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A binary option with payout \$0/\$100 is trading at \$30 with 12 hours to expiration.

Assuming the underlying follows a geometric Brownian motion (hence volatility remains constant), what probability distribution describes the option's maximum price between now and expiration?

I'm looking for a generic "formula". Even though I used price and expiration, I'm assuming the generic formula is a function of volatility (of course, price and expiration determine volatility).

More concretely:

Assume short time to expiry and hence null interest rates and dividends are null.

The time $t$ Black price (underlying $S_t$ is a GBM $dS_t = \sigma S_t dW_t$, $\sigma>0$, constant number) of a $K$-strike cash-or-nothing binary (digital) call option paying $1_{\{S_T>K\}}$ dollars at expiry time $T$ is $$P_t\triangleq \Phi\left(\frac{\ln(S_t/K)-0.5\sigma^2(T-t)}{\sigma\sqrt{T-t}}\right).$$

We are interested in the distribution (or just time $0$ expectation) of the variable:

$$\max_{t\in[0,T]} P_t, $$ (with fixed $T$, $K$ and $\sigma$) much like one is interested in the distribution (or just time $0$ expectation) of $$\max_{t\in[0,T]} S_t.$$

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You'll have to assume the dynamics of the underlying asset. You have a model in mind? GBM? It could help somebody posting an answer. –  SRKX Oct 16 '11 at 6:38
    
Good point. I was thinking Black-Scholes. –  barrycarter Oct 16 '11 at 11:26
    
I'd bet there is no closed-form solution. You are asking for the maximum of a highly unusual random process (i.e. the option TV process). –  Brian B Oct 17 '11 at 16:11
    
I'd settle for a non-closed-form solution or approximation. This could be modeled as a random walk with drift, but the drift itself changes with each step. –  barrycarter Oct 17 '11 at 16:41
    
Clarify the question: do you mean - if we consider price series of a binary option, what is the PDF of maximum of the price of such option? –  onlyvix.blogspot.com May 27 '12 at 0:03
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