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.

  • $\begingroup$ The value of an asset-or-nothing option,$S\Phi(d_1)$ is correct if the dividend yield is zero. $\endgroup$ – Dhamnekar Winod May 11 at 14:40
  • $\begingroup$ Thank you. So the only difference between the derivation would be the final condition? $\endgroup$ – amir ahun May 11 at 14:48
  • $\begingroup$ An asset-or-nothing option pays one share of stock if $S(T)>K$ and nothing otherwise. $K$ is the strike price and $S(T)$ is the stock price at option expiry. $\endgroup$ – Dhamnekar Winod May 11 at 15:17
  • $\begingroup$ It's worth learning the "change of numeraire" concept/technique at some point. You'll see it greatly simplifies certain types of pricing problems, including the asset or or nothing. $\endgroup$ – ilovevolatility May 11 at 19:14

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