Consider two stocks, $S_1$ and $S_2$, with marginal pdfs $f_{S_1}$ and $f_{S_2}$. Assume $F(S_1,S_2)$ is the joint CDF. I'm trying work out a semi-analytic formula for the price of the barrier call option $C_T = (S_1-K)1_{S_1>K}1_{L<S_2<U}$. I am able to easily solve the put barrier option formula $P_T = (K-S_1)1_{S_1<K}1_{L<S_2<U}$, but I get an undefined value for the call option.

Here is the proof for the put option: $$P_0 = \int_0^{K}\int_L^{U}{(K - S_1)\dfrac{d^2F(S_1, S_2)}{dS_1dS_2}}dS_2dS_1 \\ = \int_0^{K}{(K - S_1)(\dfrac{dF}{dS_1}(U) - \dfrac{dF}{dS_1}(L))}dS_1 \\ = \int_0^{K}{(K - S_1)\dfrac{dF}{dS_1}(U)dS_1 - \int_0^{K}(K - S_1)\dfrac{dF}{dS_1}(L)}dS_1 \\ = \int_0^{K}{(K - S_1)\dfrac{dF}{dS_1}(U)dS_1 - \int_0^{K}(K - S_1)\dfrac{dF}{dS_1}(L)}dS_1 \\ = -\int_0^{K}F(S_1,U)+\int_0^{K}F(S_1,L) \\ = \int_0^{K}(F(S_1,L)-F(S_1,U))dS_1 $$ where the second last line follows by doing integration by parts. All good so far.

As for the call option, I do the same sort of logic, but the IGP gets me to an undefined value.

$$C_0 = \int_K^{\infty}\int_L^{U}{(S_1-K)\dfrac{d^2F}{dS_1dS_2}}dS_2dS_1 \\ = \int_K^{\infty}{(S_1-K)(\dfrac{dF}{dS_1}(U) - \dfrac{dF}{dS_1}(L))}dS_1 \\ = \int_K^{\infty}{(S_1-K)\dfrac{dF}{dS_1}(U)dS_1 - \int_K^{\infty}(S_1-K)\dfrac{dF}{dS_1}(L)}dS_1 \\ = \int_K^{\infty}{(S_1-K)\dfrac{dF}{dS_1}(U)dS_1 - \int_K^{\infty}(S_1-K)\dfrac{dF}{dS_1}(L)}dS_1 $$ If I do IGP now, I get an undefined value $(\infty - \infty)$.

Using "reverse engineering", I found that the actual answer is below, but I would like to understand why. $$C_0 = \int_K^{\infty}(F(1-S_1,1-U)-F(1-S_1,1-L))dS_1 $$

Do you know why the answer is this? My guess is that the result follows obviously from some sort of reflection imaging principle, but I don't see it.


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