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Why the Normal CDF Appears in Black–Scholes Call Pricing

Article Quant Q&A · Author: Paul P M

Summary

The document explains the appearance of the standard normal cumulative distribution function in the Black–Scholes price of a European call. The model assumes the stock follows geometric Brownian motion, so its terminal log price is normally distributed. Under risk-neutral pricing, the expected discounted payoff is calculated over outcomes where the stock finishes above the strike.

Evaluating that expectation yields two normal-CDF terms: one weights the stock-price contribution and the other weights the discounted strike payment. The response interprets the relevant cutoff as the boundary between in-the-money and out-of-the-money terminal outcomes. This explanation relies on the Black–Scholes assumptions, including constant volatility and the geometric Brownian motion model; the document does not address departures from those assumptions or alternative pricing models.

Key ideas

  • Geometric Brownian motion makes the stock’s terminal log price normally distributed in the model.
  • Risk-neutral option value is the expected discounted payoff at expiration.
  • The normal CDF arises when integrating the payoff over terminal outcomes above the strike.
  • The two CDF terms weight the stock-price and strike-price components of the call value.

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


# 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.

## Answer by fni (score 4, accepted)

https://quant.stackexchange.com/a/15084

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.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.