$e^\frac{t}{2}\int_0^te^{W_s - \frac s 2}dW_s = \int_0^t\Theta_sdWs$

Question

I want to find such $\Theta_s$ for such the following equality holds

$$e^\frac{t}{2}\int_0^te^{W_s - \frac s 2}dW_s = \int_0^t\Theta_sdWs$$

My main question is:

Should I just choose $\Theta_s = e^\frac{t}{2}(e^{W_t-\frac t 2})$, because $e^\frac t 2$ does not depend on $W_s$ so I can bring it out of integral ?

Answer 1

I assume you meant $$e^\frac{t}{2}\int_0^te^{W_s - \frac s 2}dW_s = \int_0^t\Theta_sdW_s.$$ Set $$Y_t = \int_0^te^{W_s-\frac s2}dW_s.$$ Let's apply Ito formula to the equation, we obtain $$\Theta_tdW_t =\frac12e^{\frac t2}Y_tdt+e^{W_t}dW_t.$$ Since $Y\neq0$, it's well known that this equation doens't have a solution.

This is because $\int_0^t\Theta_s-e^{W_s}dW_s$ is a local martingale and $\frac12\int_0^te^{\frac s2}Y_sds$ is one iff $Y=0$.