Exercise: MC delta — bump-and-revalue vs pathwise

The delta of a European call is $\Delta = \partial C/\partial S_0$ . Two MC techniques to estimate it:

Bump-and-revalue (finite difference):

\Delta \approx \frac{C(S_0 + h) - C(S_0 - h)}{2h}.

Pathwise (path-by-path differentiation):

\Delta = e^{-rT}\mathbb{E}^{\mathbb{Q}}\!\left[\mathbb{1}_{\{S_T > K\}} \frac{S_T}{S_0}\right].

(Derivation: differentiate $(S_T - K)^+$ in $S_0$ , noting $\partial S_T/\partial S_0 = S_T/S_0$ in BS.)

Same parameters as before: $S_0 = K = 100$ , $r = 0.05$ , $\sigma = 0.2$ , $T = 1$ . True delta: $\Phi(d_1) = 0.6368$ .

Tasks

Implement both estimators with $N = 10^5$ paths. Use a bump size $h = 1$ for finite differences. Use common random numbers in the bump-and-revalue: same $Z$ seed for $S_0 + h$ and $S_0 - h$ .
Compare the variances of the two estimators across 100 independent runs.
Effect of bump size. Vary $h \in \{0.01, 0.1, 1, 10\}$ for the bump method with common random numbers and without (independent draws for + and -). Tabulate variance and bias.
Why pathwise wins. Argue why pathwise has lower variance: it's exact (no $h$ -bias) and uses the same sample to estimate $\Delta$ directly without subtraction.
When pathwise fails. The pathwise method requires the payoff to be almost-surely differentiable in $S_0$ . Why doesn't it work for a digital option $\mathbb{1}_{\{S_T > K\}}$ ?