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
    $\begingroup$ If you assume log-normality for $S(T)$, then $\log S(T)$ is normal. This will become a simple exercise. $\endgroup$
    – Gordon
    Feb 6, 2019 at 20:36
  • $\begingroup$ It’s called Bachelier model, there is explicit formula $\endgroup$
    – Ezy
    Feb 6, 2019 at 23:44
  • $\begingroup$ Since you haven't accepted this, is ZRH's answer sufficient or do you need more clarification? $\endgroup$
    – Sanjay
    Feb 10, 2019 at 2:36

1 Answer 1


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:


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 Wheels
    Feb 6, 2019 at 20:21
  • $\begingroup$ Sorry I was a bit sloppy before, edited my answer $\endgroup$
    – ZRH
    Feb 6, 2019 at 20:53

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