A Distribution-Based Route to the Black–Scholes Call Formula
Summary
The document presents a learner's attempt to derive a call option price by integrating the positive payoff over a lognormal terminal-price distribution, then substituting the risk-neutral expected price to determine the distribution's location parameter. The attempt separates the stock-price and strike components of the payoff, but arrives at an unfamiliar expression and asks where the derivation goes wrong. It also notes an informal substitution involving volatility and elapsed time.
The answer gives directional guidance rather than a complete correction: begin with the payoff and the risk-neutral terminal stock price expressed through a Brownian increment; standardize that increment into a normal random variable; and change variables so parts of the expectation become normal cumulative distribution functions. This approach helps organize the payoff threshold and derive the familiar d-plus and d-minus terms. A stochastic calculus text is cited for a full derivation. The exchange does not work through the learner's algebra, so it leaves specific density, scaling, and parameterization errors unresolved.
Key ideas
- A call price can be approached as the risk-neutral expectation of its terminal payoff.
- The terminal stock price is modeled using a Brownian-motion increment under the risk-neutral measure.
- Standardizing the increment produces a normal random variable that helps express the payoff threshold.
- A change of variables turns parts of the expectation into cumulative distribution functions.
- The answer offers pointers rather than diagnosing every algebraic issue in the attempted derivation.
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Full text
# Trouble arriving at Black-Scholes Formula
# Trouble arriving at Black-Scholes Formula
I am attempting to arrive at the Black-Scholes formula for my own understanding. I can accept one can use the risk-free distribution & rate, so I am attempting to use the distrution to arrive at the result rather than PDEs. Here is what I have: $$\lim\limits_{b \to \infty}\int_a^b pdf(s)(s-K)ds$$
Since the initial price is $S_0$ I popped that in with $s$.
$$\lim\limits_{b \to \infty}\int_a^b pdf(s*S_0)(s*S_0-K)ds$$
Where $a = K / S_0$.
And since B-S formula choose the log normal I will use that for the pdf:
$$\lim\limits_{b \to \infty}\int_a^b \dfrac{1}{\sigma\sqrt{2\pi}}* e^{\dfrac{-(ln(s*S_0)-\mu)^2}{2\sigma^2}} - \int_a^b\dfrac{K}{S_0s\sigma\sqrt{2\pi}} * e^{\dfrac{-(ln(s*S_0)-\mu)^2}{2\sigma^2}}$$
I think I'm ok up to here but I could be off with where I'm putting my $S_0$'s... anyways the first integral looks scary but I found some help here: https://stats.stackexchange.com/questions/9501/is-it-possible-to-analytically-integrate-x-multiplied-by-the-lognormal-probabi so I'm pretty sure we're going to end up with something like this (for the first integral):
$$e^{\mu+\frac{1}{2}\sigma}(\Phi(\beta)-\Phi(\alpha))$$
where $\beta = (ln(b) - \mu-\sigma^2)/\sigma$ and $\alpha = (ln(K/S_0) - \mu-\sigma^2)/\sigma$
I should probably mention here that I have been lazily mentally substituting $\sigma\sqrt{t}$ anytime I see $\sigma$.
Since $b$ is going to $\infty$,$\Phi(\beta)$ is going to $1$. At this point we should also try getting rid of $\mu$. Since $S_0e^{rt}$ is the expected forward price we can set that equal to $e^{\mu+\sigma^2/2}$ and get:
$$\mu = ln(S_0) + rt - \sigma^2/2$$
As for the second integral, we are just going to get a difference of two log-normal CDFs (I think) times $K$. The first will again be $1$ since $b$ is going to $\infty$, so we get $K(1-LNCDF(a))$.
My result is thus $S_0e^{rt}(1-\Phi(a)) - K(1-LNCDF(a))$ This doesn't look too familiar so I am wondering what I am doing wrong. Thanks!
## Answer by Matt Wolf (score 6, accepted)
https://quant.stackexchange.com/a/9362
Here couple pointers to push you back on the right path (so I hope):
- Start with the payoff function and hence $S(T)$, which consists of $(W(T)-W(t))$ , $W$ being a Brownian Motion under the risk neutral measure)
- you can greatly simplify by working with a standard normal random variable:
$$Y = \frac{-(W(T)-W(t))}{\sqrt{T-t}}$$, which helps to get rid of the indicator function and to derive d+ and d-,
- you need to perform a change of variable to express part of your result as a function of the cdf.
You can find the full derivation in Steven Shreve, Stochastic Calculus for Finance II, 2004 edition, pp.218Shown 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.