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 $$\max_{t\in[0,T]} S_t.$$

  • 4
    $\begingroup$ 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. $\endgroup$ – SRKX Oct 16 '11 at 6:38
  • $\begingroup$ Good point. I was thinking Black-Scholes. $\endgroup$ – user59 Oct 16 '11 at 11:26
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    $\begingroup$ 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). $\endgroup$ – Brian B Oct 17 '11 at 16:11
  • $\begingroup$ 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. $\endgroup$ – user59 Oct 17 '11 at 16:41
  • $\begingroup$ 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? $\endgroup$ – onlyvix.blogspot.com May 27 '12 at 0:03

I believe this can be solved using the reflection theorem: $$P(\max S_t > x) = 2 P (S_T > x)$$ Hence the required densities can be obtained solely from the distribution of $S_T$.

There is a one to one correspondence between $\max P_t$ and $\max S_t$, so that $$P (\max P_t < y) = P (\max S_t < g(y) )$$ where the function $g$ is the inverse of the function for $P_t$ in terms of $S_t$ given in the OP.

Continuing that logic I get for the final answer

$$P (\max P_t < y) = 1 - 2 N \left[ \frac{\ln(S_0/K)}{ \sigma \sqrt{T}} - InvN(y) - \sigma \sqrt{T}\right]$$

where $N[]$ is the cumulative normal distribution, $InvN[]$ is its inverse, $S_0$ is the stock price today.

Have to say I'm not 100pct confident of the algebra.

| improve this answer | |
  • $\begingroup$ I think the question is for the distribution of $\max_{t \in [0, T]}P_t$. $\endgroup$ – Gordon Dec 1 '15 at 18:38
  • $\begingroup$ Please double check the formula for $P_t$. The identity $prob (\max P_t < y) = 1 - 2 N [(ln(S_0/K) /\sigma \sqrt(T)) - InvN(y) - \sigma \sqrt(T)]$ do not appear correct. $\endgroup$ – Gordon Dec 1 '15 at 21:38

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