Let
- $f\in C^3(\mathbb R)$ be positive
- $g:=\ln f$
- $d\in\mathbb N$, $$p_d(x):=\prod_{i=1}^df(x_i)\;\;\;\text{for }x\in\mathbb R^d$$ and $\lambda^d$ denote the Lebesgue measure on $\mathcal B(\mathbb R^d)$
- $\ell>0$, $\sigma_d:=\ell d^{-\alpha}$ for some $\alpha\in[0,1]$ and $$Q_d(x,\;\cdot\;):=\mathcal N(x,\sigma_d^2I_d)\;\;\;\text{for }x\in\mathbb R^d$$
- $X$ be a $\mathbb R^d$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$ with $$X_\ast\operatorname P=p_d\lambda^d$$
- $Y$ be a $\mathbb R^d$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$ with $$\operatorname P\left[Y\in B\mid X\right]=Q_d(X,B)\;\;\;\text{almost surely for all }B\in\mathcal B(\mathbb R^d)\tag0$$
Note that, by $(0)$, $(X,Y)_\ast\operatorname P=X_\ast\operatorname P\otimes\:Q_d$ is the product of the distribution $X_\ast\operatorname P$ of $X$ under $\operatorname P$ and the Markov kernel $Q_d$. Moreover, there is a $\mathbb R^d$-valued random variable $Z$ on $(\Omega,\mathcal A,\operatorname P)$ with $Z_\ast\operatorname P=\mathcal N_d(0,I_d)$ and $Y=X+\sigma_dZ$. It's easy to see that $X$ and $Y-X$ are independent.
Assume $$I:=\int f|g'|^2\:{\rm d}\lambda^1<\infty.$$
Are we able to show $$S_d:=\frac1d\sum_{i=1}^dg''(X_i)(Y_i-X_i)^2\xrightarrow{d\to\infty}-I\;\;\;\text{almost surely?}\tag1$$
In this paper, at the beginning of page 3, it is claimed that $(1)$ holds "under appropriate technical conditions".
$(1)$ seems wrong to me. From the strong law of large numbers, we should obtain $$\frac1d\sum_{i=1}^d\frac{f''(X_i)}{f(X_i)}Z_i^2\xrightarrow{d\to\infty}\int f''\:{\rm d}\lambda^1\;\;\;\text{almost surely}\tag2$$ and $$\frac1d\sum_{i=1}^d{g'(X_i)}^2Z_i^2\xrightarrow{d\to\infty}I\;\;\;\text{almost surely}\tag3.$$ Noting that $$g''=\frac{f''}f-|g'|^2\tag4,$$ we should have $$S_d\xrightarrow{d\to\infty}\int f''\:{\rm d}\lambda^1-I\;\;\;\text{almost surely}\tag5$$ instead of $(1)$. What am I missing? It seems like the only "technical condition" that would yield the claim is that the integral on the right-hand side of $(2)$ is $0$.
Best Answer
I guess the claim is wrong in general. However, if we assume that $g'$ is Lipschitz continuous, we're able to conclude $f(x)\xrightarrow{|x|\to\infty}0$ and $f'(x)\xrightarrow{|x|\to\infty}0$. If we now further assume that $$\int|f''|\:{\rm d}\lambda^1<\infty\tag6,$$ we obtain $$\int fg''\:{\rm d}\lambda^1\xleftarrow{r\to\infty}f'(r)-f'(-r)-\int_{-r}^rf'(x)g'(x)\:{\rm d}x\xrightarrow{r\to\infty}-I\tag7$$ by Lebesgue's dominated convergence theorem and partial integration. In particular, $$\int f''\:{\rm d}\lambda^1=0\tag8.$$