Quotient of two geometric Brownian motions Given $X=X_t$, $Y=Y_t$ and
\begin{align}
      dX_t &= μX_t\,dt +  σX_t\,dB_t,
\\
      dY_t &= σY_t\,dt + μY_t\,dB_t,
\\  
      V_t &= \frac{X_t}{Y_t}
\end{align}
How to calculate $dV_t$?
I'm new to SDE and calculus as well, so I'm not sure if I solved it wrong or not...
Please help, thank you so much!
 A: It is often more simple to compute the derivative in a product formula instead of in a quotient formula. In the product formula for $X=YV$ you get, in infinitesimal increments,
$$
dX_t=d(V_tY_t)=(V_t+dV_t)(Y_t+dY_t)-V_tY_t=V_t\,dY_t+Y_t\,dV_t+dV_t\,dY_t
$$
While the last term as infinitesimal increment is random, this randomness is vanishingly small around a trend $d\langle V,Y\rangle_t$, which is the quadratic variation or correlation of the two processes. If $dV_t=a(V)dt+b(V)dB_t$, then $d⟨V,Y⟩_t=b(V)μY_t\,dt$. So let's insert all known equations
$$
X_t(μ\,dt+σ\,dB_t)=V_tY_t(σ\,dt+μ\,dBt)+Y_t\,dV_t+b(V)μY_t\,dt\\
V_t(μ-σ)(dt-dB_t)=a(V_t)\,dt+b(V_t)\,dB_t+μb(V)\,dt
$$
Comparing coefficients gives
\begin{align}
a(V_t)+μb(V_t)&=(μ-σ)V_t
\\
b(V_t)&=-(μ-σ)V_t
\\
\implies
a(V_t)&=(μ-σ)(1+μ)V_t
\end{align}

For a formal derivation of these results use the Ito theorem on transformations of stochastic processes which gives the same formula without mucking around with infinitesimals.

One could also use the solution formulas for geometric Brownian motions
$$
X_t=X_0e^{(μ+σ^2/2)t+σB_t}\\
Y_t=Y_0e^{(σ+μ^2/2)t+μB_t}\\
V_t=V_0e^{((μ-σ)-(μ^2-σ^2)/2)t-(μ-σ)B_t}
$$
which directly implies, as again a geometric Brownian motion,
$$
dV_t=[(μ-σ)(1-(μ+σ)/2)-(μ-σ)^2/2]V_t\,dt-(μ-σ)V_t\,dB_t.
$$
A: Note\begin{align}
dV &= \frac{dV}{dX}dX+ \frac{dV}{dY}dY 
+\frac12 \frac{d^2V}{dX^2}(dX)^2
 +\frac12 \frac{d^2V}{dY^2}(dY)^2
+\frac{d^2V}{dXdY}(dXdY)\\
&= \frac{1}{Y}dX- \frac{X}{Y^2}dY 
+\frac12 \cdot0\cdot(dX)^2
 +\frac{X}{Y^3}(dY)^2
- \frac{1}{Y^2}(dXdY)\\
\end{align}
Substitute $dX/X= μdt+  σdB$ and $dY/Y =σdt+ μdB$ into above to obtain
$$dV = V[\mu dt+\sigma dB - \sigma dt-\mu dB
+\mu^2(dB)^2- \mu\sigma (dB)(dB)] \\
= (\mu-\sigma+\mu^2-\mu\sigma)Vdt + (\sigma-\mu)VdB
$$
