Let $(W)_{t\ge0}$ and $(B)_{t\ge0}$ be independent Brownian motions.
Consider $X_t=\exp(-W_t)\left(x_0+\int_0^t\exp(W_s)dB_s\right)$.
Can anyone explain to me how to obtain
$dX_t=dB_t+X_t\left(\frac{1}{2}dt-dW_t\right)$
via Itô's formula?
Application of Itô's formula
1
$\begingroup$
probability
stochastic-processes
stochastic-calculus
brownian-motion
stochastic-analysis
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0Hint: write $Y_t=exp(-W_t)$ and $Z_t=\left(x_0+\int_0^t\exp(W_s)dB_s\right)$ and use the Ito's formula for a product. – 2017-02-17
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0Is $\langle Y,Z \rangle_t=\langle \int_0^t -\exp(-W_s)dW_s+\int_0^t \exp(-W_s)ds,x+\int_0^t \exp(W_s) dB_s\rangle=-\int_0^t d\langle B,W\rangle_t=-\langle B,W\rangle_t=0$ – 2017-02-17
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0I cannot post a comment with less than 15 character, nevertheless the answer is yes. – 2017-02-17