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):


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!

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    $\begingroup$ Please do not use the HTML tags in your body. You can use this site's features to format your posts optimally. $\endgroup$ – SRKX Nov 5 '13 at 11:01

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.218


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