Is it possible to calculate analytically $\Bbb E(\mathbb{I}_{\{S_{1,T}>S_{2,T}>K \}})$, using the 2-dimensional normal probability function $\Phi_2$, where $S_{1,T}$ and $S_{2,T}$ follow geometric Brownian motion with corrrelation $\rho$.

In fact, calculate $\Bbb E(\mathbb{I}_{\{S_{1,T}>S_{2,T}>K \}})$ is equal to calculate $$\Bbb P(W_{1,T} + a>W_{2,T} +b > K)$$ where $a,b, K$ are scalar, $K>0$ and $W_{1,T},W_{2,T}$ are two Brownian motions of $S_{1,T}$ and $S_{2,T}$.

I believe it is possible but have not found yet the formula!

Finally, I found the closed form expression for this. It suffices to re-write the integration region as $\{W_{1,T}-W_{2,T}>b-a, W_{2,T}>K-b \}$. As the bivariate $(W_{1,T}-W_{2,T},W_{2,T} )$ follows a bivariate normal distribution, we can obtain easily the probability via the bivariate normal probability function $\Phi_2$.

  • 2
    $\begingroup$ Yes; you’ll have to simply integrate the bivariate normal density accordingly. Are you able to do that, numerically? You can use R, MATLAB, Python… $\endgroup$ Oct 29 '21 at 13:06
  • $\begingroup$ @Kermittfrog Thank you for your comment. Finally I found the closed form expression for the probability above. I modified the answer by adding the solution sketch. $\endgroup$
    – NN2
    Oct 29 '21 at 17:17

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