I am looking at a paper by V. Piterbarg, Funding beyond discounting: collateral agreements and derivatives pricing, that you can download on the following link, in which the author adapts the Black-Scholes pricing framework to introduce collateral and funding at a non-risk-free rate.

Letting $V_t \equiv V(t)$ and $C_t \equiv C(t)$, I am having trouble to go from equation $(3)$...

$$ V_t = E_t \left[ e^{-\int_t^Tr_F(u)du}V_T+\int_t^Te^{-\int_t^ur_F(v)dv}\left( r_F(u)-r_C(u)\right) C_u \ du \right]$$

... to equation $(5)$:

$$ V_t = E_t \left[ e^{-\int_t^Tr_C(u)du}V_T\right]-E_t \left[\int_t^Te^{-\int_t^ur_C(v)dv}\left( r_F(u)-r_C(u)\right) \left(V_u-C_u\right)du \right]$$

According to the author, to go from $(3)$ to $(5)$ we only need to "rearrange terms".

Can anybody show how to go from one to the other?


One derivation is to replace $V_u$ in Equation $(5)$ using the expression given by Equation $(3)$ and then work out to reach $(5)$; see Appendix A in this paper for more details. Here, we provide another derivation. See also this question.

We recall that, from $(2)$ of Piterbarg, \begin{align*} V_t = \Delta (t) S(t) + \gamma(t), \end{align*} where $\Delta (t)= \frac{\partial V(t)}{\partial S}$, and $\gamma(t)$ is the cash account that satisfies \begin{align*} d\gamma(t) &= \big[r_C(t) C(t) + r_F(t)(V(t)-C(t))-(r_R(t)-r_D(t))\Delta(t)S(t) \big]dt\\ &=\big[r_F(t)V(t) + (r_C(t)-r_F(t)) C(t)-(r_R(t)-r_D(t))\Delta(t)S(t) \big]dt. \end{align*} Moreover, based on Equation $(4)$ in the paper, \begin{align*} dS(t)/S(t) = (r_R(t)-r_D(t))dt + \sigma_S(t) dW_S(t). \end{align*} Then, from the self-financing condition, \begin{align*} dV_t &= \Delta (t) dS(t) + d\gamma(t)\\ &=\big[r_F(t)V_t + (r_C(t)-r_F(t)) C(t)\big]dt + \Delta (t)S(t)\sigma_S(t) dW_S(t). \tag{*} \end{align*}

From $(*)$, \begin{align*} d\left(e^{-\int_0^t r_F(v)dv}V_t \right) &=-r_F(t)e^{-\int_0^t r_F(v)dv}V_tdt + e^{-\int_0^t r_F(v)dv}dV_t\\ &=e^{-\int_0^t r_F(v)dv}\big[(r_C(t)-r_F(t)) C(t)dt + \Delta (t)S(t)\sigma_S(t) dW_S(t)\big]. \end{align*} Therefore, \begin{align*} e^{-\int_0^T r_F(v)dv}V_T-e^{-\int_0^t r_F(v)dv}V_t &=\int_t^Te^{-\int_0^u r_F(v)dv}\big[(r_C(u)-r_F(u)) C(u)du\\ &\qquad + \int_t^T\Delta (u)S(u)\sigma_S(u) dW_S(u). \end{align*} Taking conditional expectation with respect to $\mathscr{F}_t$ on both sides, we obtain that \begin{align*} E_t\left(e^{-\int_0^T r_F(v)dv}V_T \right)-e^{-\int_0^t r_F(v)dv}V_t &=E_t\left(\int_t^Te^{-\int_0^u r_F(v)dv}\big[(r_C(u)-r_F(u)) C(u)du\right), \end{align*} which leads to Equation $(3)$ in Piterbarg, that is, \begin{align*} V_t &= E_t\left(e^{-\int_t^T r_F(v)dv}V_T + \int_t^Te^{-\int_t^u r_F(v)dv}\big[(r_F(u)-r_C(u)) C(u)du\right)\tag{3} \end{align*}

$$$$ To derive Equation $(5)$, we note that, from $(*)$ above, by rearranging terms, \begin{align*} dV_t &= \big[r_F(t)V_t + (r_C(t)-r_F(t)) C(t)\big]dt + \Delta (t)S(t)\sigma_S(t) dW_S(t)\\ &=\big[r_C(t)V_t + (r_F(t)-r_C(t))(V_t -C(t))\big]dt + \Delta (t)S(t)\sigma_S(t) dW_S(t).\tag{**} \end{align*} As above, \begin{align*} d\left(e^{-\int_0^t r_C(v)dv}V_t \right) &=-r_C(t)e^{-\int_0^t r_C(v)dv}V_tdt + e^{-\int_0^t r_C(v)dv}dV_t\\ &=e^{-\int_0^t r_C(v)dv}\big[(r_F(t)-r_C(t))(V_t -C(t))dt + \Delta (t)S(t)\sigma_S(t) dW_S(t) \big], \end{align*} and, consequently, \begin{align*} E_t\left(e^{-\int_0^T r_C(v)dv}V_T \right) -e^{-\int_0^t r_C(v)dv}V_t&=E_t\left( \int_t^Te^{-\int_0^u r_C(v)dv}\big[(r_F(u)-r_C(u))(V_u -C(u))du\right), \end{align*} which leads to Equation $(5)$ in Piterbarg immediately, that is, \begin{align*} V_t =E_t\left(e^{-\int_t^T r_C(v)dv}V_T \right) - E_t\left(\int_t^Te^{-\int_t^u r_C(v)dv}\big[(r_F(u)-r_C(u))(V_u -C(u))du \right). \tag{5} \end{align*}


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.