the displayed discounted expectation. For a terminal rate-dependent payoff $H(r_T)$, the same equation has terminal condition $H$ instead of one. A coupon bond is a sum of such unit-bond prices weighted by the coupon payments.

The common market price of risk can also be derived from trading, instead of being introduced arbitrarily for each maturity. Under the physical measure a bond's diffusion exposure is $\eta P_r$ and its excess drift above bank financing is $P_t+\mu P_r+{\eta }^2{P}_{rr}/2-rP$. For two nondegenerate traded bonds, combine their holdings to cancel their Brownian exposure. Absence of arbitrage requires that combination to earn the bank rate; hence their excess-drift-to-exposure ratios must coincide. Calling that common ratio $\lambda$ yields the pricing equation with ${\mu }_Q=\mu -\eta \lambda$ for every maturity. A derivative $V(t,r)$ is then hedged locally by holding $V_r/P_r$ units of a nondegenerate bond and financing the remaining value through the bank account. Their Brownian exposures agree, and substituting both pricing equations proves the drift gain identity. This is short-rate diffusion hedging; it requires $\eta P_r\ne 0$, rather than a bare assertion that one factor automatically gives completeness everywhere.

The entire yield curve is generated consistently from these bond prices. The continuously compounded yield and the instantaneous forward rate areThese encode different summaries: the yield averages the forward rate over the remaining lifetime. Bond prices must satisfy $P(t,t)=1$ and be positive; they need not be below one in models permitting negative rates.

The Vasicek model has constant pricing dynamicsSolving the linear equation gives $r_{t+u}=b+(r_t-b)e^{-au}+\eta {\int }_t^{t+u}{e}^{-a(t+u-v)}d{W}_v^Q$. Consequently, with $\tau =T-t$ and $B(\tau )=(1-e^{-a\tau })/a$,This conditional normal random variable has variance ${\eta }^2{\int }_0^{\tau }B(u{)}^{2}du$. Its normal exponential moment yieldsEvaluating the elementary integral givesThis explicit form makes estimation and valuation convenient. The rate is Gaussian, its conditional mean reverts to $b$, and its long-run variance is ${\eta }^2/(2a)$. Its disadvantage, when nonnegative rates are desired, is a positive probability of negative rates.

The CIR model replaces constant volatility by square-root volatility:The standard solution stays nonnegative. The Feller positivity condition for the CIR model is $2ab\ge {\eta }^2$ in this parameterization; with a positive initial rate it makes zero inaccessible. Below this threshold zero can be reached, although the process does not cross into negative rates. For an affine diffusion bond pricing ansatz $P=A(\tau )e^{-B(\tau )r}$, substitution into the bond equation and matching its constant and rate coefficients givesWriting $d=\sqrt{a^2+2{\eta }^2}$, factor the quadratic in the first equation and integrate from $B=0$; integrating the second equation then gives the CIR bond pricing formulasThe square-root form produces an affine term structure while respecting nonnegative rates. These parameters are pricing-measure parameters; historical mean-reversion estimates alone do not determine them without a risk-premium specification.

Both preceding constant-parameter models constrain the family of initial curves. The Hull-White model allows a deterministic time-dependent drift:For a smooth desired initial forward curve $f_0(t)$, its exact forward-curve fit in the Hull-White model is obtained by takingTo derive this rather than merely assert calibration, write $r_t=x_t+\varphi (t)$, where $d{x}_t=-a{x}_{t}dt+\eta d{W}_t^Q$, $x_0=0$. The conditional Gaussian-integral calculation at time zero givesso its forward curve is $\varphi (T)-{\eta }^{2}B(T{)}^2/2$. Choose $\varphi (T)=f_0(T)+{\eta }^{2}B(T{)}^2/2$. Differentiating this shift and adding $a\varphi (t)$ gives exactly the displayed $\beta (t)$. Thus $P(0,T)=\exp (-{\int }_0^T{f}_0(u)du)$ for every maturity, while $a$ and $\eta$ govern the stochastic fluctuations. This Gaussian model still allows negative rates.

Finally, one-factor models are economical and often give explicit bond prices, but their single Brownian driver is restrictive. Bond diffusion exposures $v_i=\eta P_{i,r}$ give the instantaneous covariance matrix $v{v}^T$, of rank at most one. Nondegenerate maturity changes are instantaneously perfectly correlated, with correlation $+1$ or $-1$, though their finite-horizon correlations need not equal one. A single factor therefore cannot independently describe level, slope and curvature shocks to the whole yield curve. Multiple factors or a full forward-rate model offer greater flexibility. A valid one-factor model must specify the pricing drift, fit the initial curve to the required accuracy, and handle boundary and martingale conditions; naming a rate diffusion alone is not a complete pricing model.

