Pricing a Vasicek Zero-Coupon Bond Call with Forward Measures

# Analytic ZCB call option under Vasicek


# Analytic ZCB call option under Vasicek












The call option with strike $X$ and maturity $T$ on a ZCB maturing at time $S$, where $T\le S$, is $$ZBO(t,T,S,X)=E_t[e^{-\int_t^Tr_sds}(P(T,S)-X)^+]$$ The ZCB price is denoted by $$P(t,T)=E_t[e^{-\int_t^Tr_sds}]$$ I am familiar with the Black-Scholes metholodogy of deriving the call option price from first principles, and I am interested in applying the methodology here to compute the bond option price. The Vasicek model allows the bond price to be computed analytically. The SDE for the short rate is $$dr_t=k(\theta-r_t)dt+\sigma dW_t$$ and it can be shown that the bond price is $$P(t,T)=A(t,T)e^{-B(t,T)r_t}$$ where $$B(t,T)=\frac{1}{k}(1-e^{-k(T-t)})$$ and $$A(t,T)=\exp{\bigg[\Big(\theta-\frac{\sigma^2}{2k^2}\Big)[B(t,T)-T+t]-\frac{\sigma^2}{4k}B(t,T)^2}\bigg]$$ My initial steps are to express $ZBO(t,T,S,X)$ in terms of indicator functions, as follows \begin{align} ZBO(t,T,S,X)&=E_t\Big(e^{-\int_t^Tr_sds}(P(T,S)-X)\mathbb{1}_{P(T,S)>K}\Big)\\ &=E_t\Big(e^{-\int_t^Tr_sds}P(T,S)\mathbb{1}_{P(T,S)>K}\Big)-XE_t\Big(e^{-\int_t^Tr_sds}\mathbb{1}_{P(T,S)>K}\Big)\\ &=E_t\Big(e^{-\int_t^Tr_sds}E_t[e^{-\int_T^Sr_sds}]\mathbb{1}_{P(T,S)>K}\Big)-XE_t\Big(e^{-\int_t^Tr_sds}\mathbb{1}_{P(T,S)>K}\Big)\\ &=E_t\Big(e^{-\int_t^Sr_sds}\mathbb{1}_{P(T,S)>K}\Big)-XE_t\Big(e^{-\int_t^Tr_sds}\mathbb{1}_{P(T,S)>K}\Big)\\ &=E_t\Big(P(t,S)\mathbb{1}_{P(T,S)>K}\Big)-XE_t\Big(P(t,T)\mathbb{1}_{P(T,S)>K}\Big)\\ \end{align} The main issues are that I have assumed $e^{-\int_t^Tr_sds}E_t[e^{-\int_T^Sr_sds}]=e^{-\int_t^Sr_sds}$ and that $P(t,T)$ appears to be a random variable, even though it is a $\mathbb{Q}$ expectation and thus a constant. Furthermore, the Black-Scholes methodology can only be applied after changing numeraires. If there is a Radon-Nikodym derivative that can do the following $$E_t\Big(P(t,S)\mathbb{1}_{P(T,S)>K}\Big)-XE_t\Big(P(t,T)\mathbb{1}_{P(T,S)>K}\Big)=E_t\Big(P(T,S)\mathbb{1}_{P(T,S)>K}\Big)-XE_t\Big(P(T,S)\mathbb{1}_{P(T,S)>K}\Big)$$ then the bond price can easily be found.

Any help is appreciated.

## Answer by user16651 (score 2, accepted)

https://quant.stackexchange.com/a/30789

Hint

Note $$X=(P(T,S)-K)^+=(P(T,S)-K)1_{P(T,S)>K}\tag 1$$ We know $$V(t,T,S,K)=\mathbb{E^Q}_t\left[\frac{B(t)}{B(T)}X\right]\tag 2$$ $(1)$ and $(2)$, we have $$V(t,T,S,K)=\underbrace{\mathbb{E}^Q_t\left[\frac{B(t)}{B(T)}P(T,S)1_{P(T,S)>K}\right]}_{I}-K\,\underbrace{\mathbb{E}_t^Q\left[\frac{B(t)}{B(T)}1_{P(T,S)>K}\right]}_{J}\tag 3$$ Now we should calculate $I$ and $J$. By application of Bayes formula, we have $$\mathbb{E}_t^{Q_S}\left[1_{P(T,S)>K}\right]=\frac{\mathbb{E}_t^Q\left[\frac{B(t)}{B(T)}P(T,S)1_{P(T,S)>K}\right]}{P(t,S)}$$ thus $$I=\mathbb{E}_t^Q\left[\frac{B(t)}{B(T)}P(T,S)1_{P(T,S)>K}\right]=P(t,S)\mathbb{E}_t^{Q_S}\left[1_{P(T,S)>K}\right]\tag 4$$ on the other hand $$\mathbb{E}_t^{Q_S}\left[1_{P(T,S)>K}\right]=Q_{S}\left(P(T,S)>K\Big{|}\,\mathcal{F}_t\right)\tag 5$$ $(4)$ and $(5)$ $$I=P(t,S)Q_{S}\left(P(T,S)>K\Big{|}\,\mathcal{F}_t\right)\tag 6$$ Similarly, we have $$\mathbb{E}_t^{Q_T}\left[1_{P(T,S)>K}\right]=\frac{\mathbb{E}_t^Q\left[\frac{P(T,T)}{B(T)P(0,T)}1_{P(T,S)>K}\right]}{\frac{P(t,T)}{B(t)P(0,T)}}=\frac{\mathbb{E}_t^Q\left[\frac{B(t)}{B(T)}1_{P(T,S)>K}\right]}{P(t,T)}$$ therefore $$J=P(t,T)\mathbb{E}_t^{Q_T}\left[1_{P(T,S)>K}\right]=P(t,T)\,Q_{T}\left(P(T,S)>K\Big{|}\,\mathcal{F}_t\right)\tag 7$$ $(3)$ , $(6)$ and $(7)$ $$V(t,T,S,K)=P(t,S)Q_{S}\left(P(T,S)>K\Big{|}\,\mathcal{F}_t\right)-KP(t,T)\,Q_{T}\left(P(T,S)>K\Big{|}\,\mathcal{F}_t\right)$$ Note

- $B(t)$ is Money account.

- $Q_T$ and $Q_S$ are forward measures.