Pricing a European Asset-or-Nothing Call with Black–Scholes
Summary
The document presents a derivation question for a European asset-or-nothing call, whose terminal payoff is the asset price when that price exceeds the strike and zero otherwise. The author transforms the Black–Scholes pricing equation into a heat equation, applies its Gaussian integral solution, and reports obtaining an expression proportional to the asset price times the normal cumulative distribution evaluated at d1. The question is whether the boundary or terminal conditions, or the derivation, explain a discrepancy with a formula that includes discounting.
The text contains the setup and an attempted calculation, but no accepted answer or independent verification. In the standard no-dividend Black–Scholes setting, the asset-or-nothing call value is the spot price times Φ(d1); discounting the cash-or-nothing payoff is a different case. The document’s stated discounted expression should therefore be checked against the payoff and model assumptions. Its derivation also uses scaled variables whose definitions are not fully supplied here.
Key ideas
- An asset-or-nothing call pays the underlying asset when its terminal price exceeds the strike.
- The author transforms the pricing equation into a heat equation and evaluates its Gaussian solution.
- The attempted result is the spot price multiplied by the normal cumulative distribution at d1.
- The stated discounted formula should be checked against the payoff and pricing assumptions.
- The document supplies no answer verifying its boundary conditions or derivation.
Tags
Full text
# Deriving the black-scholes formula for the European asset-or-nothing call option
# Deriving the black-scholes formula for the European asset-or-nothing call option
I would like to find out what boundary/final conditions i should be using to find the formula for a European asset-or-nothing call option, as i feel that is where I'm making my mistake.
I've read that the formula should be: $$ Se^{-r (T-t)}\Phi(d_1) $$ however, I've ended up with, $$ S\Phi(d_1) $$
I have final payoff function:
$$ C_{AoN}(S,T) = \left\{ \begin{array}{lll} 0 & \mbox{S$<$E};\\ S/2 & \mbox{S=E};\\ S & \mbox{S$>$E }\end{array} \right. $$
and I've set final and boundary conditions;
$$ u(x,0) =\hat{C}(\hat{S},1)=max(\hat{S},0)=max(e^x,0), \:\: x \in \mathbb{R}, $$
$$ \left. \ \begin{array}{cc} & u(x,\tau)\rightarrow 0 \:as \: x\rightarrow -\infty \\ & u(x,\tau) \sim e^x \: as \: x \rightarrow \infty \end{array} \right \} 0<\tau<\alpha$$
I've used the transformation: $$u(x,\tau)=e^{\lambda x +\mu \tau}v(x,\tau)$$ with, $$ \left \{ \begin{array}{cc} & \lambda = \frac{1-\nu}{2} \\ & \mu = -\frac{(\nu +1)^2}{4} \end{array} \right. $$
This part is where i think I've gone wrong;
I now try to solve the heat equation: $$ v_{\tau}-v_{xx}=0, \:\: x \in \mathbb{R},\: 0<\tau<\alpha $$ with the initial condition: $$v(x,0)= e^{-\lambda x}u(x,0)=e^{-\lambda x}\max(e^x,0)$$ $$ = \max(e^{-\lambda x}e^{ x},0)= \max(e^{\frac{\nu +1}{2}x},0)=:v_0(x) $$
The solution to out equation is given by
$$v(x,\tau) = \frac{1}{\sqrt{4 \pi\tau}}\int_{-\infty}^{\infty}e^{\frac{-(x-y)^2}{4 \tau}}v_0(y)dy$$
Using $ z=\frac{y-x}{\sqrt{2 \tau}}$, we have $$ v(x,\tau) =\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-\frac{z^2}{2}}v_0(x+\sqrt{2 \tau}z)dz.$$
From here, carrying on with black scholes formula proof, I have ended up with $$ S\Phi(d_1) $$.
Any advice would be appreciated. This is a new subject for me.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.