# Formula for variance of European call/put in Black Scholes

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.

-
Why do you need that for? Are you trying to compute the volatility of the value of the call? – SRKX Jun 27 '13 at 15:31
Just to be sure: If $X:=\exp{(-rT)}(S_T-K)_+$ you want to know $Var(X)$ ? – user8 Jun 28 '13 at 9:25
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. – Amin Jun 28 '13 at 9:47

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.
my $S_T$ is the discounted one and also my $K$. – user8 Jun 28 '13 at 10:02