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SDE 1 is S1 = S10 exp( (r1-sigma^2/2) * dt + sigma dW1 )

S2 = S20 exp( (r2-sigma2^2/2) * dt + sigma2 dW2 )

E[dW1 dW2] = rho

I want to price an option on S1 x S2 I know I need to use the SDE's to find the SDE for d(S1 S2) using Ito...but what if I use this

S = S1 x S2 using two analytical solutions of SDE? Why the drift term comes out incorrectly? Why can't I multiply two SDE solutions to get an equation for the third....?

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    $\begingroup$ Please use latex for formulas. $\endgroup$ – Richard Dec 3 '14 at 8:22
  • $\begingroup$ The reason it doesn't work is because you need the Ito correction term. In a nutshell: do a 2dim taylor expansion and keep the $dw^2 =dt$ terms. $\endgroup$ – Kian Dec 4 '14 at 11:56
  • $\begingroup$ You say that the drift term is incorrect - what is your result of the term. It is unclear what you are asking. Please use latex and ask a clear question. Of course it is is the same whether you multiply 2 solutions or you write down a correct (!) SDE for the product and solve it afterwards. $\endgroup$ – Richard Jan 5 '15 at 14:18
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You can use Ito's product rule $d(X \, Y) = dX \, Y + X \, dY + dX \, dY$. In your case, you have $$ dS_{1,t} = S_{1,t} \left( r_1 dt + \sigma_1 dW_{1,t} \right) $$ and $$ dS_{2,t} = S_{2,t} \left( r_2 dt + \sigma_1 dW_{2,t} \right) $$

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By all means, if $dS_i = \mu(S^i_t)\mathrm dt + \sigma(S^i_t)\mathrm dW^i_t$ for $i=1,2$ and you know explicit formulas for $S_i$ then their product satisfies the SDE you derive using the Ito lemma for a product. Could you elaborate, on what is the issue you have with the drift term?

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    $\begingroup$ The brownian motions are correlated ($\rho$), so $S\neq S_1S_2$. $\endgroup$ – emcor Dec 3 '14 at 11:11
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    $\begingroup$ @emcor: can you be more precise, $S = S_1\cdot S_2$ by definition of $S$ $\endgroup$ – Ulysses Dec 3 '14 at 13:56
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    $\begingroup$ Their joint distribution is different, therefore the solutions cannot be multiplied as individually. $\endgroup$ – emcor Dec 3 '14 at 14:11
  • $\begingroup$ Of course they can be multiplied! One just has to take care for all probabilistic aspects but $S=S_1* S_2$ always holds. one just has to carfully analyse the distribution of $S$ and/or its SDE. $\endgroup$ – Richard Jan 5 '15 at 14:19
  • $\begingroup$ The covariation term $[ X, Y]$ will handle the correlated brownians, giving you $\rho \sigma_1 \sigma_2 dt$ $\endgroup$ – Drew Jan 5 '15 at 14:51

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