# Monte carlo delta calculation for Worst/Best Of Option

I try to calculate the Delta for WO by finite difference.

For example, $$K = 1.$$

$$S_t = S_0 e^{(r - d_1 - \frac{\sigma_1^2}{2})t + \sigma_1 W_t^1}$$ $$F_t = F_0 e^{(r - d_2 - \frac{\sigma_2^2}{2})t + \sigma_2 W_t^2}$$

$$Payoff = (\min\{ \frac{S_t}{S_0}, \frac{F_t}{F_0}\} - K )_{+}$$

For the partial delta calculation I shift the spot and rerun monte carlo, such that, my the bumped forwards is following:

$$S_t^{up} = (S_0 + S_0 * 0.01) e^{(r - d_1 - \frac{\sigma_1^2}{2})t + \sigma_1 W_t^1}$$
$$F_t = F_0 e^{(r - d_2 - \frac{\sigma_2^2}{2})t + \sigma_2 W_t^2}$$

$$Payoff^{up} = (\min\{ \frac{S_t^{up}}{S_0}, \frac{F_t}{F_0}\} - K )_{+}$$ Then I calculate the simple difference: $$\varDelta_{proxy} = Payoff^{up} - Payoff$$

As result I get the partial sensitivity, but arises the problem with explanation, when I shift initial spots by the shift size, due to I use the ratio in payoff, my shifted forward is divided into shifted spot and the price of option unchanged.

About the monte carlo engine, please don't care. I have a semantic error related to the payoff.

Can someone explain to me where I`m wrong with my unchanged price?

$$S_0$$ should remain unchanged as it is defined in the contract terms. In order to compute delta by finite difference, you should shift the price from $$S_t$$ to $$S_t+shift$$, generate the price at maturity then compute the payoff using generated price WITHOUT FORGETTING TO KEEP $$S_0$$ IN THE DENOMINATOR UNCHANGED.
• Thank you man! The question regarding the explain your delta. So we get the sensitivity WITHOUT FORGETTING TO KEEP $S_0$ IN THE DENOMINATOR UNCHANGED. How to shift the initial spot to check the calculated delta? Feb 9, 2020 at 20:19