3
$\begingroup$

I have a quite basic question, but I can't find a reference with it.

Recall that we can use the Black-Scholes formula to price a European call or put for a market consisting when:

  • the underlying asset following geometric Brownian motion;
  • the risk free interest rate is considered constant;
  • the volatility of the underlying asset returns is constant.

In deriving this, one writes the call/put as expectations of discounted payoffs, e.g. $C=E^Q[\exp^{-rT}(S_T-K)_+]$ for call ($Q=$risk neutral prob.), where $(S_t)$ is follows geometric Brownian motion.

My question is : what is the variance of what lies in the bracket ? I ask this for calls and puts.

$\endgroup$
3
  • $\begingroup$ Why do you need that for? Are you trying to compute the volatility of the value of the call? $\endgroup$
    – SRKX
    Jun 27, 2013 at 15:31
  • $\begingroup$ Just to be sure: If $X:=\exp{(-rT)}(S_T-K)_+$ you want to know $Var(X)$ ? $\endgroup$
    – math
    Jun 28, 2013 at 9:25
  • $\begingroup$ Yes. I am pretty sure that this follows the same derivation as for B-S formula, but curiously, I can't find any reference with it. $\endgroup$
    – Amin
    Jun 28, 2013 at 9:47

1 Answer 1

1
$\begingroup$

I just sketch how you can derive the formula. In your setting we have $S_T=s_0e^{\sigma W_T-\frac{1}{2}\sigma^2T}$. W.l.o.g we put $s_0=1$. Hence $S_T$ is lognormal distributed and $S_T^2=e^{2\sigma W_T-\sigma^2T}$, which is still lognormal. $Var(X)=E[X^2]-E[X]^2$. We just have to compute $E[X^2]$. Let $A:=\{S_T>K\}$

$$E[X^2]=E[(S_T-K)_+^2]=E[S_T^2\mathbf1_A]-2KE[S_T\mathbf1_A]+K^2Q[A]$$

Now the two last terms you have already calculated in the usual derivation of Black Scholes. For the first one you use the same techniques as for $E[S_T\mathbf1_A]$ with a different lognormal distribution.

$\endgroup$
3
  • $\begingroup$ my $S_T$ is the discounted one and also my $K$. $\endgroup$
    – math
    Jun 28, 2013 at 10:02
  • $\begingroup$ OK, I restate my question, as I realize it wasn't written the right way : do you know if in the end, we get a 'nice' formula ? Again, if there is a reference (which I guess exists), that would be fine. $\endgroup$
    – Amin
    Jun 28, 2013 at 10:05
  • $\begingroup$ @Amin you get a similar but of course longer expression as in the bs formula. $\endgroup$
    – math
    Jun 28, 2013 at 12:02

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.