# Brownian Bridges

A Brownian bridge can be defined as standard Brownian motion conditioned on hitting zero at a fixed future time T, or as any continuous process with the same distribution as this. Rather than conditioning, a slightly easier approach is to subtract a linear term from the Brownian motion, chosen such that the resulting process hits zero at the time T. This is equivalent, but has the added benefit of being independent of the original Brownian motion at all later times.

Lemma 1 Let X be a standard Brownian motion and ${T > 0}$ be a fixed time. Then, the process

 $\displaystyle B_t = X_t - \frac tTX_T$ (1)

over ${0\le t\le T}$ is independent from ${\{X_t\}_{t\ge T}}$.

Proof: As the processes are joint normal, it is sufficient that there is zero covariance between them. So, for times ${s\le T\le t}$, we just need to show that ${{\mathbb E}[B_sX_t]}$ is zero. Using the covariance structure ${{\mathbb E}[X_sX_t]=s\wedge t}$ we obtain,

 $\displaystyle {\mathbb E}[B_sX_t]={\mathbb E}[X_sX_t]-\frac sT{\mathbb E}[X_TX_t]=s-\frac sTT=0$

as required. ⬜

This leads us to the definition of a Brownian bridge.

Definition 2 A continuous process ${\{B_t\}_{t\in[0,T]}}$ is a Brownian bridge on the interval ${[0,T]}$ if and only it has the same distribution as ${X_t-\frac tTX_T}$ for a standard Brownian motion X.

In case that ${T=1}$, then B is called a standard Brownian bridge.

There are actually many different ways in which Brownian bridges can be defined, which all lead to the same result.

• As a Brownian motion minus a linear term so that it hits zero at T. This is definition 2.
• As a Brownian motion X scaled as ${tT^{-1/2}X_{T/t-1}}$. See lemma 9 below.
• As a joint normal process with prescribed covariances. See lemma 7 below.
• As a Brownian motion conditioned on hitting zero at T. See lemma 14 below.
• As a Brownian motion restricted to the times before it last hits zero before a fixed positive time T, and rescaled to fit a fixed time interval. See lemma 15 below.
• As a Markov process. See lemma 13 below.
• As a solution to a stochastic differential equation with drift term forcing it to hit zero at T. See lemma 18 below.

There are other constructions beyond these, such as in terms of limits of random walks, although I will not cover those in this post.

It is well known that Brownian motion ${X_t}$ has scaling invariance, so that it has the same distribution as ${S^{-1/2}X_{tS}}$ for any fixed positive real S. Brownian bridges inherit the same invariance property.

Lemma 3 If ${{\it B}}$ is a Brownian bridge on the interval ${[0,T]}$ then ${S^{-1/2}B_{tS}}$ is a Brownian bridge on the interval ${[0,T/S]}$. In particular, ${T^{-1/2}B_{tT}}$ is a standard Brownian bridge.

Proof: We may suppose that ${B_t=X_t-(t/T)X_T}$ for a standard Brownian motion X. As ${\tilde X_t=S^{-1/2}X_{tS}}$ is also a standard Brownian motion, then

 $\displaystyle S^{-1/2}B_{tS}=S^{-1/2}\left(X_{tS}-\frac{tS}TX_{T}\right)=\tilde X_t-\frac{t}{T/S}\tilde X_{T/S}$

is a Brownian bridge, as required. ⬜

In particular, this means that every Brownian bridge can be written as a scaled version of a standard Brownian bridge. We could have restricted to ${T=1}$ in the definition above, without any real loss of generality. However, it is sometimes convenient to have Brownian bridges defined on non-unit intervals, so I did not make this restriction. I did, however, restrict to intervals of the form ${[0,T]}$, although even this is not really necessary. Given any interval of the form ${[S,T]\subseteq{\mathbb R}_+}$ and a standard Brownian motion X, we can construct a Brownian bridge by subtracting off a linear term which makes the process go to zero at each of the endpoints S and T.

Lemma 4 Let X be standard Brownian motion and ${S < T}$ be fixed nonnegative times. Then, the process

 $\displaystyle B_t = X_{S+t} - X_S-\frac{t}{T-S}(X_T-X_S)$

is a Brownian bridge on the interval ${[0,T-S]}$, and is independent of ${X_t}$ over the range ${t\in{\mathbb R}_+\setminus(S,T)}$.

