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 $$C(0)$$.

• When you say BS equation, you mean the PDE? Then, your payout at time $T$ can be formulated as a boundary condition, so the equation stays the same. Mar 11, 2020 at 4:13
• You have a question, why don't you write it? Why did you remove it? -1 Jul 6, 2020 at 2:59

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.)