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So I have done this exercise and the proof holds, but I really don't believe it can be correct because the proof is worth twice as much as other exercises. I am also not 100% sure if $d_s d_r=d_rdW_s=d_sdW_r=dW_r dW_s=0$.

Feel free to give any feedback or correct me, thanks a lot!


The exercise says:

Let $X_{t}$ be an Ito process, given by $X_{t} = X_{0} + \int_{0}^{t} \mu_s ds+\int_{0}^{t} \sigma_s dW_s$, where $X_0\in\mathbb{R}$, and let $Y_t$ be another Ito process given by $Y_{t} = Y_{0} + \int_{0}^{t} b_s ds+\int_{0}^{t} h_s dW_s$, where $Y_0\in\mathbb{R}$.

Show that $X_tY_t=X_0Y_0+\int_0^t X_s dYs+\int_0^t Y_s dX_s+[X,Y](t)$, where $[X,Y](t)=\int_0^t\sigma_s h_s ds$.


And my answer is...

Consider $dX_t=\mu_t dt+\sigma_t dW_t$ and $dY_t=b_t dt+h_t dW_t$ and let $LHS=X_tY_t$ and $RHS=X_0Y_0+\int_0^t X_s dYs+\int_0^t Y_s dX_s+\int_0^t\sigma_s h_s ds$.

This gives: $$LHS=X_{t} Y_{t}=\left(X_{0} +\int_{0}^{t} \mu_s ds+\int_{0}^{t} \sigma_s dW_s\right)\left(Y_{0} + \int_{0}^{t} b_s ds+\int_{0}^{t} h_s dW_s\right)=X_0 Y_0+X_0\int_0^t b_s ds+X_0\int_0^t h_s dW_s+Y_0\int_0^t \mu_s ds+\int_0^t\mu_s b_s (ds)^2+\int_0^t \mu_s h_s ds dW_s+Y_0\int_0^t \sigma_s dW_s+\int_0^t\sigma_s b_s dW_s ds+\int_0^t \sigma_s h_s (dW_s)^2=X_0 Y_0+X_0\int_0^t b_s ds+X_0\int_0^t h_s dW_s+Y_0\int_0^t \mu_s ds+Y_0\int_0^t \sigma_s dW_s+\int_0^t \sigma_s h_s ds,$$ since $dW_s ds=(ds)^2=0$ and $(dW_s)^2=ds$.

Now the right-hand side (RHS) gives:

$$RHS=X_0Y_0+\int_0^tX_sdYs+\int_0^tY_sdX_s+\int_0^t\sigma_s h_s ds= X_0Y_0+\int_0^t\left(X_{0} + \int_{0}^{s} \mu_r dr+\int_{0}^{s} \sigma_r dW_r\right)\left(b_s ds+h_s dW_s\right)+\int_0^t\left(Y_{0} + \int_{0}^{s} b_r dr+\int_{0}^{s} h_r dW_r\right)\left(\mu_s ds+\sigma_s dW_s\right)+\int_0^t\sigma_s h_s ds=$$ And using $d_s d_r=d_rdW_s=d_sdW_r=dW_r dW_s=0$, we get $LHS=RHS$.

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  • $\begingroup$ Also, if it is correct, feel free to tell me so :) $\endgroup$ – jmmatias1985 Aug 10 '15 at 16:22
  • $\begingroup$ No, your calculations are not at all correct. Do you really think that $$\int_0^t \mu(s) \, ds \cdot \int_0^t b(s) \, ds = \int_0^t b(s) \mu(s) (ds)^2 = 0$$ for any $b$ and $\mu$? The first as well as the second "=" is (in general) not correct. $\endgroup$ – saz Aug 10 '15 at 17:46
  • $\begingroup$ I think I get it now, I have to use Ito's product rule, right? $\endgroup$ – jmmatias1985 Aug 10 '15 at 23:17
  • $\begingroup$ The thing is that I know how to prove it in a general case (it is in my book), but without using the formulas given and I find it really straightforward... But I wanted to try to do it using the formulas given. $\endgroup$ – jmmatias1985 Aug 11 '15 at 0:18
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    $\begingroup$ Yes, you have to use Itô's product rule (or Itô's formula). $\endgroup$ – saz Aug 11 '15 at 4:52
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Ito product formula is direct from Ito lemma so you can try to derive that first. Once you are ok with the Ito product formula, you start by writing $$d(X_tY_t)=dX_tY_t+X_tdY_t+dX_tdY_t = (Y_0+\int_0^tb_sds+\int_0^th_sdW_s)(\mu_tdt+\sigma_tdW_t)+(X_0+\int_0^t\mu_sds+\int_0^t\sigma_sdW_s)(b_tdt+h_tdW_t)+(\mu_tdt+\sigma_tdW_t)(b_tdt+h_tdW_t)$$ then apply the multiplication rules ($dtdt=0...)$, integrate on each sides (not forgetting that on the RHS $X_0Y_0$ appears), and you will be done.

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I really dislike writing things like "dt dt = 0". One direct way is if you know the Ito rule for processes in $R^n$. In that case, apply it to the $R^2$ process $Z_t = (X_t, Y_t)$ and the function $f(x,y) = xy$

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