# Are we allowed to apply a (measure-theoretic version of) Taylor's theorem here?

Let

• $$f$$ 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 $$n\ge2$$
• $$c:=\int{f'(x)}^2{f(x)}\:{\rm d}x=1$$
• $$F_n:=\left\{x\in\mathbb R^n:\max(|r_n(x)-c|,|s_n(x)-c|) for $$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.