Proof: As ${\tilde X_t=X_{S+t}-X_S}$ is a Brownian motion and,

 $\displaystyle B_t = \tilde X_t - \frac{t}{T-S}\tilde X_{T-S}$

it follows from the definition that B is a Brownian bridge which, by lemma 1, is independent of ${X_t-X_S}$ over ${t\ge T}$. It only remains to show that it is also independent of ${X_t}$ over ${t\le S}$. However, ${\{X_t\}_{t\le S}}$ is independent of ${\tilde X}$ by the independent increments property, so is independent of B. ⬜

Lemma 4 extends in a straightforward way to an arbitrary collection of non-overlapping time intervals. This gives an independent collection of Brownian bridges, all of which are independent of the original Brownian motion at all times outside of these intervals. This is the situation shown in figure 1 above. The Brownian bridges are constructed on each interval of the Brownian motion by subtracting out a linear term, as shown in the top plot, and then rescaled to standard Brownian bridges in the bottom plot.

Lemma 5 Let X be standard Brownian motion and ${\{(S_i,T_i)\}_{i\in I}}$ be pairwise disjoint subintervals of ${{\mathbb R}_+}$. Then, the processes ${\{B^i_t\}_{t\in[0,T_i-S_i]}}$ defined by

 $\displaystyle B^i_t = X_{S_i+t} - X_{S_i}-\frac{t}{T_i-S_i}(X_{T_i}-X_{S_i})$ (2)

are independent Brownian bridges, and are independent of ${X_t}$ over the range ${t\in{\mathbb R}_+\setminus\bigcup_i(S_i,T_i)}$.

Proof: For each ${i\in I}$, lemma 4 says that ${B^i}$ is a Brownian bridge independent of ${X_t}$ over ${t\in{\mathbb R}_+\setminus(S_i,T_i)}$. For any ${j\not=i}$ in I, as ${B^j}$ depends only on X over this range, we conclude that ${B^i}$ is independent of all such ${B^j}$ as required. ⬜

This result provides a very useful practical technique when simulating Brownian motion. Suppose we start by simulating at a fixed finite set of times. Then, if the values are later required at points outside of this set, we can ‘interpolate’ across each of the intervals between the original set of times using independent Brownian bridges. That is, if ${S_i < t < T_i}$ then,

 $\displaystyle X_t = B^i_{t-S_i}+X_{S_i}+\frac{t-S_i}{T_i-S_i}X_{S_i}.$

Lemma 5 holds in exactly the same way if X is itself a Brownian bridge rather than a Brownian motion.

Lemma 6 Let X be a Brownian bridge on interval ${[0,T]}$ and ${\{(S_i,T_i)\}_{i\in I}}$ be pairwise disjoint subintervals of ${[0,T]}$. Then, the processes ${\{B^i_t\}_{t\in[0,T_i-S_i]}}$ defined by (2) are independent Brownian bridges, and are independent of ${X_t}$ over the range ${t\in[0,T]\setminus\bigcup_i(S_i,T_i)}$.

Proof: Suppose that ${X_t=Y_t-tY_T/T}$ for a Brownian motion Y. Then, (2) also holds with Y in place of X and, hence, lemma 5 says that ${B^i}$ are independent Brownian bridges, and independent of ${Y_t}$ over ${t\in[0,T]\setminus\bigcup_i(S_i,T_i)}$. Hence, they are also independent of X over this range as required. ⬜

An alternative method of describing a Brownian bridge is as a joint normal process, so that its distribution is completely determined by its means and covariances.

Lemma 7 A continuous real valued stochastic process ${\{B_t\}_{t\in[0,T]}}$ is a Brownian bridge if and only if it is joint normal with zero mean and covariances

 $\displaystyle {\mathbb E}[B_sB_t]=s(T-t)/T$ (3)

for ${0\le s\le t\le T}$.

Proof: We suppose that B is given in terms of Brownian motion X by (1). As X is joint normal with zero mean, the same holds for B. For times ${0\le s\le t\le1}$, the covariances are given by,

 \displaystyle \begin{aligned} {\mathbb E}[B_sB_t] &={\mathbb E}[(X_s-sX_T/T)(X_t-tX_T/T)]\\ &={\mathbb E}[X_s^2-sX_TX_t/T-tX_sX_T/T+stX_T^2/T^2]\\ &=s-st/T-ts/T+stT/T^2\\ &= s-st/T \end{aligned}

as required. ⬜

Although equation (3) only holds for ${s\le t}$, by symmetry it can be written as

 $\displaystyle {\mathbb E}[B_sB_t]=s\wedge t-\frac{st}T$

which holds for all ${s,t\in[0,T]}$, and is sometimes a more convenient form. As the covariance is invariant under replacing s and t by ${T-s}$ and ${T-t}$ respectively, we immediately see that Brownian bridges are preserved under time reversal.