For the integrable terminal bracket criterion, assume $M_0=0$, or more generally $M_0\in L^2$. This initial-value assumption is necessary if it is not already part of the notation: an integrable but non-square-integrable random constant process has zero quadratic variation and still fails square-integrability.

For the zero-starting case, choose a localizing sequence ${\tau }_n$ bounding both $|M|$ and its quadratic variation. The stopped processes are true square-integrable martingales, and the Itô isometry givesFor each fixed time these variables are uniformly integrable by their bounded second moments. Almost sure convergence as $n\to \infty$ is therefore convergence in $L^1$. Passing to the limit in the stopped conditional-expectation identity gives $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$: the local martingale is a true martingale. The Fatou lemma also givesso $M$ belongs to the global ${\mathcal{M}}_c^2$ space. For general square-integrable $M_0$, apply this argument to $M-M_0$ and use $|M_t{|}^2\le 2|M_0{|}^2+2|M_t-M_0{|}^2$.

Put $A_t=[M{]}_t$. Its continuous, strictly increasing paths start at zero and tend to infinity. Thus the inverse $T_s$ is finite and continuous, andso each $T_s$ is a stopping time. Set ${\mathcal{G}}_s={\mathcal{F}}_{T_s}$ and ${\beta }_s=M_{T_s}$. This is the inverse-clock construction in the Dambis-Dubins-Schwarz theorem.

To verify its martingale property, stop $M$ at times ${\tau }_n$ bounding its path, its bracket and time itself. The stopped $M$ and $M^2-A$ are bounded martingales. Optional sampling at $T_s\wedge {\tau }_n$ and $T_t\wedge {\tau }_n$ shows that the time-changed process, stopped at $A_{{\tau }_n}$, is a martingale and that its square minus the stopped clock $s\wedge A_{{\tau }_n}$ is a martingale. These clock stopping times tend to infinity because $A_{\infty }=\infty$. Therefore $\beta$ is a continuous local martingale with ${\beta }_0=0$ and quadratic variation $[\beta {]}_s=s$. The Lévy characterization of Brownian motion identifies it as a ${\mathcal{G}}_s$-Brownian motion.

Strict monotonicity and continuity give $A_{T_s}=s$ and $T_{A_t}=t$, soThe inverse-clock argument proves the asserted representation; the random clock need not be independent of the Brownian motion it produces.

For $t>1$, $B_{t^2}$ is not measurable with respect to the original Brownian filtration ${\mathcal{F}}_t$: its conditional variance is $t^2-t>0$. Thus it is not even adapted on the full half-line. It also fails the martingale condition before time one. With $s=1/4$, $t=3/4$,The appropriate filtration isBrownian independent increments make $M_t=B_{t^2}$ a continuous ${\mathcal{G}}_t$-martingale, with quadratic variation $[M{]}_t=t^2$. Define the stochastic integralThe value of the integrand at zero is immaterial; it is square-integrable against the bracket because ${\int }_0^t(2s{)}^{-1}d(s^2)=t$. Consequently $[W{]}_t=t$, and the Lévy characterization of Brownian motion makes $W$ a ${\mathcal{G}}_t$-Brownian motion. Associativity of stochastic integrals gives the Brownian motion under a quadratic time change representationThe function $f$ is continuous at zero. One can first verify the inverse-integrand identity on $[\epsilon ,t]$ and then let $\epsilon \downarrow 0$ using the Itô isometry, since $\mathbb{E}{M}_{\epsilon }^2={\epsilon }^2$.

Let $V_t$ be the total variation of $X$ on $[0,t]$, and set ${\tau }_n=\inf \{t:V_t\ge n\}$. Adaptation and continuity of $V$ make these stopping times; finite variation on every compact interval gives ${\tau }_n\uparrow \infty$. The stopped process $U=X^{{\tau }_n}$ satisfies $|U_t|\le n$ and has total variation at most $n$. It is a martingale by the bounded local martingale criterion, and is square-integrable.

Fix $t$ and take deterministic partitions $0=t_0<\cdots <t_m=t$ with mesh tending to zero. The martingale-difference orthogonality of the increments givesPathwise, however,by uniform continuity on $[0,t]$. These sums are bounded by $n^2$, so the dominated convergence theorem gives $\mathbb{E}{U}_t^2=0$. Hence $U_t=0$ almost surely. Take all rational $t$ on a common probability-one event and use continuity to obtain $U\equiv 0$. Then let $n\to \infty$. This proves the continuous finite-variation local martingale is constant result in its zero-starting form:The proof does not assume the uniqueness of quadratic variation that is to be established next.

