Finding a Call Option’s Profit Probability Under Geometric Brownian Motion

Summary

The document asks for the probability that a call buyer earns more than a specified profit at expiry when the stock follows geometric Brownian motion. It gives the initial stock price, drift and volatility parameters, strike, expiry, and risk-free rate, then attempts to transform the payoff condition into a threshold event for the terminal stock price. The calculation uses the lognormal distribution of the stock price and a standard normal cumulative distribution function.

The author’s attempt applies the risk-neutral drift adjustment while evaluating a probability question framed under the stated stock process, leading to a result inconsistent with the supplied answer. The key lesson is to distinguish the physical price dynamics used to calculate an investor’s outcome probability from risk-neutral dynamics used to price derivatives. The post presents the setup and a reported answer, but it does not include a complete correction or explain which probability measure the exercise intends; that interpretation matters to the result.

Key ideas

• A call buyer’s profit threshold can be rewritten as a terminal stock-price threshold.
• Under geometric Brownian motion, the log terminal price has a normal distribution.
• Probability calculations depend on whether the stated dynamics are physical or risk-neutral.
• The attempted solution and supplied answer expose a drift-measure issue but do not fully resolve it.

Tags

# Geometric brownian motion and probabilities


# Geometric brownian motion and probabilities












> A stock's price movement is described by the equations $dS_t=0.02S_tdt+0.25S_tdW_t$ and $S_0=100$. An investor buys a call option on said stock with a strike price $K=95$ which expires in $T=2$ years. What is the probability that the investor makes a profit greater than $20$ at expiry? The risk-free rate is $2\%$ (continuously compounded).

I'm also given $N(0.08144)=0.532454, N(0.434993)=0.668216,N(0.8907)=0.808041$ so no calculator will be needed.

My attempt: We know that (skipping Ito's lemma) $$S_t=S_0e^{-0.01125t+0.25W_t}$$ The probability I need to find is $P(S_2-K>20)=P(S_2>115)=P(e^{-0.01125*2+0.25W_2}>1.15)=P(\frac{W_2}{\sqrt{2}}>\frac{ln(1.15)+0.0225}{0.25\sqrt{2}})=1-N(0.458946)$

which must be wrong.

I've gone through this countless times without success, can anyone tell what I'm doing wrong? If it helps the correct answer is given as $19.2\%$ Thanks in advance.

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