# Briefly stated, why does the function N(x) appear in the European call option pricing model?

I'm aware of the the mathematical formula for the price of a European call option on a stock however I'd like to think about it in an intuitive way.

It is because Black and Scholes assume that the stock follows a geometric brownian motion, i.e. under the historical probability $\mathbb{P}$ the stock moves according to: $$\frac{dS(t)}{S(t)} = \mu dt + \sigma dW^{\mathbb{P}}(t)$$ Solving this SDE we obtain that $$S(t)=S(0)e^{(\mu - \frac{1}{2}\sigma^2)t + \sigma W^{\mathbb{P}}(t)} = S(0)e^{(\mu - \frac{1}{2}\sigma^2)t + \sigma \sqrt{t} z}$$ where $z \sim N(0,1)$. Thanks to Girsanov's theorem we can show that under the risk neutral measure, the stock still follows a Brownian Motion, in particular it can be shown that under $\mathbb{Q}$ the stock price follows $$S(t)=S(0)e^{(r-\frac{1}{2}\sigma^2)t + \sigma W^{\mathbb{Q}}(t)} = S(0)e^{(r - \frac{1}{2}\sigma^2)t + \sigma \sqrt{t} z}$$ It can be shown that the price of an option should be equal to the expected value under the risk neutral measure ($\mathbb{Q}$) of the discounted payoff at expiration, i.e. $$c(S,t) = \mathbb{E}_t^{\mathbb{Q}}[(S(T) - K)^+ e^{-r(T-t)}]$$ Now, take the expected value: $$c(s,t)= \mathbb{E}_t^{\mathbb{Q}}[(S(T) - K)^+ e^{-r(T-t)}] = \int_{S(T)>K}(S(T)-K)e^{-r(T-t)}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz =$$ $$= \int_{S(T)>K}(S(t)e^{(r - \frac{1}{2}\sigma^2)(T-t) + \sigma \sqrt{T-t} z}-K)e^{-r(T-t)}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz =$$ $$= S(t)N(d_1) - Ke^{-r(T-t)}N(d_2)$$ where $$d_1=\left[ln\frac{S(t)}{K} + (r-\frac{\sigma^2}{2}(T-t))\right]\frac{1}{\sigma\sqrt{T-t}}$$ $$d_2= d_1 - \sigma\sqrt{T-t}$$
Notice that $d_1$ just indicates the cutoff for z s.t. S(T) > K, i.e. where the option closes in the money. Summarizing, the normal cdf simply derives from the assumption of a stock diffusing like a geometric brownian motion both under the historical and the risk-neutral probability measure.
• Yes, just use Ito’s lemma on $d\log S(t)$. You have something like $$d\log S(t) = S(t)\left[\mu dt + \sigma dW(t) -\frac{1}{2}\sigma^2 dt\right]$$ which can be solved diving by $S(t)$ and integrating, i.e. $$\int_0^t \frac{d\log S(t)}{S(t)} = \int_0^t (\mu-0.5\sigma^2)dt + \int_0^t \sigma^2 dW(t)$$ $$\log S(t) = (\mu - 0.5\sigma^2)t + \sigma W(t)$$ then exponentiate to obtain $$S(t) = e^{(\mu - 0.5\sigma^2)t + \sigma W(t)}$$ – fni Oct 20 '14 at 6:26