# Closed form expression for $\Bbb E(\mathbb{I}_{\{S_{1,T}>S_{2,T}>K \}})$

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 $$\Phi_2$$.

• 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… Oct 29 '21 at 13:06
• @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.
– NN2
Oct 29 '21 at 17:17