Lemma 8 If ${B_t}$ is a Brownian bridge on the interval on the interval ${[0,T]}$, then so is ${B_{T-t}}$.

Rather than subtracting a linear term, Brownian bridges can alternatively be constructed from Brownian motion by rescaling as follows.

Lemma 9 Let X be a Brownian motion and ${T > 0}$. Then, the processes ${\{B_t,\tilde B_t\}_{t\in[0,T]}}$ defined by

 \displaystyle \begin{aligned} &B_t = tT^{-1/2}X_{T/t-1},\\ &\tilde B_t = (T-t)T^{-1/2}X_{t/(T-t)}, \end{aligned}

are Brownian bridges.

Proof: As B is clearly a joint normal process with zero mean, we just need to show that the covariances are of the form (3). For times ${0 < s\le t \le T}$ then, as ${T/t-1 \le T/s-1}$ we obtain

 \displaystyle \begin{aligned} {\mathbb E}[B_sB_t] &=st{\mathbb E}[X_{T/t-1}X_{T/s-1}]/T\\ &=st(T/t-1)/T \end{aligned}

as required. Then, ${\tilde B_t=B_{T-t}}$ is also a Brownian bridge by lemma 8. ⬜

If B is a standard Brownian bridge then, by definition, it is a continuous process with ${B_0=B_1}$. Hence, we can consider extending the time index to all of ${{\mathbb R}}$ by making it periodic with period one. To express this, use ${\{t\}}$ to denote the fractional part of time t. That is, ${\{t\}}$ is a real number in the interval ${[0,1)}$ equal to ${t-\lfloor t\rfloor}$, where ${\lfloor t\rfloor}$ is the largest integer less than or equal to t. Then, ${B_{\{t\}}}$ is periodic with period one, and equal to B on the unit interval. It would be good if this was invariant under time translations, so that ${B_{\{T+t\}}}$ has the same joint distribution as ${B_{\{t\}}}$ for each fixed real T. We know that this cannot be true, since ${B_{\{t\}}}$ hits zero at all integer times. However, we do obtain invariance if, as well as translating the time index we subtract a constant so that it is equal to zero at time zero.

Lemma 10 If B is a Brownian bridge then, for any fixed time ${T\in{\mathbb R}}$, the process

 $\displaystyle X_t=B_{\{T+t\}}-B_{\{T\}}$

over ${t\in[0,1]}$ is also a Brownian bridge.

Proof: This can be proven by computing covariances and showing that it agrees with (3). However, I will take slightly different approach. Consider splitting the process into the intervals ${[0,T]}$ and ${[T,1]}$. The process ${B_{\{T+t\}}-B_{\{T\}}}$ exchanges these intervals, and translates them in space so that it is zero at times 0 and 1 (and is continuous at time ${1-T}$, where the exchanged intervals are joined). Lemma 6 shows that this decomposes the Brownian bridge into a pair of bridges, and joins them back together in the opposite order, so gives a Brownian bridge again.

More precisely, supposing that ${0 < T < 1}$ (wlog), applying lemma 6 to the intervals ${(0,T)}$ and ${(T,1)}$, we can write

 $\displaystyle B_t = \begin{cases} tB_T/T + B^1_t,&\textrm{if }t\le T,\\ (1-t)B_T/(1-T)+B^2_{t-T},&\textrm{if }t\ge T \end{cases}$

for independent Brownian bridges ${B^1,B^2}$ on intervals ${[0,T]}$ and ${[0,1-T]}$ respectively, which are independent of ${B_T}$. Then,

 $\displaystyle B_{\{T+t\}}-B_T = \begin{cases} -tB_T/(1-T) + B^2_t,&\textrm{if }t\le 1-T,\\ (t-1)B_T/T+B^1_{t+T-1},&\textrm{if }t\ge 1-T. \end{cases}$

As ${-B_T}$ has the same distribution as ${B_{1-T}}$, lemma 6 shows that this process is again a Brownian bridge. ⬜

Another method of demonstrating this invariance is to subtract a constant value from ${B_{\{t\}}}$ so that the resulting process is invariant under time translation. This works if we subtract the average value of B, and follows easily from lemma 10.

