2
$\begingroup$

We denote discount factor $D(t),$ zero coupon bond $B(t,T),$ $E_t[X] = E[X|\mathcal{F}(t)]$ and $T$-forward measure $E_t^{T}[\ ].$

First, let me fix the Libor and Forward Libor to avoid ambiguity

Libor $L(t,T):$ $$B(t, T)\cdot \Big(1 + (T-t) L(t, T)\Big) = 1.$$ Forward Libor $F(t,T-\delta,T):$ $$\Big(1 + (T-t)F(t,T-\delta,T)\Big)B(t,T) = B(t,T-\delta)$$

Now we see the cap $$C(t;T,L^*) = \dfrac{1}{D(t)}E_t\left[D(T)\delta\Big(F(t,T-\delta,T) - L^*\Big)^+\right]$$ We can change into forward measure $$C(t;T,L^*) = \delta B(t,T)E^T_t\left[\Big(F(t,T-\delta,T) - L^*\Big)^+\right]$$ and $F(t,T-\delta,T)$ is $T$-forward martingale, the above formula become the standard Black-Scholes.

But if we choose $$C(t;T,L^*) = \dfrac{1}{D(t)}E_t\left[D(T)\delta\Big(L(T-\delta,T) - L^*\Big)^+\right]$$ then we can transform into $$C(t;T,L^*) = (1+\delta L^*)\cdot E^{T}_{t}\left[\left(\dfrac{1}{1+\delta L^*} - B(T-\delta,T)\right)^+\right]$$ it become a bond put option expiring at time $T - \delta$ maturing at time $T.$

But $B(t,T)$ is impossible log-normal under $T$-forward measure, then we can't use Black-Scholes. So how to deal with for this case?

$\endgroup$
  • $\begingroup$ As you said, the bond price $B(t,T)$ will certainly not follow Black-Scholes dynamics. However, the forward Libor could. Therefore, it depends on what you want to do. If you want to price, you can use Black-76 formula to price the cap from your formula $C(t;T,L^*) = \delta B(t,T) E^T_t \left[ \left( F(t,T-\delta,T) - L^*\right)^+ \right]$ $\endgroup$ – JejeBelfort May 31 '17 at 8:40
3
$\begingroup$

Note that \begin{align*} &\ \dfrac{1}{D(t)}E_t\left(D(T)\delta\Big(L(T-\delta,T) - L^*\Big)^+\right)\\ =&\ \dfrac{1}{D(t)}E\left(D(T-\delta) E\left(\frac{D(T)}{D(T-\delta)}\delta\Big(L(T-\delta,T) - L^*\Big)^+\mid\mathcal{F}_{T-\delta}\right) \mid \mathcal{F}_t\right)\\ =&\ \dfrac{1}{D(t)}E\left(D(T-\delta) B(T-\delta, T)\delta\Big(L(T-\delta,T) - L^*\Big)^+\mid\mathcal{F}_t\right)\\ =&\ (1+\delta L^*)\dfrac{1}{D(t)}E\left(D(T-\delta)\left(\dfrac{1}{1+\delta L^*} - B(T-\delta,T) \right)^+\mid\mathcal{F}_t\right)\tag{1}\\ =&\ (1+\delta L^*)B(t, T)E^T_t\left(\frac{D(T-\delta)}{D(T)}\left(\dfrac{1}{1+\delta L^*} - B(T-\delta,T) \right)^+\right). \end{align*} Your transformation from $$C(t;T,L^*) = \dfrac{1}{D(t)}E_t\left(D(T)\delta\Big(L(T-\delta,T) - L^*\Big)^+\right)$$ to $$C(t;T,L^*) = (1+\delta L^*)\cdot E^{T}_{t}\left(\left(\dfrac{1}{1+\delta L^*} - B(T-\delta,T)\right)^+\right)$$ does not appear correct.

We also note that $(1)$ is indeed the value of a put bond option with maturity $T-\delta$. Based on a certain short rate model such as the Hull-White model, this value can be computed analytically.

$\endgroup$

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.