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I am following a paper on how to price foreign bonds but there are some difference between my derivation and theirs. I am trying to express foreign bond in terms of domestic bond and exchange rate.

My question is: Can I treat my problem as Product of "Geometric Brownian Motion Processes in Risk-Neutral Measure" (i.e. can I derive everything starting from Q-measure, instead of change to Q-measure at the end of derivation)?

Given $B_R(t)$, $B_{R_f}(t)$, $e(t)$, and $C(t)$ stands for domestic bonds, foreign bonds, exchange rate, and risk-free cash.

$$ \begin{split} \frac{dC(t)}{C(t)} &= r(t)dt\\ \frac{de(t)}{e(t)} &= \xi(t)dt+\sigma_e[dW_e(t)+\lambda_edt]\\ &= [r(t)-r_f(t)]dt+\sigma_edW_e(t) \end{split} $$ where $$ \lambda_e=\frac{-\xi(t)+r(t)-r_f(t)}{\sigma_e} =\xi(t)dt+\sigma_edW_e^Q(t) $$ where $$dW_e^Q(t)=dW_e(t)+\lambda_edt$$


$$ \begin{split} \frac{dB_R(t)}{B_R(t)} &= r(t)dt+\sigma_R[dW_r(t)+\lambda_rdt]\\ &= r(t)dt+\sigma_RdW_r^Q(t) \end{split} $$


$$ \begin{split} \frac{d\widehat{B_{R_f}(t)}}{\widehat{B_{R_f}(t)}} &= r_f(t)dt+\sigma_{R_f}[dW_{r_f}(t)+\lambda_{r_f}dt] \\ &= r_f(t)dt+\sigma_{R_f}dW_{r_f}^{Q_f}(t) \end{split} $$


$$ Corr[dW_{r_f}^{Q_f}(t),dW_e^Q(t)] = \rho_{r_f,e}dt $$


I'd like to translate the foreign bond into domestic terms, i.e. $\widehat{B_{R_f}(t)}$ units of the foreign currency are worth $\widehat{B_{R_f}(t)} * e(t)$ in the domestic currency.

$B_{R_f}(t) = \widehat{B_{R_f}(t)} * e(t)$. By Ito lemma, I got

$dB_{R_f}(t) = \widehat{B_{R_f}(t)} * de(t)+ e(t) * d\widehat{B_{R_f}(t)} + d\widehat{B_{R_f}(t)} * de(t)\\$

$~~~~~~~~~~~~=\widehat{B_{R_f}(t)}e(t)\left[\xi(t)dt+\sigma_edW_e^Q(t)\right]+\widehat{B_{R_f}(t)}e(t) \left[r_f(t)dt+\sigma_{R_f}dW_{r_f}^{Q_f}(t) \right]+\widehat{B_{R_f}(t)}e(t)\left[\xi(t)dt+\sigma_edW_e^Q(t)\right]\left[r_f(t)dt+\sigma_{R_f}dW_{r_f}^{Q_f}(t) \right]$

$~~~~~~~~~~~~=B_{R_f}(t)\left\{ \left[\xi(t)dt+\sigma_edW_e^Q(t)\right] + \left[r_f(t)dt+\sigma_{R_f}dW_{r_f}^{Q_f}(t)\right] +\left[\xi(t)dt+\sigma_edW_e^Q(t)\right]\left[r_f(t)dt+\sigma_{R_f}dW_{r_f}^{Q_f}(t) \right] \right\}$

$\frac{dB_{R_f}}{B_{R_f}}=\left[\xi+r_f+\sigma_{R_f}\sigma_e\rho_{r_f,e}\right]dt+\sigma_{R_f}dW_{r_f}^{Q_f}+\sigma_edW_e^Q\\$

$~~~~~~~~=\left[r-\lambda_e\sigma_e+\sigma_{R_f}\sigma_e\rho_{r_f,e}\right]dt+\sigma_{R_f}dW_{r_f}^{Q_f}+\sigma_edW_e^Q~~~where~~ \xi+r_f=r-\lambda_e\sigma_e\\$

My goal is to rewrite the above equation into the following form so that I can have $dW_{r_f}^{Q}=[**dt+dW_{r_f}^{Q_f}]$.

$\frac{dB_{R_f}}{B_{R_f}}=rdt+\sigma_{R_f}\left[\left(-\frac{\lambda_e\sigma_e}{\sigma_{R_f}}+\sigma_e\rho_{r_f,e}\right)dt+dW_{r_f}^{Q_f}\right]+\sigma_edW_e^Q$

$dW_{r_f}^{Q}=\left(-\frac{\lambda_e\sigma_e}{\sigma_{R_f}}+\sigma_e\rho_{r_f,e}\right)dt+dW_{r_f}^{Q_f}$

Is there anything wrong within the derivation? I cannot get the same answer from the paper which has

$\frac{dB_{R_f}}{B_{R_f}}=rdt+\sigma_{R_f}\left[\rho_{r_f,e}dt+dW_{r_f}^{Q_f}\right]+\sigma_edW_e^Q$

Thank you very much for your help

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  • $\begingroup$ Please, explicitly mention the paper in the question. $\endgroup$ – Raskolnikov Nov 13 '19 at 8:52
  • $\begingroup$ This is a working paper that I am not supposed to post it publicly. Thanks for the suggestion! $\endgroup$ – Ying Nov 21 '19 at 5:07

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