Let $S,P$ be geometric BMs:
$$dS_t=S_t(\mu dt + \sigma dW_t^1)$$ $$dP_t=P_t(\tau dt + \beta (\rho dW_t^1+ \sqrt{1-\rho^2}dW_t^2)$$
Where $W^1$ and $W^2$ are independent standard BM.
I want to show their product is a Geometric BM.
Thanks to Did comments, I understand where I am wrong, I was working following the idea that $X_t$ is a GBM with drift $X\mu$, volatility $X\sigma$ if and only if $$d\log X_t= \mu dt + \sigma dW_t$$
while it is actually the case that $X_t$ is a GBM with drift $X\mu$ volatility $X\sigma$ iff $$d\log X_t= (\mu - \frac{\sigma^2}{2}) dt + \sigma dW_t$$
Using product rule, in fact, I get that the drift equals
$$\tau+\mu + \sigma \beta\rho$$
But if I try to proceed using logs I get a different answer: starting with
$$\tag{$\star$} d(\log(PS)) = d(\log P)+ d(\log S)$$
$$ =\frac{1}{P}dP-\frac{1}{2P^2}d[P] +\frac{1}{S}dS-\frac{1}{2S^2}d[S]$$
Ignoring the volatility part I get that the drift equals: $$ (\tau + \mu) - \frac{\beta^2 +\sigma^2}{2} $$
Which, summing the volatility squared divided by 2 gives the same result as above.