Corollary 11 Let B be a standard Brownian bridge with mean ${\bar B = \int_0^1 B_tdt}$, and define the process ${X_t=B_{\{t\}}-\bar B}$ over ${t\in{\mathbb R}}$. Then, for any fixed time ${T\in{\mathbb R}}$, the process ${X_{T+t}}$ has the same joint distribution as ${X_t}$.

Proof: Let ${B^\prime_t=B_{\{T+t\}}-B_{\{T\}}}$ which, by lemma 10, is a standard Brownian bridge with mean ${\bar B^\prime=\bar B-B_{\{T\}}}$. We can write

 $\displaystyle X_{T+t}=B_{\{T+t\}}-\bar B = B^\prime_{\{t\}}-\bar B^\prime$

which, therefore, has the same joint distribution as ${X_t}$. ⬜

Corollary 11 defines a continuous periodic process which is invariant under time translation, from which the original Brownian bridge is obtained via ${B_t=X_t-X_0}$. This invariant process can be described as joint normal with prescribed covariances.

Lemma 12 The process X defined by corollary 11 is joint normal with zero mean and covariances

 $\displaystyle {\mathbb E}[X_sX_t]=\frac1{12}-\frac12\{t-s\}\{s-t\}.$

for all times ${s,t\in{\mathbb R}}$.

Proof: As a standard Brownian bridge is joint normal with zero mean, the same holds for X. By translation invariance and periodicity,

 \displaystyle \begin{aligned} {\mathbb E}[(X_t-X_s)^2] &={\mathbb E}[(X_{t-s}-X_0)^2] ={\mathbb E}[B_{\{t-s\}}^2]\\ &=\{t-s\}(1-\{t-s\})\\ &=\{t-s\}\{s-t\}. \end{aligned}

Hence,

 \displaystyle \begin{aligned} {\mathbb E}[X_sX_t] &=\frac12\left({\mathbb E}[X_s^2]+{\mathbb E}[X_t^2]-{\mathbb E}[(X_t-X_s)^2]\right)\\ &=c-\frac12\{t-s\}\{s-t\} \end{aligned}

where ${c={\mathbb E}[X_t^2]}$ is a constant independent of t. As the integral of ${{\mathbb E}[X_0X_t]=c-t(1-t)/2}$ over the unit interval is zero, we obtain ${c=1/12}$ as required. ⬜

I now move on to the next way that Brownian bridges can be described — as a Markov process.

Lemma 13 A continuous process ${\{B_t\}_{t\in[0,T]}}$ is a Brownian bridge if and only if ${B_0=0}$ (almost surely) and it is Markov (with respect to its natural filtration ${\mathcal F_\cdot}$) such that for all ${0\le s < t\le T}$ then, conditional on ${\mathcal F_s}$, ${B_t}$ is normally distributed with mean and variance given by,

 \displaystyle \begin{aligned} &{\mathbb E}[B_t\;\vert \mathcal F_s]=\frac{T-t}{T-s}B_s,\\ &{\rm Var}(B_t\;\vert \mathcal F_s)=\frac{(t-s)(T-t)}{T-s}. \end{aligned}

Proof: Since the joint distribution of Markov processes are uniquely determined by their initial value and transition probabilities, it is sufficient to show that a Brownian bridge B is Markov as described, and the converse statement follows automatically. First, by lemma 6, ${B_{s+u}-(T-s-u)B_s/(T-s)}$ is a Brownian bridge over ${0\le u\le T-s}$ independently of ${\mathcal F_s}$. Taking ${u=t-s}$ shows that ${B_t-(T-t)B_s/(T-s)}$ is normal with zero mean and variance ${(t-s)(T-t)/(T-s)}$ independently of ${\mathcal F_s}$ as required. ⬜

Brownian bridges are commonly defined as Brownian motion conditioned on hitting zero at time T. This is a bit problematic, since the hitting zero at any fixed positive time T is a zero probability event, so cannot be conditioned on. Instead, we consider conditioning on ${X_T}$ being close to zero, rather than exactly equal to zero, and show that a Brownian bridge is obtained in the limit.

