Finding the distribution of maximum stock price under Black-Sholes model in a specified interval

Question

Stochastic differential equation of Black-Scholes model is defines as \begin{eqnarray}\label{ref9} dS_t = (r-d)S_tdt+ \sigma S_t dW_t \end{eqnarray}
where the interest rate $r$, the dividend yield $d$, and volatility $\sigma$ are assumed to be constant. Additionally, $S_0 \in (0, +\infty)$, $t \in [0, T]$, and $W_t$ is Brownian Motion stochastic process with following properties:

1- starts at zero

2-independent distributed increments

3-stationary distributed increments

4-normally distributed increments : $W_{t+s}-W_{t}\sim Normal(0,s)$

5-continuous sample paths

Another representation of Black-Scholes model is Geometric form which defined as \begin{eqnarray}\label{ref1} S_t = S_0 \exp\left\{(r-d-\frac{\sigma^2}{2})t+\sigma W_t\right\} \end{eqnarray}

My question here is: What is the distribution of \begin{eqnarray} M = max\{S(t),\, 0\leqslant t \leqslant T\} \end{eqnarray}

Answer 1

The distribution of the maximum of a 1-dimensional diffusion process is known, and for a Brownian motion with drift (the exponent of the geometric BM), the formula is fairly explicit. See "On the joint distribution of the maximum and its location for a linear diffusion" by E. Csáki, A. Földes, Antónia, and P. Salminen: http://www.numdam.org/item/?id=AIHPB_1987__23_2_179_0.

Answer 2

First of all, I would like to thank John Dawkins for his kind consideration and time. Honestly, I could not convince myself by reading the sources John preferred me. So, I decided to search more and find a solution to my question. Here, is my finding.

At the beginning, let's consider a standard Brownian motion $B(t)$. Assume that $T_a = inf\{t: B(t) = a\}$ stands for hitting time, i.e., the first time when Brownian motion hits level $a>0$, and $M(t) = \sup_{0\leqslant s\leqslant t}B(s)$. We have \begin{eqnarray} P(B(t)> a) = P(B(t)> a , M(t)> a)+P(B(t)> a , M(t)\leqslant a) \end{eqnarray} Note that $P(B(t)> a , M(t)\leqslant a)$ since $M(t)\geqslant B(t)$. Now \begin{equation*} P(B(t)> a , M(t)> a)= P(B(t)> a | M(t)> a)\times P(M(t)> a)\\ = P(B(t)> a | T_a< t)\times P(M(t)> a) \end{equation*} we have that $B(T_a+s)-a$ is a Brownian motion. Conditioned on $ T_a\leqslant t$, \begin{eqnarray} P(B(t)> a | T_a< t) = P(B(T_a+(t-T_a))-a> 0 |T_a< t) = \frac{1}{2} \end{eqnarray} since the Brownian motion satisfies $P(B(t)> 0) = \frac{1}{2}$ for every t. Applying this identity, we obtain \begin{eqnarray} P(B(t)> a) = \frac{1}{2}P(M(t)> a) \end{eqnarray} or \begin{eqnarray} P(M(t)> a) = 2P(B(t)>a) = P(|B(t)|>a) \Rightarrow P(M(t)\leqslant a)= P(|B(t)|\leqslant a) \end{eqnarray} we know that $B(t)\sim N(0,t)$, so \begin{eqnarray*} P(M(t)\leqslant a) = \frac{1}{\sqrt{2\pi t}}\int_{-a}^{a} e^{-\frac{x^2}{2t}} = \frac{2}{\sqrt{2\pi t}}\int_{0}^{a} e^{-\frac{x^2}{2t}} \end{eqnarray*} we can also write it down in this way \begin{eqnarray} P(M(t)\leqslant a) = 2P(B(t)\leqslant a) = 2\Phi(\frac{a}{\sqrt{t}}) \end{eqnarray} Now, consider a stock price $S_t = S_0 e^{\sigma W_t}$,i.e., a Brownian motion process without drift $(\mu = 0)$. Then, we conclude that \begin{eqnarray} M_0(T) = \max_{0\leqslant t \leqslant T}S_t = \max_{t\in[0,T]}S_0 e^{\sigma W_t} = S_0e^{\sigma\max_{t\in[0,T]}W_t} = S_0e^{\sigma M(T)} \end{eqnarray} Therefore, \begin{eqnarray*} P(M_0(T)\leqslant y)&=&P(S_0e^{\sigma M(T)}\leqslant y) \\ &=& P(e^{\sigma M(T)}\leqslant \frac{y}{S_0}) \\ &=& P(\sigma M(T)\leqslant log(\frac{y}{S_0})) \\ &=& 2\Phi\left(\frac{log(\frac{y}{S_0})}{\sqrt{T}}\right),\quad y>S_0 \end{eqnarray*} and for the inverse distribution function, we get this result \begin{eqnarray} F^{-1}_{M_0(T)}(x) = S_0e^{\sqrt{T}\Phi^{-1}(\frac{x}{2})} \end{eqnarray}

Generally, for the geometric Brownian motion, it is more complicated to find the exact formula. I only put the final result. Considering the following Geometric Brownian motion for a stock price,

\begin{eqnarray*} S(t) = S_0exp\left((r-d-\frac{\sigma^2}{2})t+\sigma W(t)\right) \end{eqnarray*} and maximum process of such model \begin{eqnarray} M_0(T) = \max_{0 \leqslant t \leqslant T}S(t) \end{eqnarray} we have that \begin{eqnarray} P(M_0(T)\leqslant x) = \Phi\left(\frac{-(r-d-\frac{\sigma^2}{2})T+log(\frac{x}{S_0})}{\sigma \sqrt{T}}\right)-\left(\frac{S_0}{x}\right)^{1-2\frac{(r-d)}{\sigma^2}}\Phi\left(\frac{-(r-d-\frac{\sigma^2}{2})T-log(\frac{x}{S_0})}{\sigma \sqrt{T}}\right), \quad x>S_0 \end{eqnarray}