• $f$ denote the density of $\mathcal N(0,1)$
  • $g:=\ln f$
  • $r_n(x):=\frac1{n-1}\sum_{i=2}^n{g'(x_i)}^2$ and $s_n(x):=-\frac1{n-1}\sum_{i=2}^ng''(x_i)$ for $x\in\mathbb R^n$ and $n\ge2$
  • $c:=\int{f'(x)}^2{f(x)}\:{\rm d}x=1$
  • $F_n:=\left\{x\in\mathbb R^n:\max(|r_n(x)-c|,|s_n(x)-c|)<n^{-\frac18}\right\}$ for $n\ge2$
  • $\ell>0$
  • $\sigma_n:=\ell(n-1)^{-\frac12}$ for $n\ge2$
  • $Q_n$ denote the Markov kernel on $(\mathbb R^n,\mathcal B(\mathbb R^n))$ given by $$Q_n(x,\;\cdot\;)=\mathcal N(x,\sigma_n^2I_n)\;\;\;\text{for all }x\in\mathbb R^n$$ for $n\ge2$
  • $B_n(x,y):=\sum_{i=1}^n(g(y_i)-g(x_i))$ and $\alpha_n(x,y):=\min\left(1,e^{B_n(x,\:y)}\right)$ for $x,y\in\mathbb R^n$ and $n\ge2$

Fix $\varphi\in C_c^\infty(\mathbb R)$. Now let $$A_n(x):=n\int Q_1(x_1,\:{\rm d}y_1)(\varphi(y_1)-\varphi(x_1))\int Q_{n-1}((x_2,\ldots,x_n),{\rm d}(y_2,\ldots,y_n))\alpha_n(x,y)$$ and $$\tilde A_n(x):=\ell^2(\varphi''(x_1)\int Q_{n-1}((x_2,\ldots,x_n),{\rm d}(y_2,\ldots,y_n))\alpha_{n-1}((x_2,\ldots,x_n),(y_2,\ldots,y_n))+\ell^2g'(x_1)\varphi'(x_1)\int Q_{n-1}((x_2,\ldots,x_n),{\rm d}(y_2,\ldots,y_n))e^{B_{n-1}((x_2,\:\ldots\:,\:x_n),(y_2,\:\ldots\:,\:y_n))}1_{\left\{\:e^{B_{n-1}((x_2,\:\ldots\:,\:x_n),(y_2,\:\ldots\:,\:y_n))}\:<\:0\:\right\}}$$ for $x\in\mathbb R^n$.

Are we able to show that $\sup_{x\in F_n}|A_n(x)-\tilde A_n(x)|\xrightarrow{n\to\infty}0$?

The trick should be a Taylor expansion of $\alpha_n$ with respect to $y_1$. However, I've got a hard time to figure out whether we are really allowed to apply a version of Taylor's theorem. The problem being that $\alpha_n$ is not partially differentiable with respect to $y_1$ on all of $\mathbb R$.

I was able to deduce (but don't know whether it's relevant or not) that if $X^n$ is an $\mathbb R^n$-valued standard normally distributed random variable (i.e. $X^n\sim(f\lambda)^{\otimes n}$, where $\lambda$ denotes the Lebesgue measure on $\mathcal B(\mathbb R^n)$) on a probability space $(\Omega,\mathcal A,\operatorname P)$ for $n\ge2$, then $\operatorname P\left[X^n\in F_n\right]\xrightarrow{n\to\infty}1$.

Remark: You may want to take note of the following related questions I've asked before:

EDIT: Fix $x\in F_n$. Let $h(y):=e^{B_n(x,\:y)}-1$ for $y\in\mathbb R^n$. Note that $2\alpha_n(x,y)=e^{B_n(x,\:y)}+1-|h(y)|$ for all $y\in\mathbb R^n$. By the first link above, $|h|$ is partially differentiable at $y$ with respect to $y_1$ with $$\frac{\partial|h|}{\partial y_1}(y_1)=\operatorname{sgn}(h(y))\frac{\partial h}{\partial y_1}(y)$$ for all $y\in\{h\ne0\}\cup\left\{\frac{\partial h}{\partial y_1}=0\right\}$. Now, $\{h\ne0\}=\{y\in\mathbb R^n:|x|\ne|y|\}$ and $\left\{\frac{\partial h}{\partial y_1}=0\right\}=\{y\in\mathbb R^n:y_1=0\}$. Let $Y$ be a $\mathbb R^n$-valued random variable on a probability space $(\Omega,\mathcal A,\operatorname P)$ with $Y\sim Q_n(x,\;\cdot\;)$. Note that the inner integral occuring in $A_n(x)$ is equal to $\left.\operatorname E\left[\alpha_n(x,(y_1,Y_2,\ldots,Y_n))\right]\right|_{y_1=Y_1}$.

Now I might be wrong, so don't take the following for granted: Fix $y_1\in\mathbb R$ for the moment. In order to apply Taylor, we would at least need that $|h|$ is differentiable at $(y_1,Y_2,\ldots,Y_n)$ with respect to $y_1$. If $y_1\ne0$, then this means (by the result mentioned above; at least if I'm not terribly wrong) that we should have $\operatorname P\left[|x|=|(y_1,Y_2,\ldots,Y_n)|\right]=0$. However, since $\operatorname E\left[Y_i\right]=x_i$ for all $i\in\{2,\ldots,n\}$ (and $Y_2,\ldots,Y_n)$ are mutually independent), we should even got $$\operatorname P\left[|x|=|(y_1,Y_2,\ldots,Y_n)|\right]\xrightarrow{n\to\infty}1_{\{x_1\}}(y_1)$$ (since $\sigma_n\xrightarrow{n\to\infty}0$).

Please tell me if I made any wrong conclusion above, but it seems like there is no chance to apply Taylor here.

EDIT 2: On the other hand, the projection to the first coordinate of the set on which $|h|$ is not partially differentiable with respect to $y_1$ is countable (again by the answer to the question in the first link) and hence (by definition of the product measure) the whole set should have $n$-dimensional Lebesgue measure $0$ ... I'm confused.


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