0
$\begingroup$

I am looking for an explanation of the following fact, which seems to be rather simple yet I am missing something. Say that $S_t$ is a stock following GBM $$ dS_t = r S_td_t + \sigma S_t dW_t,$$ and I want to price a derivative with payoff $max(S_T^2-K,0)$. I know how to do this using the risk-neutral expectation, but it's rather long and apparently unnecessary.

I read in a book (Mark Joshi's Concepts...) another solution which uses the forward price. If $F_T(t)$ is the forward price of $S$ at time $t$, then one can write

$$ F_T(T)^2 = (F_T(0)^2 e^{\sigma^2 T})e^{-\frac{\sigma'^2}{2} T + \sigma'\sqrt{T}N(0,1)},$$ with $\sigma'=2\sigma$, so then one can use Black's formula with forward equal to $F_T(0)^2 e^{\sigma^2T}$ and volatility $\sigma'$.

But how is Black's formula justified here exactly? I assume that the formula can be applied for a call option on the forward price of an asset, but how do we know there exists a tradable asset with volatility $\sigma'$, for instance? Or is this even necessary?

$\endgroup$
3
  • 1
    $\begingroup$ Hi @dbluedesk. A similar question was asked here: quant.stackexchange.com/questions/26240/…. Gordon proposes a direct approach and I suggest another one directly inspired by Mark Joshi's work $\endgroup$
    – Quantuple
    Feb 21, 2017 at 14:45
  • $\begingroup$ Hello @Quantuple, thank you! I think my question is: when you say in the linked answer that $N_{t,T}$ is an asset, how do you get this? I always thought you could use any self-financing portfolio as a numeraire, but I don't see how you can reproduce your $N_{t,T}$ $\endgroup$
    – dbluesk
    Feb 21, 2017 at 15:03
  • 1
    $\begingroup$ A numéraire is essentially a (tradable) asset whose price is always positive. $N(t,T)$ is merely the $t$-value of a contingent claim with payoff $S_T^2$. $\endgroup$
    – Quantuple
    Feb 21, 2017 at 15:41

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Browse other questions tagged or ask your own question.