# Law of large numbers for functions of perturbed random variables

Let $$\Omega$$ be a random vector in $$\mathbb{R}^k$$. Suppose $$f:\mathbb{R}^k \to \mathbb{R}$$ is a continuous function and $$\mathbb{E}\left[ \left\lvert f(\Omega) \right\rvert \right] < +\infty$$. Let $$\{\omega_i\}$$ be an i.i.d. sample of $$\Omega$$, and suppose $$\{\delta_i\}$$ is a fixed sequence in $$\mathbb{R}^k$$ converging to zero. Under what conditions do we have (either almost surely, or in probability) $$\frac{1}{n} \sum_{i=1}^{n} f(\omega_i + \delta_i) \to \mathbb{E}\left[ f(\Omega) \right]?$$

I know that $$\frac{1}{n} \sum_{i=1}^{n} f(\omega_i) \to \mathbb{E}\left[ f(\Omega) \right]$$ in probability/almost surely by the weak/strong law of large numbers. I'm hoping that the result carries over without any additional assumptions on $$\Omega$$ and the sequence of perturbations $$\{\delta_i\}$$.

More generally, I'm looking for techniques that can be used to approach these kinds of questions. In particular, I'm interested in analyzing extensions where the sequence $$\{\delta_i\}$$ may itself be random (not necessarily i.i.d.).

Edit: I don't want to assume that the support of $$\Omega$$ is compact.

• Lipschitz continuity of $f$ would be enough for prob 1 convergence. Though the result should also hold under more general assumptions. – Michael Jun 17 '19 at 18:26
• @Michael When the support of Ω is unbounded, requiring Lipschitz continuity (instead of local Lipschitz continuity) seems like a strong assumption... – madnessweasley Jun 17 '19 at 18:39
• It fails for the nonLipschitz function $f(x,y)=xy$. Consider iid $(X_i,0)$ but $\delta_i=(0,1/i)$. – Michael Jun 17 '19 at 18:48

Some observations on i.i.d. vectors $$\{X_i\}_{i=1}^{\infty}$$ through a function $$f:\mathbb{R}^k\rightarrow\mathbb{R}$$.

1) As in my above comments:

• Lipschitz-like property: If there is a real-valued constant $$L>0$$ such that: $$||f(X_i+\delta_i)-f(X_i)|| \leq L||\delta_i||\quad \forall i \in \{1, 2, 3, ...\}$$ then the desired probability 1 convergence holds.

• It fails for the non-Lipschitz function $$f(x,y)=xy$$. Consider i.i.d. $$\{(Z_i,0)\}_{i=1}^{\infty}$$ and $$\delta_i=(0,1/i)$$ for $$i\in\{1, 2, 3, ...\}$$. Then $$f(Z_i,0) = 0$$ for all $$i$$ but if we assume $$P[Z_1 \geq n^2]=1/n$$ then by Borel-Cantelli we see $$\frac{1}{n}\sum_{i=1}^n Z_i/i$$ does not converge since $$Z_n/n\geq n$$ infinitely often.

2) You can use a generalized LLN result such as this:

Theorem: If $$\{Y_i\}_{i=1}^{\infty}$$ are mutually independent with $$E[Y_i]=0$$ for all $$i$$ and: $$\sum_{i=1}^{\infty} \frac{E[Y_i^2]}{i^2} < \infty \quad (Eq. 1)$$ then $$\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n Y_i = 0 \quad \mbox{(with prob 1)}$$

Proof: [Y. S. Chow, “On a strong law of large numbers for martingales,” Annals of Mathematical Statistics, vol. 38, no. 2, article 610, 1967.]

So if $$\{X_i+\delta_i\}_{i=1}^{\infty}$$ are mutually independent you can define $$Y_i=f(X_i+\delta_i)-E[f(X_i+\delta_i)]$$, assume (Eq. 1) holds, and also assume $$\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^nE[f(X_i+\delta_i)] = E[f(X_1)]$$

• Thanks for your efforts. I'll spend some time looking into your methods to see if I can say something more – madnessweasley Jun 18 '19 at 1:17