Lemma 14 Let ${\{X_t\}_{t\in[0,T]}}$ be a standard Brownian motion and ${S_n}$ be a sequence of measurable subsets of ${{\mathbb R}}$ with positive measure. We suppose that, for every ${\epsilon > 0}$, ${S_n\subseteq(-\epsilon,\epsilon)}$ for sufficiently large n. Then, the distribution of ${X}$ conditioned on ${X_T\in S_n}$ tends weakly to that of a Brownian bridge as n goes to infinity.

Proof: Let ${\mathcal C}$ be the space of continuous functions ${z\colon[0,T]\rightarrow{\mathbb R}}$ and ${Z_t\colon \mathcal C\rightarrow{\mathbb R}}$ be the coordinate process defined by ${Z_t(z)=z_t}$ over ${0\le t\le T}$. We also let ${\mathcal E}$ be the sigma-algebra on ${\mathcal C}$ generated by ${\{Z_t\colon t\in[0,T]\}}$, which is just the Borel sigma-algebra corresponding to the topology of uniform convergence on ${\mathcal C}$. If ${\mu}$ is the probability measure on ${(\mathcal C,\mathcal E)}$ with respect to which Z is a Brownian motion, then weak convergence of distributions, as stated in the lemma, just means that

 $\displaystyle {\mathbb E}[f(X)\;\vert X_T\in S_n]\rightarrow \mu(f)$ (4)

as ${n\rightarrow\infty}$ for each bounded continuous function ${f\colon\mathcal C\rightarrow{\mathbb R}}$.

By lemma 1, ${X_t=B_t+tX_T/T}$ for a Brownian bridge B independent of ${X_T}$. Hence, writing ${Y_t=tX_T/T}$,

 $\displaystyle {\mathbb E}[f(X)\;\vert X_T\in S_n]=\int {\mathbb E}[f(z+Y)\;\vert X_T\in S_n]d\mu(z).$

For any fixed ${\epsilon > 0}$, we have ${\lvert Y\rvert < \epsilon}$ on the event ${X_T\in S_n}$ for all sufficiently large n. So, by continuity of ${f}$, ${{\mathbb E}[f(z+Y)\;\vert X_T\in S_n]}$ tends to ${f(z)}$ as ${n\rightarrow\infty}$. Hence, (4) follows from bounded convergence. ⬜

Next, Brownian bridges can be constructed by a restricting a standard Brownian motion X to the time interval ${[0,\tau]}$, where ${\tau}$ is the last time in a fixed interval ${[0,T]}$ at which X hits zero, and rescaling to the unit interval.

Lemma 15 Let X be a Brownian motion, T be a fixed positive time and

 $\displaystyle \tau=\sup\left\{t\in[0,T]\colon X_t=0\right\}.$

Then, the process ${B_t=\tau^{-1/2}X_{t\tau}}$ over ${0\le t\le1}$ is a standard Brownian bridge independent of ${1_{[\tau,\infty)}X}$.

We note that ${\tau}$ is not a stopping time since, for any time ${t\in(0,T)}$, the event ${\{\tau\le t\}}$ is equivalent to the event that ${X}$ does not hit zero on ${(t,T]}$. This is not measurable with respect to ${\{X_s\colon s\le t\}}$ and, so, ${\tau}$ is not a stopping time for the natural filtration of X. Instead, the event mentioned is in the sigma-algebra generated by ${\{X_s\colon s\ge t\}}$ meaning that it is a stopping time in the reversed time direction. In fact, lemma 15 is just a special case of a more general result for such ‘time-reversed stopping times’, which I prove now.

Lemma 16 Let X be a Brownian motion and ${\tau\colon\Omega\rightarrow(0,\infty)}$ be a random time such that ${\{\tau \ge t\}}$ is measurable w.r.t. ${\{X_s\colon s\ge t\}}$ for each ${t > 0}$. Then, the process ${\{B_t\}_{t\in[0,1]}}$ defined by

 $\displaystyle B_t=\tau^{-1/2}(X_{t\tau}-tX_\tau)$

is a standard Brownian bridge independent of ${1_{[\tau,\infty)}X}$.

