Show that the probability of this sequence of random variables being a root of this function tends to $0$

Question

Let

• $f\in C^3(\mathbb R)$ with $f>0$, $$\int f(x)\:{\rm d}x=1$$ and such that $(\ln f)'$ is Lipschitz continuous, $$p_n(x):=\prod_{i=1}^nf(x_i)\;\;\;\text{for }x\in\mathbb R^n$$ and $$h_n^{(x)}(z):=\frac{p_n(x+z)}{p_n(x)}-1\;\;\;\text{for }x,z\in\mathbb R^n$$ for $n\in\mathbb N$
• $(\sigma_n)_{n\in\mathbb N}\subseteq(0,\infty)$ be decreasing with $\sigma_n\xrightarrow{n\to\infty}0$
• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $X^{(n)},Y^{(n)}$ be $\mathbb R^d$-valued random variables with $X^{(n)}\sim p_n\lambda^n$ ($\lambda^n$ denoting the Lebesgue measure on $\mathcal B(\mathbb R^n)$) and $$\operatorname P\left[Y^{(n)}\in B\mid X^{(n)}\right]=\mathcal N_d\left(X^{(n)},\sigma_n^2I_n\right)\;\;\;\text{almost surely for all }B\in\mathcal B(\mathbb R^n)\tag1$$ ($I_n$ denoting the $n\times n$ identity matrix) for $n\in\mathbb N$

Note that by $(1)$ there is a $\mathbb R^d$-valued random variable $Z^{(n)}$ on $(\Omega,\mathcal A,\operatorname P)$ independent of $X^{(n)}$ with $Z^{(n)}\sim\mathcal N_d(0,\sigma_n^2I_n)$ and $$Y^{(n)}=X^{(n)}+Z^{(n)}\tag2$$ for all $n\in\mathbb N$.

I want to show that $$\operatorname P\left[h_n^{\left(X^{(n)}\right)}\left(Z^{(n)}\right)=0\right]\xrightarrow{n\to\infty}0\tag3.$$

Obviously, $$h_n^{(x)}(z)=0\Leftrightarrow\prod_{i=1}^n\frac{f(x_i+z_i)}{f(x_i)}=1\;\;\;\text{for all }x,z\in\mathbb R^d.\tag4$$

We may note that the Lipschitz continuity (with Lipschitz constant $c$, say) of the derivative of $g:=\ln f$ implies the following:

1. $g(y)-g(x)-g'(x)(y-x)\ge-\frac c2|y-x|^2$ for all $x,y\in\mathbb R$
2. $f(y)\ge f(x)e^{\frac{|g'(x)|^2}{2c}}e^{-\frac c2\left|y-x-\frac{g'(x)}c\right|^2}\ge f(x)e^{-\frac c2\left|y-x-\frac{g'(x)}c\right|^2}$ for all $x,y\in\mathbb R$
3. $f(x)\le\sqrt{\frac c{2\pi}}e^{-\frac{|g'(x)|^2}{2c}}\le\sqrt{\frac c{2\pi}}$ for all $x\in\mathbb R$
4. $f$ is Lipschitz continuous with constant $\frac c{\sqrt{2\pi e}}$

Answer 1

My intuition is that if you choose $f$ constant on a neighborhood of $0$, then there will be a growing probability that the coordinates of $X^{(n)}$ and $Y^{(n)}$ lay both in this neighborhood (as the variance of the Gaussian goes to zero), meaning that the probability you are interested in does not go to zero in this case.