Black-Scholes Pricing for Fixed Payout Derivative I have a question about how to derive a Black-Scholes-esque pricing model, given a fixed payout rather than one that varies with stock price.
More specifically, if given a derivative which pays out a fixed \$x if the share price $S(T)$ is greater than or equal to an agreed amount $K$ at expiration time $t = T$. If $S(T) < K$, the payout is \$0.
I recognize and understand the Black-Scholes equation for a traditional European call option, but I'm wondering if there's some simple way to plug this into the equation, or work-backwards as to derive an equation for $C(0)$.
 A: This is called a digital option.
As Raskolnikov indicates in the comments, the Black Scholes PDE is intended for much more than just vanilla options. It works the same way for basically any derivative of the underlying price: you can put any terminal payoff function as your final condition and then solve for the initial price.
However, of course changing this terminal payoff function changes the price you get.
If you're comfortable with probability, it's easier to think about it in terms of a pricing distribution (often called a risk-neutral distribution). The Black Scholes assumptions imply that the price is given by the expected discounted value of the claim, where the distribution for the underlying is geometric brownian motion with a growth rate equal to the risk-free rate. 
Under these assumptions, the distribution of the underlying value at time to expiration $T$ has lognormal density where $$\log(S_T/S_0)\sim N(r'T, \sigma^2T)$$ where $r'=r-\frac{1}{2}\sigma^2.$ So the expected value of the terminal payment is $$ E(x1_{S> K})= xP(S>K) \\= xP(\log S/S_0>\log K/S_0)\\=x P\left(\frac{\log S/S_0-r'T}{\sqrt{\sigma^2 T}} > \frac{\log K/S_0-r'T}{\sqrt{\sigma^2 T}}\right)\\=xN\left(\frac{\log S_0/K+r'T}{\sqrt{\sigma^2 T}}\right)\\= x N(d_2)$$ where we use the conventional abbreviation from the formula for vanilla options in the last line. So since the discount is constant, we can just multiply by the discount $e^{-rT}$ to get the value $C(0)=x e^{-rT}N(d_2).$
(It's helpful to remember in any case that $N(d_2)$ is the (risk-neutral) probability that the underlying finishes above the strike.)
