How does one prove asymptotic normality of the Non-linear least squares from First order conditions? Our model is $Y=X(\beta_0)+u$, where $u\sim IID(0,\sigma_0^2I)$, and $X(\beta)$ is a non-linear function of the beta.
When trying to minimize the $SSR(\beta)$ we get the following FOC:
$\nabla X(\beta)^T(Y-X(\beta))=0$, where $\nabla X(\beta)$ is the gradient.
How does one prove the asymptotic normality of $\hat\beta$ from this FOC?
Any help would be appreciated.
Edit:
If we apply a taylor expansion of the first order to each component $X_t(\beta)$ of $X(\beta)$, we obtain $X_t(\beta)=X_t(\beta_0)+\nabla X(\bar\beta_{(t)})^T(\beta-\beta_0)$, where $\bar\beta_{(t)}$ is a point in the line segment that joins $\beta$ and $\beta_0$. This point may be different for each taylor expansion we do, and that's why it's indexed by $t$.
Inserting the taylor expansion in the FOC:
$n^{(-1/2)}(\nabla X(\beta)^T(u-\nabla \bar X^T(\beta-\beta_0))=0$, where $\nabla \bar X$ is the matrix with $\nabla X(\bar\beta_{(i)})$ as each i-th column.
Are all of the above calculations correct? I ask this because in this book, the authors state in page 225 that we should obtain a term with second derivatives of $X(\beta)$... I do not understand why this is.
Any help would be appreciated
 A: The first order condition is
$$
g(b):=\frac{\partial S(\beta)}{\partial \beta}\bigg|_{\beta=b}=0
$$
by construction, as we define $b$ as the solution (we may here assume unique minimum, nice parameter space). Let $\beta_0$ be the true value and $\bar\beta$ a vector with elements satisfying $\bar{\beta}_i\in [\min\{b_i, \beta_{0, i}\}, \max\{b_i, \beta_{0, i}\}]$. Then
$$
g(b)=g(\beta_0)+\frac{\partial g(\beta)}{\partial \beta'}\bigg|_{\beta=\bar\beta}(b-\beta_0)=0.
$$
Note here that
$$
\frac{\partial g(\beta)}{\partial \beta}\bigg|_{\beta=\bar\beta}=\frac{\partial^2 S(\beta)}{\partial \beta\partial\beta'}\bigg|_{\beta=\bar\beta}
$$
a $p\times p$ matrix (where $p$ is the length of $\beta$). Rearrange the terms
\begin{align}
g(\beta_0)+\frac{\partial g(\beta)}{\partial \beta'}\bigg|_{\beta=\bar\beta}(b-\beta_0)&=0\\
\left(\frac{\partial g(\beta)}{\partial \beta'}\bigg|_{\beta=\bar\beta}\right)^{-1}g(\beta_0)+(b-\beta_0)&=0\\
\sqrt n(b-\beta_0)&=-\left(n^{-1}\frac{\partial g(\beta)}{\partial \beta'}\bigg|_{\beta=\bar\beta}\right)^{-1}\frac{1}{\sqrt{n}}g(\beta_0)
\end{align}
Usually, it is assumed that
\begin{align}
n^{-1}\frac{\partial g(\beta)}{\partial \beta'}\bigg|_{\beta=\beta_0}&\overset{p}\to Q \, (\text{pos. def.})\\
\frac{1}{\sqrt{n}}g(\beta_0) & \overset{d}{\to} N(0, \sigma^2Q).
\end{align}
Since $\bar{\beta}$ is sandwiched between $b$ and $\beta_0$ and $b\overset{p}{\to}\beta_0$, we may replace the evaluation at $\bar\beta$ by $\beta_0$ in the asymptotic analysis. By the continuous mapping theorem the inverse of the matrix of second derivatives will tend to $Q^{-1}$ and by Slutsky's theorem the asymptotic distribution of $\sqrt{n}(b-\beta_0)$ is the same as that of
\begin{align}
-\left(n^{-1}\frac{\partial g(\beta)}{\partial \beta'}\bigg|_{\beta=\beta_0}\right)^{-1}\frac{1}{\sqrt{n}}g(\beta_0) \overset{d}{\to} N(0, \sigma^2Q^{-1}QQ^{-1})=N(0, \sigma^2Q^{-1}).
\end{align}
