stochastic integral equation

Question

For $0 \leq t \leq T$, define

$$Z_t:=\exp {\left\lbrace \int_0^t X_sdW_s - \frac 12 \int_0^t X_s^2ds \right\rbrace }$$

Show that this process satisfies the stochastic integral equation

$$Z_t=1+\int_0^tZ_s X_s dW_s , \qquad 0≤t≤T$$

Answer 1

You have $Z_t = f(Y_t)$ where $f(y) = \mathrm e^y$ and $$ \mathrm dY_t = -\frac12 X^2_t\mathrm dt + X_t\mathrm dW_t. $$ Just apply Ito's lemma to $f(Y_t).$

Answer 2

here is solution with details : indeed For $0 \leq t \leq T$ let $$Z_t:=\exp {Y_t }$$ with $$Y_t:= \int_0^t X_sdW_s - \frac 12 \int_0^t X_s^2ds $$ let $ f \in C^2(R) $ such that $ Z_t=f(Y_t) $ so by Ito's formula we get : $$ f(Y_t)=f(Y_0)+\int_0^t \frac{\partial f}{\partial x}(Y_s)dY_s+ \frac{1}{2}\int_0^t \frac{\partial^2f}{\partial x^2}(Y_s)d<Y>_s $$ to calculate the quadratic variation of process Y
we have $$\mathrm dY_t = -\frac12 X^2_t\mathrm dt + X_t\mathrm dW_t.$$ then $$d<Y>_s=X^2_sd<W>_s=X^2_sds $$ and we have also $$ Z_s=f(Y_s)=exp(Y_s)=\frac{\partial f}{\partial y}(Y_s)=\frac{\partial^2f}{\partial y^2}(Y_s) $$ $$ f(Y_0)=exp(Y_0)=exp{\left\lbrace \int_0^0 X_sdW_s - \frac 12 \int_0^0 X_s^2ds\right\rbrace }=exp(0)=1 $$ then we can write : $$Z_t=f(Y_t)=1+\int_0^t Z_sdY_s+ \frac{1}{2}\int_0^t Z_sX^2_sds $$ or $$ \mathrm dY_s = -\frac12 X^2_s\mathrm ds + X_s\mathrm dW_s $$ then $$ Z_t=f(Y_t)=1+\int_0^t Z_s(-\frac12 X^2_s\mathrm ds + X_s\mathrm dW_s) + \frac{1}{2}\int_0^t Z_sX^2_sds $$

hence $$Z_t=1+\int_0^tZ_s X_s dW_s , \qquad 0≤t≤T$$