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?

  • $\begingroup$ The reasoning seems right. I did not have a deep look into the computations though. For instance, are you sure about your expression for $S_t^{(2)}$ ? Looks awkward to me although you indeed have a martingale. Same for your application of Girsanov (irrespective of the rest). $\endgroup$
    – Quantuple
    Jul 17, 2017 at 7:42
  • $\begingroup$ @Quantuple thanks for your feedback. I found few errors. I think now it should look better. $\endgroup$
    – lemontree
    Jul 17, 2017 at 15:14
  • 1
    $\begingroup$ Still some problems IMO. In your expression of $S_t^{(2)}$ it should be squared volatilities in the drift term. Also you should best not introduce $Z_t$ as it puts you on a wrong path when applying Girsanov (your application of Girsanov is wrong, because $W_t^1$ and $Z_t$ are not independent.) $\endgroup$
    – Quantuple
    Jul 17, 2017 at 16:17
  • $\begingroup$ oh yes I see that. But then I think my whole approach is for nothing. $\endgroup$
    – lemontree
    Jul 17, 2017 at 17:13
  • 1
    $\begingroup$ Basically, you need to two-dimensional Girsanov transformation. $\endgroup$
    – Gordon
    Jul 17, 2017 at 18:02


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.