= Solution
Each $N_i$ is geometric on $\{1,2,\ldots\}$ with success probability $a=Z_p/(MZ_q)$. Hence
$$
\boxed{\mathbb E[N_i]=\frac{MZ_q}{Z_p},\qquad\widehat{Z_q/Z_p}=\frac1{MN}\sum_{i=1}^Nn_i.}
$$
Linearity of <expectation> proves the estimator is unbiased. Moreover the joint probability factors:
$$
P(N_i=k,X_i=x)=(1-a)^{k-1}\frac{\tilde p(x)}{MZ_q}=[(1-a)^{k-1}a]p(x).
$$
Thus $\boxed{N_i\text{ and }X_i\text{ are independent}}$, not merely uncorrelated. Fresh <independent> trials make the simulated cycles <independent> as well.
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