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Stochastic Taylor's Expansion (Ito Rule)

gnob

New member
Dec 11, 2012
11
Good day.

For $k\geq0$ a continuous function on $\mathbb{R}_+$ and $\{W_t\}$ a standard Brownian motion, could you help me find the Taylor's expansion of the following exponential: $e^{-\int_0^t k(s) dW_s}.$

For the case where $e^{-k W_t}$ where $k>0$ is a constant, I was able to recompute its Taylor's expansion as
$$
e^{-k W_t} = 1 - k\int_0^t e^{-kW_s}dW_s + \frac{1}{2}k^2\int_0^t e^{-kW_s}ds.
$$
But in the first exponential above, we have an integral exponent with a nonconstant $k,$ but a function. Please help me on this.

Thanks a lot in advance.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Good day.

For $k\geq0$ a continuous function on $\mathbb{R}_+$ and $\{W_t\}$ a standard Brownian motion, could you help me find the Taylor's expansion of the following exponential: $e^{-\int_0^t k(s) dW_s}.$

For the case where $e^{-k W_t}$ where $k>0$ is a constant, I was able to recompute its Taylor's expansion as
$$
e^{-k W_t} = 1 - k\int_0^t e^{-kW_s}dW_s + \frac{1}{2}k^2\int_0^t e^{-kW_s}ds.
$$
But in the first exponential above, we have an integral exponent with a nonconstant $k,$ but a function. Please help me on this.

Thanks a lot in advance.
Honestly I'm not sure to give the correct answer to the question but if we consider a stochastic integral like...

$\displaystyle \int_{0}^{t} f \{W(s)\}\ d W(s)\ (1)$

... if f(x) is analytic around x=0, i.e. is...

$\displaystyle f \{W(s)\} = \sum_{n=0}^{\infty} a_{n}\ W^{n} (s)\ (2)$

... then the following formula can be applied...

$\displaystyle \int_{0}^{t} W^{n}(s)\ d W (s) = \frac{1}{n+1}\ W^{n+1} (t) - \frac{n}{2}\ \int_{0}^{t} W^{n-1} (s)\ d s\ (3)$

Kind regards

$\chi$ $\sigma$