The [[Delta Method]] allows us to make asymptotic statements about $Y$, given we know the asymptotic behavior of $X$ and given that $Y$ can be expressed as a function $g(X)$. In a multivariate setup, where both $\bar X_n$ and $\mu$ are $\in \mathbb R^d$, and $\Sigma$ is a $(d \times d)$ [[Covariance Matrix]], the [[Multivariate Central Limit Theorem]] tells us the following.. $ \sqrt n \, (\bar X_n- \mu) \xrightarrow[]{}\mathcal N(0, \Sigma) $ The function $g$, which maps $X \mapsto Y$ can be a scalar-valued or vector-valued function with $k$ dimensional output. $ g(X): \mathbb R^d \mapsto \mathbb R^k $ **Univariate Delta Method:** We can state a convergence to a Gaussian, where the [[Variance]] is scaled by the squared slope $g^\prime(\theta)$. $ \sqrt n \big(g(\bar X_n)-g(\theta)\big) \xrightarrow[n \to \infty]{(d)}\mathcal N \Big(0, g^\prime(\theta)^2*\sigma^2\Big) $ **Multivariate Delta Method:** Instead of the derivative $g^\prime(\theta)$, we need to work with $\nabla g(\theta)$. $ \sqrt n \big(g(\bar X_n)-g(\theta)\big) \xrightarrow[n \to \infty]{(d)} \mathcal N \Big(0, \nabla g(\theta)^T*\Sigma* \nabla g(\theta)\Big) $ When $g$ is a vector-valued function, then $\nabla g$ is a [[Jacobian#Jacobian Matrix|Jacobian Matrix]] $\mathbf J$, where the gradients of each output element $g_i$ are column vectors that are glued together from left to right. $ \nabla g=\mathbf J= \begin{bmatrix} \frac{\partial g_1}{\partial x_1}& \ldots & \frac{\partial g_k}{\partial x_1} \\ \vdots & \ddots & \vdots \\ \frac{\partial g_1}{\partial x_d}& \ldots & \frac{\partial g_k}{\partial x_d} \end{bmatrix} $