= Solution
Here the <negative binomial distribution> has mean parameter $\mu>0$ and size parameter $\theta>0$. Its support in the original PDF is all nonnegative integers; a finite endpoint appearing in the TeX aid would be incorrect. The <probability generating function>, obtained from the generalized binomial series, is
$$
G(s)=\left(\frac{\theta}{\theta+\mu(1-s)}\right)^\theta.
$$
Differentiating at one gives $G'(1)=\mu$ and $G''(1)=\mu^2(1+1/\theta)$. Since the <variance> is $G''(1)+G'(1)-G'(1)^2$, this proves
$$
\boxed{\mathbb E Y=\mu,\qquad\operatorname{Var}(Y)=\mu+\mu^2/\theta.}
$$
The extra positive term explains its <overdispersion> relative to a <Poisson distribution>.
For one observation, write the <log-likelihood> as
$$
\ell_y=\log\Gamma(\theta+y)-\log\Gamma(\theta)-\log(y!)+y\log\mu+\theta\log\theta-(\theta+y)\log(\mu+\theta).
$$
The mean <score function> and its cross derivative are
$$
u_\mu=\frac{y}{\mu}-\frac{\theta+y}{\mu+\theta}=\frac{\theta(y-\mu)}{\mu(\mu+\theta)},\qquad \frac{\partial u_\mu}{\partial\theta}=\frac{y-\mu}{(\mu+\theta)^2}.
$$
Taking <expectations> gives $I_{\mu\theta}=-\mathbb E(\partial_\theta u_\mu)=0$. In addition,
$$
I_{\mu\mu}=\frac{\theta}{\mu(\mu+\theta)}=\frac1{\mu+\mu^2/\theta}.
$$
For reference, with $\psi$ the <digamma function>, the size <score function> is
$$
u_\theta=\psi(\theta+y)-\psi(\theta)+\log\theta+1-\log(\mu+\theta)-\frac{\theta+y}{\mu+\theta}.
$$
Its <variance> $I_{\theta\theta}$ is finite and positive for finite positive $\mu,\theta$. Thus <negative binomial mean-size parameter orthogonality> makes the expected <Fisher information matrix> diagonal. For an independent identically distributed sample and a regular identifiable interior true parameter, <asymptotic normality of a maximum likelihood estimator> gives
$$
\sqrt n\begin{pmatrix}\widehat\mu-\mu\\\widehat\theta-\theta\end{pmatrix}\xrightarrow d N_2\left(0,\begin{pmatrix}\mu+\mu^2/\theta&0\\0&I_{\theta\theta}^{-1}\end{pmatrix}\right).
$$
Hence \b[the asymptotic correlation is zero]. This is first-order asymptotic independence, not a claim of exact finite-sample independence. Indeed the mean score yields $\widehat\mu=\bar y$ at an interior fit, while the size fit also depends on the observed dispersion. The argument excludes a true boundary such as the Poisson limit $\theta=\infty$.
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