Given the stock price process dS(t) = σ dW(t) with initial stock level S(0) and a barrier H such that H > S(0), what is the probability that the barrier is

 

The stock price at time T is normally distributed with mean equal to the initial stock price S(0) (since there is no drift in the process) and variance equal to σ²T. The probability of breaching the barrier H at any time up to T can be approximated by considering the probability of S(T) exceeding H, which is calculated as:

 

P(breaching H between 0 and T) ≈ 1 - Phi((H - S(0)) / (σ * sqrt(T)))

 

Here, Phi represents the CDF of the standard normal distribution, and ((H - S(0)) / (σ * sqrt(T))) standardizes the threshold level in terms of the standard normal distribution. a) Probability the Barrier is Breached:

 

To calculate the probability that the barrier H is breached by time T in a Brownian motion with volatility σ and initial stock price S(0), we can use the reflection principle (https://www.finance-tutoring.fr/the-reflection-principle/)

 

The probability that a Brownian motion will exceed a certain level H for the first time by T is twice the probability that it will be above H at time T. The relevant probability can be found using the cumulative distribution function (CDF) of the standard normal distribution.

 

The probability P that the barrier H is breached by time T is:

 

P(barrier breached by time T) = 2P(S(T) >= H)

 

Since S(t) is a Brownian motion with zero drift, the stock price at time T is normally distributed with a mean of S(0) and variance σ^2T. The probability of breaching the barrier H is:

 

P(S(T) >= H) = 1 - Φ((H - S(0)) / (σ√T))

 

where Φ is the CDF of the standard normal distribution. Therefore, the probability of the barrier being breached is:

 

P(barrier breached by time T) = 2(1 - Φ((H - S(0)) / (σ√T)))