Determining the No Arbitrage price of max[B(T), S(T)]

Question

Following is given,

$dB(t)=rB(t)dt$

$dS(t)= (r-\delta)S(t)dt+\sigma S(t)dW(t)$

where, $r$ is the risk-free interest rate, $\delta$ the continous dividend yield $\sigma$ is the stock asset volatility and $W$ brownian motion.

By solving the SDE that is the money account B and applying Itô's lemma on the stock dynamic S I get,

$B(T)=e^{r(T-t)}$ and

$S(T)=S(t)e^{(r-\delta-\sigma^2/2)+\sigma(W(T)-W(t))}$

I know further that the No Arbitrage price is defined as,

$\Pi(t;X)=1/B(T)*E^Q_t[max[B(T),S(T)]]$

Question

Can I somehow cancel out B(T) since it is deterministic? How do I calculate the expectation of a maximum with a brownian motion W?

I am new to probability theory, financial mathematics and stochastic calculus. Would appreciate step-wise guidance. Thank you!

Bob Jansen · Accepted Answer · 2019-12-10 20:29:08Z

8

$\max(B_T,S_T)=\max(0,S_T-B_T)+B_T,$ so this is just a call option (with strike $B_T$) plus $B_T.$

Bob Jansen♦

8,6206 gold badges39 silver badges60 bronze badges

answered Dec 10, 2019 at 20:17

user39119

$\begingroup$ Hi, is it an integral that i have to solve afterwards? $\endgroup$
– Dreason94
Commented Dec 11, 2019 at 20:56
$\begingroup$ Yes, an integral, but Black and Scholes already solved it in 1973 ;) . Just use the Black Scholes formula for a call with $B_T$ replacing the strike price $X$ and with an additional $B_T$ term added. $\endgroup$
– Alex C
Commented Dec 11, 2019 at 21:11
$\begingroup$ Thank you very much. Say that it would be an exam question. Can I just type out N[d_1] and N[d_2] and just say the are normal cumulative density functions? $\endgroup$
– Dreason94
Commented Dec 11, 2019 at 22:04

Add a comment |

1 Answer 1