> [!abstract]
> **The Delta method** estimates the mean and variance of some function $f(X)$. If $X$ is asymptotically Gaussian, so is the estimate.
Suppose there is a distribution $D$ with mean $\mu$ and variance $\sigma^{2}$, and we want to estimate some function $g(\mu)$ from an iid. sample $X_{1}, \dots, X_{n}\sim D$.
The Taylor expansion of $g(\bar{X})$ around $\mu$ is: $g(\bar{X})=g(\mu)+ (\bar{X}-\mu)g'(\mu)+O((\bar{X}-\mu)^{2})$where $(\bar{X}-\mu)=O(n^{-1 / 2})$ by the CLT.
The **delta method** linearizes the expansion to be $g(\bar{X})\approx g(\mu)+ (\bar{X}-\mu)g'(\mu)$ and gives the estimates $\begin{align*}
\mathbb{E}[g(\bar{X})]&\approx g(\mu)\\
\mathrm{Var}(g(\bar{X}))&\approx \mathrm{Var}\left((\bar{X}-\mu)g'(\mu)\right)\\
&= \mathrm{Var}(\bar{X})g'(\mu)^{2}\\
&= \frac{\sigma^{2}}{n}g'(\mu)^{2}
\end{align*}$
Since the estimate of $g(\bar{X})$ is a linear transformation of $(\bar{X}-\mu)$, and $(\bar{X}-\mu) \overset{D}{\approx} N(0, \sigma^{2} / n)$ by the CLT, the **asymptotic distribution** of $g(\bar{X})$ is also normal: $g(\bar{X}) \overset{D}{\approx} N\left(g(\mu),\frac{\sigma^{2}}{n}g'(\mu)^{2}\right)$and the parameters are the **asymptotic mean/variance**.
> [!algorithm] The Delta Method
> In general, given a variable $Y$ with $\mathbb{E}[Y]=\mu$, the delta method estimates its function $g(Y)$ to have $\begin{align*}
\mathbb{E}[g(Y)]&\approx g(\mu)\\
\mathrm{Var}(g(Y))&\approx \mathrm{Var}(Y)g'(\mu)^{2}
\end{align*}$this does not assume that $Y$ is normal (even asymptotically).
>
> Furthermore, if $Y \overset{D}{\approx} N(\mu,\sigma^{2})$, then the asymptotic distribution of its function $g(Y)$ is $g(Y) \overset{D}{\approx} N\left(g(\mu),\frac{\sigma^{2}}{n}g'(\mu)^{2}\right)$