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3

You know that Brownian motion {W(t)} is a stochastic process with the following properties: (Independence of increments) W(t) − W(s) , for t > s , is independent of the past, that is, of W(u) , 0 ≤ u ≤ s, or of $F_s$ , the σ-field generated by W(u), u ≤ s. (Normal increments) W(t) − W(s) has Normal distribution with mean 0 and variance t − s. This implies ...

5

You ask 2 questions and I try to answer: 1) Why do we use geometric Brownian motion ($\ln S_t-\ln S_0$ is normally distributed)? In this case you have $$S_t = S_0 \exp( (\mu-\sigma^2/2) t + \sigma B_t),$$ which means that you model positive prices. Furthermore the log-return $$\ln(S_t/S_0) = (\mu-\sigma^2/2) t + \sigma B_t,$$ is normally distributed. ...

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Given its price today, the stock price at time T is lognormally distributed, whereas $lnS_T$ is normally distributed, that is $lnS_T$ ~ $N \Bigr(lnS_0 + (\mu- \frac{\sigma^2}{2}T),\sigma^2T \Bigl)$ see for example Hull - Options, Futures, and other Derivatives. Plugging in the numbers you get $lnS_T$ ~ $N(3.981291519,0.16875)$ Then the probability you ...

2

As the stock price process $S$ follows a geometric Brownian motion, we have that \begin{align*} S_T &= S_0 e^{(\mu-\frac{1}{2}\sigma^2)\, T + \sigma\, W_T}\\ &= S_0 e^{(\mu-\frac{1}{2}\sigma^2)\, T + \sigma\, \sqrt{T}\, \xi}, \end{align*} where $\xi$ is a standard normal random variable. Then, we have the probability \begin{align*} P(S_T > 95) ...

1

In the Black-Scholes framework, we assume the log returns are normally distributed. This is equal to saying the underlying is log-normally distributed. If you look at Geometric Brownian Motion on wikipedia, you'll see this: The above solution S_t (for any value of t) is a **log-normally distributed** random variable The wikipedia is correct.

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