Covariance of two expectation estimators that used different numbers of samples Say I have two estimates of the mean of two functions:
$$Q^1_{N_1}=\frac{1}{N_1}\sum_{i=1}^{N_1}f^1(X_i), \quad Q^2_{N_2}=\frac{1}{N_2}\sum_{i=1}^{N_2}f^2(X_i),$$
where each sample $X_i$ is identical when $i \leqslant \min(N_1, N_2)$ and is selected from the probability distribution $X$, and I want to calculate the covariance of the two estimators $\mathrm{Cov}(Q^1_{N_1}, Q^2_{N_2})$. 
Lemna 3.2 of this paper attempts to prove that
$$\mathrm{Cov}(Q^1_{N_1}, Q^2_{N_2})=\frac{\mathrm{Cov}(f^1(X), f^2(X))}{\max(N_1,N_2)}=\frac{\sqrt{\mathrm{Var(f^1(X)) \mathrm{Var}(f^2(X))}}}{\max(N_1,N_2)}\rho_{12},$$
where $\rho_{12}:=\dfrac{\mathrm{Cov}(f^1(X), f^2(X))}{\sqrt{\mathrm{Var}(f^1(X)) \mathrm{Var}(f^2(X))}}$.
I am pretty sure that the important piece there is
$$\mathrm{Cov}(Q^1_{N_1}, Q^2_{N_2}) = \frac{\mathrm{Cov}(f^1(X), f^2(X))}{\max(N_1,N_2)}.$$
This has been a hard proof for me to follow, so I am asking this community for help. I would say my biggest source of confusion is where the $\max(N_1, N_2)$ term comes from.
Here is what I have tried so far. My result is simillar to that in the paper, except I have a minimum instead of a maximum. Can anyone help give me insight as to where the maximum operator comes into play?

The covariance of $Q^1_{N_1}$ and $Q^2_{N_2}$ may be calulated as
\begin{align*}
&\mathrel{\phantom{=}}{} \mathrm{Cov}(Q^1_{N_1}, Q^2_{N_2})\\
&=\frac{1}{\min(N_1,N_2)}\sum_{i=1}^{\min(N_1, N_2)}\left(\left(f^1(x_i) - \frac{1}{N_1}\sum_{j=1}^{N_1}f^1(x_j)\right)\left(f^2(x_i) - \frac{1}{N_2}\sum_{j=1}^{N_2}f^2(x_j)\right)\right),
\end{align*}
where the minimum of $N_1$ and $N_2$ are the number of samples common to both estimators. Only $\min(N_1,N_2)$ samples are common to both estimators, so we must use at most $\min(N_1,N_2)$ samples to estimate the covariance. Assuming the number of samples is large, we can say
$$\mathrm{Cov}(Q^1_{N_1}, Q^2_{N_2})= \frac{1}{\min(N_1,N_2)}\mathrm{Cov}(f^1(X), f^2(X))=\frac{\sqrt{\mathrm{Var(f^1(X)) \mathrm{Var}(f^2(X))}}}{\min(N_1,N_2)}\rho_{12}.$$
 A: $\DeclareMathOperator{\cov}{cov}\def\peq{\mathrel{\phantom{=}}{}}$\begin{align*}
&\peq \cov(Q_1, Q_2) = E(Q_1 Q_2) - E(Q_1) E(Q_2)\\
&= \frac{1}{N_1 N_2} E\left( \left( \sum_{k = 1}^{N_1} f_1(X_k)\right) \left( \sum_{k = 1}^{N_2} f_2(X_k) \right) \right) - E(f_1(X)) E(f_2(X))\\
&= \frac{1}{N_1 N_2} \sum_{\substack{1 \leqslant k_1 \leqslant N_1\\1 \leqslant k_2 \leqslant N_2}} E(f_1(X_{k_1}) f_2(X_{k_2})) - E(f_1(X)) E(f_2(X))\\
&= \frac{1}{N_1 N_2} \Biggl( \sum_{\substack{1 \leqslant k_1 \leqslant N_1\\1 \leqslant k_2 \leqslant N_2\\k_1 = k_2}} E(f_1(X_{k_1}) f_2(X_{k_2})) + \sum_{\substack{1 \leqslant k_1 \leqslant N_1\\1 \leqslant k_2 \leqslant N_2\\k_1 ≠ k_2}} E(f_1(X_{k_1}) f_2(X_{k_2})) \Biggr)\\
&\peq - E(f_1(X)) E(f_2(X))\\
&= \frac{1}{N_1 N_2} \Biggl( E(f_1(X) f_2(X)) \sum_{\substack{1 \leqslant k_1 \leqslant N_1\\1 \leqslant k_2 \leqslant N_2\\k_1 = k_2}} 1 + E(f_1(X)) E(f_2(X)) \sum_{\substack{1 \leqslant k_1 \leqslant N_1\\1 \leqslant k_2 \leqslant N_2\\k_1 ≠ k_2}} 1 \Biggr)\\
&\peq - E(f_1(X)) E(f_2(X))\\
&= \frac{1}{N_1 N_2} (\min(N_1, N_2) E(f_1(X) f_2(X)) + (N_1 N_2 - \min(N_1, N_2)) E(f_1(X)) E(f_2(X)))\\
&\peq - E(f_1(X)) E(f_2(X))\\
&= \frac{\min(N_1, N_2)}{N_1 N_2} (E(f_1(X) f_2(X)) - E(f_1(X)) E(f_2(X)))\\
&= \frac{1}{\max(N_1, N_2)} \cov(f_1(X), f_2(X)).
\end{align*}
