Pricing a Cross-Asset Digital Option with a Stock as Numeraire
Summary
The document considers a payoff equal to one stock’s terminal price when a second stock finishes above a strike. The stocks follow correlated geometric Brownian dynamics, with their correlation represented by a shared Brownian motion, and the risk-free rate is set to zero. The proposed derivation uses the paying stock as numeraire: changing measure turns the option value into that stock’s current price multiplied by the probability that the other stock exceeds the strike under the new measure.
A Girsanov shift changes the drift of the threshold stock under this measure, allowing the event probability to be written using a normal distribution. The calculation illustrates how cross-asset dependence affects a digital-style payoff. The setup is idealized, with constant volatility, specified diffusion dynamics, and no dividends or other market frictions; the document presents a proposed derivation rather than a general pricing treatment.
Key ideas
- The payoff combines one stock’s value with a threshold event on another stock.
- Using the paying stock as numeraire converts valuation into a probability under a changed measure.
- The shared Brownian component shifts the threshold stock’s drift under the new measure.
- The resulting probability is normal under the assumed constant-volatility diffusion model.
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Full text
# pricing option with two stocks
# pricing option with two stocks
> Let $\left(S_t^{(1)}\right)_{t\ge0}$ and $\left(S_t^{(2)}\right)_{t\ge0}$ be the price processes of two stocks with dynamics $$ \begin{align} & dS_t^{(1)}=\sigma_{11}S_t^{(1)}dW_t^{(1)} \\[6pt] & dS_t^{(2)}=\sigma_{21}S_t^{(2)}dW_t^{(1)}+\sigma_{22}S_t^{(2)}dW_t^{(2)} \end{align} $$ where $W_t^{(1)}$ and $W_t^{(2)}$ are independent Brownian motions. Further let the riskfree interest rate $r=0$. I want to price a option with a payoff $X(T)= S_T^{(2)}1_{\{S_T^{(1)}>K\}}$ at maturity T.
First we observe:
$$ \begin{align} & S_t^{(1)}=S_0^{(1)}\exp(-\dfrac{\sigma_{11}^2}{2}t+\sigma_{11}W_t^{(1)}) \\[6pt] & S_t^{(2)}=S_0^{(2)}\exp(-\dfrac{1}{2}(\sigma_{21}^2+\sigma_{22}^2)t+\sigma_{21}W_t^{(1)}+\sigma_{22}W_t^{(2)}). \end{align} $$
So using the risk-neutral valuation formula the price of this option at time $0$ is:
$$ V_0=e^{-rT}\mathbb E_{\mathbb Q}[X(T)]=\mathbb E_{\mathbb Q}[S_T^{(2)}1_{\{S_T^{(1)}>K\}}].$$
Now using $S_t^{(2)}$ as numéraire gives the new measure:
$$ \dfrac{d\mathbb{Q^N}}{d\mathbb Q}=\dfrac{S_T^{(2)}}{S_0^{(2)}}=\exp\left(-\pmatrix{-\sigma_{21} & -\sigma_{22}}\pmatrix{W_T^{(1)} \\W_T^{(2)}}-\dfrac{1}{2}(\sigma_{21}^2+\sigma_{22}^2)T\right) $$
Now by Girsanov's theorem $\hat W_t^{(1)}:=W_t^{(1)}-t\sigma_{21}$ is a $\mathbb Q^N$- Brownian motion. So we get:
$$ \begin{align} V_0 = e^{-rT}\mathbb E_{\mathbb Q}[X(T)] & = \mathbb E_{\mathbb Q}[S_T^{(2)}1_{\{S_T^{(1)}>K\}}] \\[6pt] & = \mathbb E_{\mathbb Q^N}\left[S_T^{(2)}1_{\{S_T^{(1)}>K\}}\left(\dfrac{d\mathbb{Q^N}}{d\mathbb Q}\right)^{-1}\right] \\[6pt] & = S_0^{(2)}\mathbb Q^N\left(S_T^{(1)}>K\right) \\[6pt] & = S_0^{(2)}\mathbb Q^N\left(S_0^{(1)}\exp(-\dfrac{\sigma_{11}^2}{2}T+\sigma_{11}W_T^{(1)})>K\right) \\[6pt] & = S_0^{(2)}\mathbb Q^N\left(S_0^{(1)}\exp\left(-\dfrac{\sigma_{11}^2}{2}T+\sigma_{11}(\hat W_T^{(1)}+\sigma_{21}T)\right)>K\right) \\[6pt] & = S_0^{(2)}\Phi\left(\dfrac{\log(S_0^{(1)}/K)-\left(\dfrac{\sigma_{11}^2}{2}-\sigma_{11}\sigma_{21}\right)T}{\sigma_{11}\sqrt T}\right). \end{align} $$
Am I right? Is there another, maybe simpler way to do it?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.