Pricing an Asset-or-Nothing Option on an Independent Asset

# Pricing for an Odd Type of Asset or Nothing Option


# Pricing for an Odd Type of Asset or Nothing Option












Trying to derive the pricing function for a derivative on two assets $S^1$ and $S^2$ with the following payoff function:

$$\Phi(S^1_T,S^2_T)=S_T^1 \, \unicode{x1D7D9}\{S_T^2\le K\}$$

where I'm simply using $\unicode{x1D7D9}$ as the indicator function. Also, importantly, the two assets are driven by independent Wiener processes. So, effectively, we have a binary put option on $S_T^2$, where the payoff is $S_T^1$. So, I know a few things from the start. If the payoff was some fixed amount $K$, the pricing function would be $Ke^{-rT}N(-d_2)$. On the other hand, if the payoff was the asset $S_T^2$ itself, the pricing function would be $S_0^2N(-d_1)$.

So, given that we have independent Weiner processes, I feel like the pricing should be more similar to a fixed payoff binary. Furthermore, the payoff should be the risk-neutral expectation of the payoff asset. That is,

\begin{align} &\pi(t)=S_t^1e^{r(T-t)}e^{-r(T-t)}N(-d_2)\\ \iff& \pi(t)=S_t^1N(-d_2) \end{align}

with $d_2$ defined as it would be in the standard B-S model. I'm hoping someone could confirm/deny this and perhaps provide a more rigorous derivation. Thanks.

## Answer by spaceisdarkgreen (score 4, accepted)

https://quant.stackexchange.com/a/36211

The price is, under the risk-neutral measure, $$ P_t = e^{-r(T-t)}\mathbb E[S_T^1 \mathbb 1(S_T^2\le K)\mid \mathcal F_t].$$ Since the risk-neutral asset processes are independent geometric brownian motions, $S_T^1$ and $S_T^2$ are conditionally independent given $\mathcal F_t.$

So the conditional expectation factors and you get $$ P_t = e^{-r(T-t)}\mathbb E[S_T^1\mid \mathcal F_t]\mathbb E[\mathbb 1(S_T^2\le K)\mid \mathcal F_t] = S^1_t N(-d_2)$$ (where "$d_2$" is of course relative to the $S^2_t$ process) just like you say.