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Similarly to the Black and Scholes formula, I am looking to replicate Bachelier's caplet formula with two digital options: (1) asset-or-nothing (forward rate in this case) and (2) cash-or-nothing. For reference, Bachelier's caplet formula is: $$c(t,T_{i-1},T_i) = \delta*P(t,T_i)*\Bigl((F(t,T_{i-1},T_i)-K)*\Phi(D)+\sigma*\sqrt{T_{i-1}-T_i}*\phi(D)\Bigr)$$ $$where$$$\delta=$ frequence factor, $P(t,T_i)$ is the discount factor, $D={F(t,T_{i-1},T_i)-K}/{\sigma*\sqrt{T_{i-1}-T_i}}$, $\Phi$ is the cumulative distribution function and $\phi$ is the probability density function

For reference, in the BS formula, the part $$S*N(d_1)$$ is for the asset-or-nothing and the part $$K*e^{-rt}*N(d_2)$$ is for the cash-or-nothing.

My take is that, as $\Phi(D)$ represents the probability to be in-the-money, a digital caplet cash-or-nothing is value as: $$D_{cash}(t,T_{i-1},T_i) = \delta*P(t,T_i)*K*\Phi(D)$$ and a digital caplet asset-or-nothing is value as: $$D_{asset}(t,T_{i-1},T_i) = \delta*P(t,T_i)*\Bigl(F(t,T_{i-1},T_i)*\Phi(D)+\sigma*\sqrt{T_{i-1}-T_i}*\phi(D)\Bigr)$$

Many thanks!

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  • $\begingroup$ This may be helpful for you. $\endgroup$
    – Gordon
    Jan 18, 2019 at 19:03
  • $\begingroup$ Thanks for the link unfortunately, unless I am missing something, I don't see how this replies to my question. $\endgroup$
    – qbodart
    Jan 21, 2019 at 17:06

2 Answers 2

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The Bachelier digital caplet formula is just $ \delta*P(t,T_i)*\Phi(D).$ For a derivation see this answer with the $d_2$ there replaced by your $D$.

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Assume that you are working on a measure $\mathbb{Q}$ under which your underlying $S$ is a martingale (e.g., $S$ is a forward price/LIBOR rate under the forward measure, numeraire of which is the discount factor to your derivative's payment date). Just notice that

\begin{align} \mathbb{E}^\mathbb{Q} \left[\left(S_T - K\right)^+\right] & = \mathbb{E}^\mathbb{Q} \left[\left(S_T - K\right) \mathbf{1}_{S_T > K}\right] \\ & = \mathbb{E}^\mathbb{Q} \left(S_T \mathbf{1}_{S_T > K}\right) - K \mathbb{Q} \left(\left\{S_T > K\right\}\right) \\ & = S_0 \mathbb{E}^\mathbb{Q} \left(\frac{S_T}{S_0} \mathbf{1}_{S_T > K}\right) - K \mathbb{Q} \left(\left\{S_T > K\right\}\right) \end{align}

Because $S$ is a martingale, $\frac{S_T}{S_0}$ is a random variable with expected value 1 (not necessarily positive for a Gaussian underlying… Let us say that if $S_0$ and $K$ are positive, the indicator function saves you :) ), and thus defines a measure change. Let us define $\mathbb{Q}^S$ by $\frac{\mathrm{d}\mathbb{Q}^S}{\mathrm{d}\mathbb{Q}} = \frac{S_T}{S_0}$. The corresponding numeraire is $S$ itself. You then have $$ \mathbb{E}^\mathbb{Q} \left[\left(S_T - K\right)^+\right] = S_0 \mathbb{Q}^S \left(\left\{S_T > K\right\}\right) - K \mathbb{Q} \left(\left\{S_T > K\right\}\right) $$

This gives you exactly your decomposition in the asset-or-nothing option and $K$ digital options!

So in the Bachelier formula (just carry out the integration), the asset-or-nothing option value is $$ S_0 \Phi \left(D\right) + \sigma \sqrt{T} \phi \left(D\right) $$ and the binary option term is $$ K \Phi \left(D\right) $$

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