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A 15 mark past paper question essentially ask s me to derive the Black Scholes formula for pricing options.

Let $S_t=S_0e^{(r-\frac{\sigma^2}{2})t+\sigma B_t}$ where $B_t$ is a standard Brownian Motion. $r$ is the risk free rate of interest and $\sigma$ is the stock volatility. Show that the price of a call option with payoff $h=max(S_T-K,0)$, maturity $T$ strike price $K$ satisfying $S_t$ is given its price by

$$e^{-rT}\mathbb{E}[h]= S_0\Phi(x)-Ke^{-rT}\Phi(x-\sigma \sqrt{T})$$

where $x=\frac{\text{ln}(\frac{S_0}{K})+(r+\frac{\sigma^2}{2})T}{\sigma \sqrt{T}}$ and $\Phi$ denotes the distribution function of a standard normal variate.

Where I got to and what I understand is only the following.

I need to find $\mathbb{E}[h]$ so this is, $\mathbb{E}[(S_T-k)I_{S_T>K}]$ where $I$ is the indicator function. This reduces to $\mathbb{E}[S_TI_{S_T>K}]-KP(S_T>K)$.

What I don't understand(The solution),

1.for $P(S_T>K)$, $P(S_T>K)=P(\text{ln}S_T>\text{ln}K)$ so

$$P\bigg(\frac{\text{ln}\frac{S_T}{S_0}-(r-\frac{\sigma^2}{2})T}{\sigma \sqrt{T}} > \frac{\text{ln}\frac{K}{S_0}-(r-\frac{\sigma^2}{2})T}{\sigma \sqrt{T}}\bigg)$$ $$P\bigg(Z > - \frac{\text{ln}\frac{S_0}{K}+(r+\frac{\sigma^2}{2})T}{\sigma \sqrt{T}}+\sigma\sqrt{T}\bigg) = \Phi(x-\sigma \sqrt{T})$$

since $\text{ln}(S_T)$ is a normal variate with mean $\text{ln}(S_0)+(r-\frac{\sigma^2}{2})T)$ and variance $\sigma^2T$. QUESTION: Why ln in particular? It just comes out of nowhere, it just suddenly says $\text{ln}S_T$ is a normal blah blah, and I don't get why it could just pop up like that. Where did it come from? And, how can I check the mean/variance of $\text{ln}S_T$ claimed here? I thought $\mathbb{E}[\text{ln}S_T]=\frac{1}{\sqrt{2 \pi t}}\int_{\infty}^{\infty}ln(x)e^{-\frac{x^2}{2t}}dx$ but I don't know how to calculate this integral. Also, what is $Z$?

  1. One also observes that $\mathbb{E}[S_TI_{S_T>K}]=\mathbb{E}[(S_0e^{(r-\frac{\sigma^2}{2})T+\sigma B_t})I_{S_T>K}] = \mathbb{E}[(S_0e^{(r-\frac{\sigma^2}{2})T+\sigma \sqrt{T}Z})I_{Z>-x+\sigma\sqrt{T}}]$. QUESTION; Simply, what is going on with this $Z$ again? Is this an interchange of variables?I don't immediately see what $Z$ is, and thus why there is a $Z$ in the LHS of the inequality above. Is this using the same $Z$ as in $1$ above?

Sorry it is tedious but I'm struggling to understand what's going on, please kindly help! Your time is very much appreciated,

Thank you in advance

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  • $\begingroup$ The logarithm of $S$ follows a Wiener process - this implies that $\log S$ is normally distributed. $\endgroup$ – Mark Viola Dec 6 '15 at 19:30
  • $\begingroup$ Hi again, thanks for answering! Right, though, specifically, how do I proceed with the integral...which leads to finding the mean...? $\endgroup$ – John Trail Dec 6 '15 at 19:37

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