# Pricing caplet with Bachelier (normal dynamic) using forward measure

I'm trying to price caplet with Bachelier under forward measure, but I can't find any solution. Remind that Bachelier assumed rates follow a normal dynamic. So here what I was doing :

$C_t(T,T+d)$ represents the price of a Caplet at time t, for a Caplet between T and T+d. \

$B(t,T)$ represents the ZCB between at time t, of maturity T. \

$r_t(T,T+d)$ represents the forward rate at time t, between T and T+d. \

$\Big( \dfrac{C_t(T,T+d)}{B(t,T+d)} \Big)_{t \in [0,T]}$ ($\mathbb{F}, \mathbb{P}^{T+d}$)-martingale, and : \begin{align} C_t(T,T+d) &= B(t,T+d) \cdot \mathbb{E}^{T+d} \Big[ \, \dfrac{C_{T+d}(T,T+d)}{B(T+d,T+d)} \; | \; \mathcal{F}_t \Big] \\ &= B(t,T+d) \cdot d \cdot \mathbb{E}^{T+d} \Big[ \, (r_T(T,T+d)-K)^+ \; | \; \mathcal{F}_t \Big] \end{align}

To calculate this, we must find $B^{T+d}$ under $\mathbb{P}^{T+d}$ : \begin{align} \begin{cases} P^{T+d} = L^{T+d}_{T+d} \cdot \mathbb{P}^* \qquad &\Big( L^{T+d}_{T+d} = \dfrac{\mathrm{d}\mathbb{P}^{T+d}}{\mathrm{d}\mathbb{P}^*} \Big) \\ L_t^{T+d} = \dfrac{B(t,T+d)}{P_t \cdot B(0,T+d)} \qquad &L^{T+d}_{T+d} = \dfrac{1}{P_{T+d} \cdot B(0,T+d)} \end{cases} \end{align}

$\sigma(t,T+d) \in \mathcal{F}_t$ represents the local volatility of $B(t,T+d)$, one get : \begin{align} \mathrm{d}B(t,T+d) = r_t.B(t,T+d).\mathrm{d}t + \sigma(t,T+d).\mathrm{d}B^*_t \end{align}

Then, with Girsanov's theorem : \begin{align} B_t^{T+d} = B^*_t - \displaystyle{\int_{0}^{t}} \sigma(s,T+d). \mathstrut{d}s \qquad \mbox{un } (\mathbb{F}, \mathbb{P}^{T+d})\mbox{ - MB} \end{align}

One get : \begin{align*} \begin{cases} X_t &= B(t,T) - B(t,T+d) \\ \mathrm{d}X_t &= r_t.X_t.\mathrm{d}t + \sigma_t^X . \mathrm{d}B_t^* \qquad \mbox{avec } \sigma_t^X = \sigma(t,T) - \sigma(t,T+d) \end{cases} \end{align*}

Then we should apply Ito's formula : \begin{align*} \mathrm{d}r_t(T,T+d) &= \dfrac{1}{d} \cdot \mathrm{d}\Big( \dfrac{X_t}{B(t,T+d)} \Big) \\ &= \rho_t(T,T+d).\mathrm{d}B_t^{T+d} \end{align*}

But here's my problem : when I apply it, I don't find the good Brownian motion, but $\mathrm{d}B_t^{T+d} = dB^*_t-\dfrac{\sigma(t,T+d)}{B(t,T+d)}dt$. \

Has anyone ever done this pricing ? Or do you have any indication to realize it ? I know the derivative of the forward rate, but I can't demonstrate it.

Thank you very much for reading me !

Romain

• No one has any idea ? – Romain Aug 25 '18 at 15:27