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Let $W = (W_t;F_t)$, $t \leq 0$ be a standard Wiener process, and let $(X_t)_{0 \leq t \leq 1}$ satisfy the stochastic differential equation $$ dX_t =- \frac{X_t}{1-t}dt+dW_t,\quad 0 \leq t \leq 1,\quad X_0=0 $$

Find $(X_t)_{0 \leq t \leq 1}$ and verify that

$$ L^2\text{-}\lim_{t \to 1}X_t=0$$ $X$ is called Brownian Bridge.

Paul
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user123
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    I was going to ask what is $L$? But then I realized that you must be asking about the limit in $L^2$. What horrible notation. (But presumably not your fault.) Never mind that, though. What have you tried? – Harald Hanche-Olsen Jun 01 '13 at 17:07
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    You're committing a standard newbie faux pas on math.stackexchange.com: You're phrasing the question in a manner appropriate to an instructor assigning a problem. This can arouse suspicions that you're stenographically copying a question that you don't understand, as opposed to actually asking a question of your own. Giving us some of your own thoughts on this can help. – Michael Hardy Jun 01 '13 at 17:10
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    @HaraldHanche-Olsen : I've now changed the minus sign to a hyphen: where it said $\displaystyle L^2-\lim_{t\to1}$, it now says $\displaystyle L^2\text{-}\lim_{t\to1}$. ${{}\qquad{}}$ – Michael Hardy Jun 01 '13 at 17:12

2 Answers2

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I will give you an answer on the general brownian bridge case.

Consider the SDE

$$dX_t =\frac{b-X_t}{1-t}dt+dW_t,\ X_0=a$$

for $t\in [0,1]$ with $a,b \in \mathbb R $.

An approach to solve this SDE can be obtained by the constant variation method. Indeed, consider the following ODE

$$ x'(t)= \frac{b-x(t)}{(1-t)} + f(t), \ x(0)=a$$ for $t\in [0,1]$. We can easily obtain by constant variation method the solution of this equation which is given by

$$x(t) = a(1-t) +bt +(1-t)\int_0^t \frac {f(s)}{1-s}ds$$

At this point, we would like to simply apply this result with $f(t)=dW_t /ds$, which is formally not possible. But let allow us to consider this formula anyway and then prove that

$$X_t = a(1-t) +bt +(1-t)\int_0^t \frac {1}{1-s}dW_s\tag{1}$$

is a solution to our SDE. By effect, a stochastic integration by parts give us that

$$X_t = a(1-t) +bt +W_t-(1-t)\int_0^t \frac {W_s}{(1-s)^2}ds.$$

Then, if we consider $Y_t = X_t-W_t$, we have $$ dY_t = \frac{b-Y_t}{1-t}dt - \frac{W_t}{1-t}dt$$ so we can easely conclude the wanted result.

Now, knowing that the brownian bridge have the form given by $(1)$ we are able to calculate it's $L^2-$limit. Note that $X$ defined by $(1)$ is a gaussian process with mean $a(1-t)+bt$. Using Ito's isometrie, we can straight forward calculate $$ \mathbb E \{ X_t^2\}=\left[a(1-t)+bt\right]^2 + (1-t)^2\int_0^t \frac{1}{(1-s)^2}ds=\left[a(1-t)+bt\right]^2 +t(1-t)\underset{t\rightarrow 1}{\longrightarrow }b^2$$

wich show us that $L^2-\lim_{t \rightarrow 1} X_t =b$ (with $b=0$ in your case).

Actually this limit holds even $\mathbb P-\text{almost ever}$.

Paul
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A shorter answer: The trick is remove the denominator $(1-t)$, by defining $X_t = Y_t(1-t)$, then $$dX_t = (1-t)dY_t - dY_t = -dY_t + dW_t \implies dY_t = \frac{1}{1-t} dW_t,$$ hence $$X_t = \int_0^t \frac{1}{1-u}dW_u.$$

Community
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wenlu sun
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