1
$\begingroup$

Suppose that $$\sqrt{n}(X_n - \theta)\xrightarrow{d} X,$$ according to the delta method, we have $$\sqrt{n}(g(X_n)-g(X))\xrightarrow{d} g'(\theta)X$$ when $g$ is differentiable.

My question is, if $$\sqrt{n} (g_m(X_n)-g_m(\theta))\xrightarrow{d}g_m'(\theta)X, \quad m=1,2,\ldots $$ and $g_m(x)$ converges to $g(x)$ in some sense, are there any results indicating when we can expect$$\sqrt{n}(g(X_n)-g(\theta))\xrightarrow{d}g'(\theta)X$$

Improve this question
$\endgroup$
1
  • $\begingroup$ a usual condition having some uniformity or equicontinuity in $g_{m}\to g$, because then you can do a diagonalization argument as in the proof of Arzela-Ascoli. $\endgroup$
    – Thomas Kojar
    Commented Jul 19, 2023 at 15:15

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .