Quotient of two geometric Brownian motions

Question

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!

Answer 1

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}

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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.

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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, $$ dV_t=[(μ-σ)(1-(μ+σ)/2)-(μ-σ)^2/2]V_t\,dt-(μ-σ)V_t\,dB_t. $$

Answer 2

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 $$dV = V[\mu dt+\sigma dB - \sigma dt-\mu dB +\mu^2(dB)^2- \mu\sigma (dB)(dB)] \\ = (\mu-\sigma+\mu^2-\mu\sigma)Vdt + (\sigma-\mu)VdB $$