For two candidates $A,C$, their difference is a continuous adapted finite-variation process. It is also a local martingale, sinceApply part (b) after subtracting its initial value. ConsequentlyThis proves the uniqueness of an increasing square compensator when the initial value is prescribed, in particular under the standard quadratic variation normalization $A_0=C_0=0$.

The printed assertion omits that normalization and is literally false. For standard Brownian motion $Y$, both $A_t=t$ and $C_t=t+1$ are continuous adapted increasing processes; both $Y_t^2-t$ and $Y_t^2-t-1$ are martingales. Thus uniqueness holds after fixing the initial value; otherwise it holds only up to an initial constant.

For a simple predictable process, define its stochastic integral byIt is constant after $t_n$, so the symbol $(H\cdot M{)}_{\infty }$ means its value there. Put ${\Delta }_{k}M=M_{t_{k+1}}-M_{t_k}$. If $k<\ell$, then $Z_k{\Delta }_{k}M{Z}_{\ell }$ is ${\mathcal{F}}_{t_{\ell }}$-measurable, and the conditional expectation of ${\Delta }_{\ell }M$ given that sigma-algebra is zero. Boundedness of the coefficients and square-integrability of $M$ justify the expectations, so all cross terms vanish:For each diagonal term, the martingale increment identity also givesIndeed, the omitted term $2Z_k^2{M}_{t_k}{\Delta }_{k}M$ has expectation zero. Apply the assumed martingale property of $M^2-[M]$ to replace this difference of squares by the corresponding quadratic variation increment. ThusThe intervals $(t_k,t_{k+1}]$ in the Lebesgue–Stieltjes integral exactly match these increments, including possible jumps of $M$.

Assume, as in the Bayesian model, that the unknown drift $\alpha$ is independent of the driving Brownian motion, with known positive prior precision ${\tau }_0$. Recover the observed signal from the positive stock price:For a fixed time $t$, the centered observed path $Y_s-(s/t)Y_t$ is a Brownian bridge independent of $(\alpha ,Y_t)$: it is independent of the prior and has zero covariance with $W_t$ in their joint Gaussian law. Thus the whole observed history has the same posterior information about $\alpha$ as its endpoint. This proves Brownian endpoint sufficiency for a constant drift rather than assuming it.

The likelihood of that endpoint, as a function of $\alpha$, is proportional to $\exp (\alpha Y_t-{\alpha }^{2}t/2)$. Multiplying by the normal distribution prior and completing the square gives the Gaussian Brownian drift filterDefine the observed innovation process ${\hat{W}}_t=Y_t-{\int }_0^t{\hat{\alpha }}_{s}ds$. It is a continuous observed-filtration martingale: for $s<u$, independence of future noise and the tower property of conditional expectation give $\mathbb{E}[\alpha -{\hat{\alpha }}_u\mid {\mathcal{F}}_s^S]=0$, so the conditional mean of its increment vanishes. Its quadratic variation is $t$, since subtracting the finite-variation drift from $Y$ does not change quadratic variation. The Lévy characterization of Brownian motion makes $\hat{W}$ an observed-filtration Brownian motion. Differentiating the posterior mean givesConversely, the last equation determines $\hat{\alpha }$ from $\hat{W}$ with deterministic coefficients and known initial value, and $Y=\hat{W}+\int \hat{\alpha }dt$. Thus the observed and innovation filtrations coincide, supplying the natural Brownian information needed for replication. Consequently the observed stock dynamics are $d{S}_t/S_t=\sigma d{\hat{W}}_t+\sigma {\hat{\alpha }}_{t}dt$. Its observed market price of risk is ${\kappa }_t={\hat{\alpha }}_t-r/\sigma$, and its normalized state-price density solvesThe posterior mean is unbounded, so bounded-coefficient Novikov condition reasoning from Question 2 is not available automatically. A direct finite-horizon argument supplies true pricing. Relative to driftless Wiener measure for $Y$, the Bayesian mixture has observation densityA law with observation drift $k=r/\sigma$ instead has density $e^{k{Y}_t-k^{2}t/2}$. ThereforeThis is the finite-horizon pricing density for Gaussian drift learning. The ratio changes the physical observed law into a genuine probability law, hence has mean one. Indeed, differentiation of the explicit likelihood, using $d{Y}_t=d{\hat{W}}_t+{\hat{\alpha }}_{t}dt$, givesThe Itô formula therefore gives $d{D}_t=-D_t{\kappa }_{t}d{\hat{W}}_t$, agreeing with the stochastic exponential above. Under this pricing law the stock drift is $\sigma k=r$.

