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

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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?


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.


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*}

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  • $\begingroup$ 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

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