Deriving Black–Scholes Call Value from the Expected Payoff
Summary
The document asks how to derive a European call price from its terminal payoff, beginning with the positive part of the difference between terminal asset value and strike. The questioner attempts to reduce the calculation to the probability that the terminal asset price exceeds the strike, but incorrectly treats the payoff as a constant multiplied by that probability and tries to integrate the stochastic differential equation directly.
The answer uses the geometric Brownian motion solution to express the terminal price as a lognormal random variable. Taking logarithms turns the exercise event into a threshold condition on a standard normal variable, whose tail probability can be written using the normal cumulative distribution function. This derives the exercise probability component, but the post does not complete the expected payoff calculation or explain risk-neutral valuation and discounting in full. The exercise probability alone is not the call price; the payoff magnitude conditional on exercise must also be accounted for.
Key ideas
- A call payoff depends on both whether the terminal price exceeds the strike and by how much.
- Under geometric Brownian motion, the terminal asset price has a lognormal distribution.
- Taking logarithms converts the exercise condition into a threshold for a standard normal variable.
- The normal cumulative distribution function gives the probability that the call finishes in the money.
- Exercise probability alone does not determine the option value; expected payoff and proper discounting are required.
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# Derivation of Black Scholes using expected payoff
# Derivation of Black Scholes using expected payoff
The payoff function of a call is $f(S_T, K) = (S_T - K)^+$, so the expected payoff should allow me to value the price of this call.
$$ \mathbb{E}[f(S_T, K)] = \mathbb{E}[(S_T - K)^+] = \mathbb{E}[(S_T - K) \cdot \mathbb{1}(S_T - K > 0)] $$ $$ = e^{-rT} (S_T - K) \mathbb{E}[\mathbb{1}(S_T - K > 0)] $$ $$ = e^{-rT} (S_T - K) \mathbb{P}[(S_T - K > 0)] $$
Now the question is simplified to calculating the probability that $S_T$ would be greater than $K$. $$ \mathbb{P}[(S_T - K > 0)] = \mathbb{P}[(S_T > K)] $$ $$ = \int_K^{\infty} dx $$ where $dx = dS_T = \mu S_0 dt + \sigma S_0 dW$ $$ = \int_K^{\infty} \mu S_0 dt + \sigma S_0 dW $$
I do not think this is the correct way to go, and I would appreciate any input on this matter. Thanks.
## Answer by LocalVolatility (score 0, accepted)
https://quant.stackexchange.com/a/30710
You know the solution to the SDE
\begin{equation} \mathrm{d}S_t = \mu S_t \mathrm{d}t + \sigma S_t \mathrm{d}W_t \end{equation}
is
\begin{equation} S_T = S_0 \exp \left\{ \left( \mu - \frac{1}{2} \sigma^2 \right) T + \sigma W_T \right\} \end{equation}
Now, $W_T \sim \mathcal{N}(0, T)$, so
\begin{equation} S_T > K \qquad \Leftrightarrow \qquad S_0 \exp \left\{ \left( \mu - \frac{1}{2} \sigma^2 \right) T + \sigma \sqrt{T} X \right\} > K, \end{equation}
where $X \sim \mathcal{N}(0, 1)$. Rearranging yields
\begin{equation} X > \frac{1}{\sigma \sqrt{T}} \left( \ln \left( \frac{K}{S_0} \right) - \left( \mu - \frac{1}{2} \sigma^2 \right) T \right) := \alpha \end{equation}
Now, $\mathbb{P} \left\{ X > \alpha \right\} = \mathbb{P} \left\{ X < -\alpha \right\}$ and thus $\mathbb{P} \left\{ S_T > K \right\} = \mathbb{P} \left\{ X < -\alpha \right\} = \mathcal{N}(-\alpha)$.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.