For logarithmic utility, terminal marginal utility is $1/w_T$. The state-price budget constraint and the marginal condition imply $w_T^{\ast }=1/(y{\zeta }_T)$ and $1/y=w_0$. Since ${\zeta }_T{w}_T^{\ast }=w_0$ is constant, its conditional-expectation pricing process is constant as well. Thus the entire optimal wealth process isIndeed, for any other nonnegative admissible terminal wealth $X$, the elementary inequality $\log (X/w_T^{\ast })\le X/w_T^{\ast }-1$ givesThis directly verifies optimality. Expected candidate log wealth is finite because $\mathbb{E}{\hat{\alpha }}_t^2={\hat{\alpha }}_0^2+{\tau }_0^{-1}-({\tau }_0+t{)}^{-1}$ and its time integral is finite on every fixed horizon. Applying the Itô formula to $1/\zeta$ givesThe observed self-financing portfolio with risky dollar fraction ${\pi }_t$ has diffusion coefficient ${\pi }_t\sigma$. Matching the coefficients proves the log-optimal investment with Gaussian drift learning ruleThe remaining dollars are held in the bank account; the fraction may be negative or exceed one because the question imposes neither short-sale nor borrowing restrictions.

First justify that the logarithm is defined at all times from one onward. For planar Brownian motion started at $z\ne 0$, stop on exiting the annulus $\epsilon <|x|<R$, where $\epsilon <|z|<R$. This exit time $\rho$ is finite almost surely: the stopped identity $|B_t{|}^2-2t$ gives $\mathbb{E}(t\wedge \rho )\le R^2/2$. The function $f(x)=\log |x|$ is harmonic on the annulus, so its stopped process is a bounded martingale. Optional stopping givesHitting zero before the outer circle would require hitting every inner circle first. It therefore has probability zero. Taking a countable sequence of outer radii tending to infinity proves that planar Brownian motion avoids a fixed point. Since $B_1$ has a density and is nonzero almost surely, condition on ${\mathcal{F}}_1$ to apply this result to the future path.

For $x\ne 0$, $\nabla f(x)=x/|x{|}^2$ and $\Delta f(x)=0$ in two dimensions. The Itô formula on annuli therefore givesTo localize explicitly, let ${\tau }_n=\inf \{t\ge 1:|B_t|\notin (1/n,n)\}$. These stopping times increase to infinity almost surely: on each compact time interval the continuous path has a finite maximum and, by point avoidance, a positive minimum radius. Up to ${\tau }_n$, the stochastic integrand has norm at most $n$. Its stopped integral is square-integrable on every finite horizon. The initial value $\log |B_1|$ is integrable, as follows from the Rayleigh-density calculation in part d. Hence $X^{{\tau }_n}$ is a true martingale from time one and $X$ is a local martingale. This also gives its quadratic variation from time one as ${\int }_1^t|B_s{|}^{-2}ds$.

For every zero-starting continuous local martingale $M$, there is a unique, up to indistinguishability, continuous adapted increasing process $[M]$, with $[M{]}_0=0$, such thatThis is its quadratic variation. It also satisfies, for deterministic partitions whose mesh tends to zero,with convergence uniformly on compact time intervals in probability. The theorem includes existence and finiteness on each compact interval; monotonicity is understood pathwise outside one null set.

Uniqueness follows because the difference of two candidate increasing continuous processes is both a finite-variation process and a continuous local martingale, hence is identically zero from its zero initial value. For a continuous semimartingale $X=X_0+M+A$, its quadratic variation is $[X]=[M]$, since continuous finite-variation terms have zero quadratic variation. Polarization defines the quadratic covariation by $[M,N]=([M+N]-[M-N])/4$.

The correct quadratic variation of a stochastic integral identity isThe brackets around the final integrator are missing in the PDF. Without them the assertion is false: for $H=1$ and $M$ Brownian motion, the left side is $t$ while the printed right side is $B_t$.

For elementary predictable $H$, the integral is obtained by scaling each martingale increment by its known coefficient. Quadratic variation on each interval is therefore scaled by the square of that coefficient, giving the displayed formula. To pass to a general $L^2(M)$ integrand on $[0,T]$, take elementary $H_n\to H$ in the integrand norm and put $N_n=H_n\cdot M$, $N=H\cdot M$. The Itô isometry gives $\mathbb{E}(N_n-N{)}_T^2\to 0$, hence $\mathbb{E}[N_n-N{]}_T\to 0$. The bracket Cauchy-Schwarz boundand polarization show that $[N_n]\to [N]$ uniformly in probability. On the other side, Cauchy-Schwarz for the quadratic-variation measure givesTaking limits proves the formula in the square-integrable case. Stopping and pasting as in part b proves it for locally bounded predictable integrands and continuous local martingales.