1
$\begingroup$

Exercise 2.3.1.5: The payoff of a power option is $h(S_T)$, where the function h is given by $h(x) = x^\beta(x-K)^+$. Prove that the payoff can be written as the difference of European payoffs on the underlying assets $S^{\beta+1}$ and $S^\beta$ with strikes depending on K and $\beta$.

$\endgroup$
3
$\begingroup$

We have,
$$ h(x) = x^\beta(x-K)^+ = x^\beta (x - K) \, \mathbf{1}_{[x>K]}$$ Thus we get, $$ h(x) = x^{\beta+1}\mathbf{1}_{[x>K]} - K\,x^{\beta}\mathbf{1}_{[x>K]}$$ now $x \in [x>K]$ if and only if $ x \in [x^{\beta}>K^{\beta}]$ Therefore,
$$ h(x) = x^{\beta+1}\mathbf{1}_{[x^{\beta + 1}>K^{\beta + 1}]} - K\,x^{\beta}\mathbf{1}_{[x^{\beta}>K^{\beta}]}$$

Thus if we let $\Phi_\text{Asset-or-nothing}^\text{dig}$ the contract function for an asset or nothing binary european call. we can Write:
$$h(S_t) =\Phi_\text{Asset-or-nothing}^\text{dig}(S_{t}^{\beta + 1},K^{\beta + 1}) - K \,\Phi_\text{Asset-or-nothing}^\text{dig}(S_{t}^{\beta}, K^{\beta}) $$

$\endgroup$
  • $\begingroup$ Thank you Drmanifold for providing a great answer and thank you William for editing the answer. $\endgroup$ – Liwei Zhang Dec 20 '13 at 0:39

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.