Ignoring the bounded second-order remainder, this is the <linear regression> model
$$
Y_i-g(x_0)=hZ_i\beta+\varepsilon_i,
\qquad \beta=g'(x_0).
$$
The <ordinary least squares> estimate through the known zero intercept is
$$
\widehat\beta
=\frac{\sum_i(hZ_i)(Y_i-g(x_0))}{\sum_i(hZ_i)^2}
=\boxed{\frac1N\sum_{i=1}^N\frac{Z_i(Y_i-g(x_0))}{h}},
$$
because $Z_i^2=1$. This is the <randomized symmetric finite-difference derivative estimator>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 28K / c / Solution

Given noisy evaluations at $x_0+hZ_i$, where the $Z_i$ are independent <Rademacher random variables>, the estimator
$$
\widehat g'_N(x_0)=\frac1N\sum_{i=1}^N\frac{Z_i(Y_i-g(x_0))}{h}
$$
averages one-sided finite differences from both directions. If $|g''|\leq M$ and the noise variance is $\sigma^2$, its <mean squared error> is at most $h^2M^2/4+\sigma^2/(Nh^2)$.

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