Monte carlo delta calculation for Worst/Best Of Option

Question

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?

Answer 1

$S_0$ 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.

You can find a pricer for BO and WO basket options in my website ValoMetrics.com for test purpose.