2
$\begingroup$

Here is quanto adjustments in John Hull's book Options, Futures and Other Derivatives 9th page 699.

enter image description here

I know that we have $$E_X[V] = E_Y[VW].$$ And, $V$ and $W$ should be both martingale under $Y$-measure, so we simply write $$\dfrac{dV}{V} = \sigma_V d W_V$$ $$\dfrac{dW}{W} = \sigma_W d W_W$$ $$d W_Vd W_W = \rho d t$$ But how can we obtain the final result in the book $$E_X[V] = E_Y[V]e^{\rho \sigma_V\sigma_W T}.$$ It seems not the same result as two log-normal?

$\endgroup$
3
$\begingroup$

I believe that the dynamics for $V$, under $Y$, is not the form you provided. In particular, a measure change will change the drift of $V$. Specifically, the dynamics of $V$ is typically of the form \begin{align*} \frac{dV}{V} = -\sigma_V\sigma_W \rho dt + \sigma_V d W_V. \end{align*} You can now check that the final result holds.

Addendum

We assume that, under $X$, $V$ satisfies an SDE of the form \begin{align*} \frac{dV}{V} = \sigma_V d \widetilde{W}_V. \end{align*} Moreover, the Radon-Nikodym derivative $\eta = \frac{dY}{dX}$ satisfies \begin{align*} \frac{d\eta}{\eta} = \sigma_W d \widetilde{W}_W, \end{align*} where $d\langle \widetilde{W}_V, \widetilde{W}_W\rangle = \tilde{\rho} dt$. Then, by Cholesky decomposition, \begin{align*} \frac{d\eta}{\eta} = \sigma_W d \left(\tilde{\rho}\widetilde{W}_V+ \sqrt{1-\tilde{\rho}^2} \widetilde{B}_W\right). \end{align*} where $\widetilde{W}_V$ and $\widetilde{B}_W$ are two independent standard Brownian motions. Moreover, by Girsanov transformation, \begin{align*} W_V &= \widetilde{W}_V - \sigma_W \tilde{\rho} t\, \mbox{ and}\\ B_W &= \widetilde{B}_W - \sigma_W \sqrt{1-\tilde{\rho}^2} t \end{align*} are two standard Brownian motions under $Y$. Let $W= \eta^{-1} = \left(\frac{dY}{dX}\right)^{-1}$. Then, under $Y$, \begin{align*} \frac{dV}{V} &= \sigma_V\sigma_W\tilde{\rho} dt +\sigma_V d W_V,\\ \frac{dW}{W} &=\eta d\left(\frac{1}{\eta}\right)\\ &= -\frac{d\eta}{\eta}+\frac{1}{\eta^2} d\langle \eta,\eta\rangle\\ &=\sigma_W^2 dt -\sigma_W d \left(\tilde{\rho}\widetilde{W}_V+ \sqrt{1-\tilde{\rho}^2} \widetilde{B}_W\right)\\ &=-\sigma_W d \left(\tilde{\rho}W_V+ \sqrt{1-\tilde{\rho}^2} B_W\right). \end{align*} Let $\rho=-\tilde{\rho}$ and $W_W=\tilde{\rho}W_V+ \sqrt{1-\tilde{\rho}^2} B_W$. Then, under $Y$, \begin{align*} \frac{dV}{V} &= -\sigma_V\sigma_W\rho dt +\sigma_V d W_V,\\ \frac{dW}{W} &=\sigma_W d W_W, \end{align*} where $d\langle W_V, W_W\rangle_t = \rho dt.$ Moreover, \begin{align*} E_X(V) &= E_Y\left(\frac{dX}{dY} V \right)\\ &=E_Y\left(\left(\frac{dY}{dX}\right)^{-1} V \right)\\ &=E_Y(VW). \end{align*}

$\endgroup$
  • $\begingroup$ but if $V$ is not martingale under $Y,$ then it doesn't make sense? $\endgroup$ – A.Oreo Aug 15 '17 at 14:22
  • $\begingroup$ $V$ does not have to be martingale. The purpose of measure change is to make the computation feasible, rather than to make the quantity (e.g., the payoff of an option) a martingale, though $W$ should be a martingale, and the discounted value of $V$ should be a martinagle. $\endgroup$ – Gordon Aug 15 '17 at 14:28
  • $\begingroup$ It's seem you under the world of $W,$ $\dfrac{dV}{V} = \sigma_V\sigma_W\rho d t + \sigma_V(dW_V - \sigma_Wd t).$ Sorry, could you show detail, I may confused here. $\endgroup$ – A.Oreo Aug 15 '17 at 14:39
  • $\begingroup$ I will add some details to the answer. $\endgroup$ – Gordon Aug 15 '17 at 14:53
  • $\begingroup$ See added addendum. $\endgroup$ – Gordon Aug 15 '17 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.