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If $dS_1(t)=S_1(t)\mu dt+S_1(t)\sigma_1dW_1(t)$ and $dS_2(t)=S_2(t)\mu dt+S_2(t)\sigma_2dW_2(t)$ where $W_1(t), W_2(t)$ not necessarily independent, is the $S_1+2S_2$ still geometric Brownian motion?

I tried and get $d(S_1+2S_2)=(S_1\mu_1+2S_2\mu_2)dt+S_1(t)\sigma_1dW_1(t)+2S_2(t)\sigma_2dW_2(t)$, then I do not know how to deal with the last two terms. Could someone kindly help? Thanks!

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Actually, $S=S_1+2S_2$ is such that $dS(t)=S(t)\mu dt+S(t)dY(t)$ where $$dY(t)=S(t)^{-1}(S_1(t)\sigma_1dW_1(t)+2S_2(t)\sigma_2dW_2(t))$$ hence $$d\langle Y\rangle_t=S(t)^{-2}[(S_1(t)^2\sigma_1^2+4S_2(t)^2\sigma_2^2)dt+4\sigma_1\sigma_2S_1(t)S_2(t)d\langle W_1,W_2\rangle_t]$$ and $d\langle Y\rangle_t=\sigma^2dt$ for some constant $\sigma^2$ if and only if $$(S_1(t)^2\sigma_1^2+4S_2(t)^2\sigma_2^2)dt+4\sigma_1\sigma_2S_1(t)S_2(t)d\langle W_1,W_2\rangle_t=\sigma^2S(t)^2dt$$ that is, $d\langle W_1,W_2\rangle_t=c(t)dt$ with $$S_1(t)^2\sigma_1^2+4S_2(t)^2\sigma_2^2+4\sigma_1\sigma_2S_1(t)S_2(t)c(t)=\sigma^2S(t)^2$$ Furthermore, one knows that $|c(t)|\leqslant1$ almost surely hence this requires that $$|\sigma^2(S_1(t)+2S_2(t))^2-S_1(t)^2\sigma_1^2-4S_2(t)^2\sigma_2^2|\leqslant4\sigma_1\sigma_2|S_1(t)S_2(t)|$$ almost surely. Except if $c(t)=1$ almost surely for every $t$ (that is, if $W_1=W_2$) or $c(t)=-1$ almost surely for every $t$ (that is, if $W_1=-W_2$), the distribution of $(S_1,S_2)$ has full support hence one asks that $$|\sigma^2(x+2y)^2-x^2\sigma_1^2-4y^2\sigma_2^2|\leqslant4\sigma_1\sigma_2|xy|$$ for every real numbers $(x,y)$, which is impossible except if $\sigma_1^2=\sigma_2^2$. If $\sigma_1^2=\sigma_2^2=\tau^2$, $$dY(t)=\tau^2(dW_1(t)+dW_2(t))$$ hence $Y$ is $\sqrt2\tau$ times a standard Brownian motion. To sum up, the result you try to prove holds only in a few degenerate cases.

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