Proof: The process ${\tilde X_t = tX_{\frac1t}}$ is also a standard Brownian motion, as can be determined by computing covariances and, with respect to its natural filtration, ${\tilde\tau = 1/\tau}$ is a stopping time. Hence, by the strong Markov property, the process ${Y_t=\tilde X_{\tilde\tau+t}-\tilde X_{\tilde\tau}}$ is standard Brownian motion independently of ${1_{(0,\tilde\tau]}\tilde X}$ or, equivalently, of ${1_{[\tau,\infty)}X}$. By scaling invariance of Brownian motion, ${\tilde Y_t = \tau^{1/2}Y_{t/\tau}}$ is a Brownian motion, again independent of ${1_{[\tau,\infty)}X}$. So, by lemma 9, ${B_t = t\tilde Y_{1/t-1}}$ is a Brownian bridge independent of ${1_{[\tau,\infty)}X}$. ⬜

Finally, I move on to the description of Brownian bridges as semimartingales, which decompose into martingale and finite variation terms. In the following lemma, we compute the finite variation process V such that a Brownian bridge B decomposes as ${B=W-V}$ for a local martingale W and, furthermore, show that this local martingale term is a standard Brownian motion. We work with respect to the natural filtration ${\mathcal F_t}$ of B, augmented with the zero probability events.

Lemma 17 If B is a Brownian bridge on interval ${[0,T]}$ then the process

 $\displaystyle V_t = \int_0^t\frac{B_s}{T-s}ds$

has integrable variation and,

 $\displaystyle W_t=B_t+V_t$

is a standard Brownian motion over ${0\le t\le T}$.

Proof: As ${B_s}$ is normal with zero mean and variance ${s(T-s)/T}$, we have ${{\mathbb E}[\lvert B_s\rvert]=c\sqrt{s(T-s)/T}}$ where ${c=\sqrt{2/\pi}}$ is the absolute expected value of a standard normal. Hence, the expected variation of V on the interval ${[0,T]}$ is

 \displaystyle \begin{aligned} \int_0^T\frac{{\mathbb E}[\lvert B_s\rvert]}{T-s}ds &=c\int_0^Ts^{1/2}(T-s)^{-1/2}{T}^{-1/2}ds\\ &\le c\int_0^T(T-s)^{-1/2}ds\\ &=2c\sqrt T < \infty \end{aligned}

as claimed. It only remains to show that W is a Brownian motion. We first show that it is a martingale. For times ${s < t}$ then, by lemma 13, we have

 \displaystyle \begin{aligned} {\mathbb E}[V_t-V_s\;\vert\mathcal F_s] &=\int_s^t\frac{(T-u)B_s/(T-s)}{T-u}du\\ &=\frac{t-s}{T-s}B_s\\ &={\mathbb E}[B_s-B_t\;\vert\mathcal F_s]. \end{aligned}

So, ${W_t=B_t+V_t}$ is a martingale. Next, as definition 2 expresses a Brownian bridge as standard Brownian motion plus a linear term, its quadratic variation is the same as for standard Brownian motion, ${[B]_t=t}$. Then, as V is continuous with finite variation, the quadratic variation of W is also given by ${[W]_t=t}$ so, by Lévy’s characterization, is standard Brownian motion. ⬜

Lemma 17 shows that the Brownian bridge solves the stochastic differential equation

 $\displaystyle dB_t=dW_t-\frac{B_t}{T-t}dt$ (5)

for a Brownian motion W. By the Lipschitz condition, this SDE always has a unique solution for any Brownian motion W, at least over ${t < T}$, and provides our final characterization of Brownian bridges.

Lemma 18 A continuous process ${\{B_t\}_{t\in[0,T]}}$ is a Brownian bridge if and only if ${B_0=0}$ (almost surely) and it solves the SDE (5) for a standard Brownian motion ${\{W_t\}_{t\in[0,T]}}$.

Proof: If B is a Brownian bridge, then lemma 17 states that it satisfies an SDE of the form (5). Conversely, suppose that (5) is satisfied for some Brownian motion W and that ${B_0=0}$. If we set ${M_t=B_t/(T-t)}$ then integration by parts gives ${dM_t=(T-t)^{-1}dW_t}$. Since the integrand is deterministic, this shows that ${M_t-M_s}$ is independent of ${\mathcal F_s}$ for any ${0\le s\le t < T}$, and normal with zero mean and variance

 $\displaystyle \int_s^t(T-u)^{-2}du=(t-s)(T-s)^{-1}(T-t)^{-1}.$

Hence, conditional on ${\mathcal F_s}$, ${B_t=(T-t)M_t}$ is normal with mean ${(T-t)B_s/(T-s)}$ and variance ${(t-s)(T-t)/(T-s)}$, so lemma 13 shows that B is a Brownian bridge. ⬜