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Let $B$ be a standard Brownian motion and $$ W_t=(1-t)\int_0^t \frac{1}{1-s}dB_s $$ be a Brownian bridge.
Calculate $dW_t$.

To apply Ito formula define $$ f(t,B_t)=(1-t) \int_0^t\frac{1}{1-s}dB_s $$ Then \begin{align} f_t(t,B_t)&=-\int_0^t\frac{1}{1-s}dB_s+(?) \\ f_{B_t}(t,B_t)&=(?) \\ f_{B_t,B_t}(t,B_t)&=(?) \end{align}

What are the (?) and how do we get them?

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You might want to define precisely what you call $\partial_{B_t}W_t$ in this context. – Did Nov 7 '12 at 16:36
@did I precised the items. Thank you for pointing this. – Nicolas Essis-Breton Nov 8 '12 at 12:11
up vote 4 down vote accepted

Let's define $X_t = \int_0^t \frac{\mathrm{d}B_s}{1-s}$, which is to say $\mathrm{d}X_t = \frac{1}{1-t} \mathrm{d}B_t$. Its sde thus has zero drift coefficient.

Then we are faced with using Ito formula for $W_t = (1-t) X_t$. $$ \mathrm{d} W_t = \left(\frac{\partial ((1-t)X_t))}{\partial t} + \underbrace{0}_{\text{drift}} \frac{\partial ((1-t)X_t))}{\partial X_t} + \frac{1}{2} \left(\underbrace{\frac{1}{1-t}}_\text{diffusion}\right)^2 \underbrace{\frac{\partial^2 ((1-t)X_t))}{\partial X_t^2}}_0 \right) \mathrm{d}t + \frac{\partial ((1-t)X_t)}{\partial X_t} \mathrm{d}X_t $$ Thus $$ \mathrm{d}W_t = -X_t \mathrm{d}t + (1-t) \mathrm{d} X_t = -\frac{W_t}{1-t} \mathrm{d}t + \mathrm{d} B_t $$

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Your answer smooth the trip from standard calculus to stochastic calculus. Thank you. – Nicolas Essis-Breton Nov 13 '12 at 14:22

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