Compute the price of the payoff $(2\log(S(T))-K)^+$. Before I do any algebra, I want to make sure I understand. To solve this problem, I need to solve the Black Scholes PDE with boundary condition $C(S,T)=(2\log(S(T))-K)^+$ instead of $C(S,T)=(S-K)^+$. Then I will be done. Is there another way to do it?
3
-
3$\begingroup$ If you assume log-normality for $S(T)$, then $\log S(T)$ is normal. This will become a simple exercise. $\endgroup$– GordonFeb 6, 2019 at 20:36
-
$\begingroup$ It’s called Bachelier model, there is explicit formula $\endgroup$– EzyFeb 6, 2019 at 23:44
-
$\begingroup$ Since you haven't accepted this, is ZRH's answer sufficient or do you need more clarification? $\endgroup$– SanjayFeb 10, 2019 at 2:36
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Assuming the underlying follows GBM price dynamics, I would do the following to avoid solving the PDE: $2 log(S(T))-K$ is positive for $S>e^{K/2}$. So if you take $g_{T}(\xi)$ to be the lognormal distribution of the underlying at time $T$, given an initial underlying price $S(t)$, then you should be able to obtain the solution as:
$C(S(t),t)=e^{-r(T-t)}\int_{e^{K/2}}^{\infty}g_{T}(\xi)(2log(\xi)-K)d\xi$
wrt Gordon's comment, you would apply variable substitution to solve this
-
$\begingroup$ As I understand, the problem is to compute the price of the payoff at time $t$ in terms of $S(t)$, i.e. the expected value of $\max(2\log(S(T))-K,0)$ at time $t$. $\endgroup$– L WheelsFeb 6, 2019 at 20:21
-
$\begingroup$ Sorry I was a bit sloppy before, edited my answer $\endgroup$– ZRHFeb 6, 2019 at 20:53