# Strong law of large numbers for a scaled sequence of normally distributed random variables

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 $$0$$.

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, $$\int f''\:{\rm d}\lambda^1=0\tag8.$$