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Let $H$ and $X$ be two continuous semi-martingale with $X_0=0$. Consider $$ Y_t=H_t+\int_0^t Y_sdX_s $$ Show that the solution is given by $$ Y_t=\mathcal{E}(X)_t(H_0+\int_0^t\mathcal{E}(X)_s^{-1}(dH_s-d[H,X]_s)) $$ where $\mathcal{E}(X)_t=\exp(X_t-\frac{1}{2}[X]_t).$

Note that $d\mathcal{E}(X)_t=\mathcal{E}(X)_tdX_t$. But how to use Ito formula get the solution of $Y_t$?

I try to compute $$ d(\mathcal{E}(X)_t^{-1} Y_t)=Y_t d(\mathcal{E}(X)_t^{-1})+\mathcal{E}(X)_t^{-1} dY_t+d[Y, \mathcal{E}(X)^{-1}]_t $$

My work: let $D_t=\exp(X_t-\frac{1}{2}[X]_t)$. I want to verify $$ dY_t=d(\mathcal{E}(X)_t(H_0+\int_0^t\mathcal{E}(X)_s^{-1}(dH_s-d[H,X]_s))) $$ satisfy the integral, i.e. $dY_t=dH_t+Y_tdX_t$.

That is $$ dY_t=H_0dD_t+dD_t(\int_0^t D_s^{-1}dH_s)+D_t D_t^{-1}dH_t-dD_t(\int_0^t (D_s^{-1})dH_s+dH_t... $$

Hermi
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  • What did you try so far? What happens when you use the Ito formula? – user6247850 Apr 13 '22 at 01:24
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    Well this seems close to the solution of linear SDEs you should check and adapt the proof which unless mistaken is treated here : https://math.stackexchange.com/questions/1788853/solution-to-general-linear-sde – TheBridge Apr 15 '22 at 12:27

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