# Law of large numbers for a sequence of random variables

Suppose we have a sequence of random variables $$X^M$$ which converges almost surely to a random variable $$X^0$$ and let $$(X_1^M, \ldots, X_M^M)$$ be iid samples from $$X^M$$ for $$M \in \mathbb{N}$$.

Under which conditions does the law of large numbers hold uniformly in the sense that

$$|M^{-1} \sum_{m=1}^M X_m^M - \mathbb{E}(X^0)| \xrightarrow{M \to \infty} 0 \; \text{ almost surely?}$$

I am happy to assume that the $$X^m$$ are uniformly bounded, i.e., that there is a constant $$K$$ such that

$$|X^m| \leq K, \; \text{almost surely for all } m \in \mathbb{N}_0.$$

I looked into uniform LLNs but they generally do not seem to fit the setting above.

• Under the condition that $X^m$ have integrable dominant, this follows from Uniform law of large numbers. Please write if you have problems applying it. Jun 3 '19 at 20:13
• @zhoraster: I looked at these kind of results. In the case above, the discrete parameter $M$ would probably take the role of $\theta$. However, I have problems with the condition that $M \mapsto X^M(\omega)$ is continuous, since it is only defined at the natural numbers. It would be great if you could provide more details.
– dawo
Jun 4 '19 at 5:04
• @zhoraster: What could maybe work is to define the function $f(\omega, M^{-1}) = X^M(\omega)$. Then $\lim_{M\to \infty} f(\omega, M^{-1}) = f(\omega, 0) = X^0(\omega)$ as required. To extend the function to a continuous function on [0,1] one could linearly interpolate between the points $M^{-1}$. Is that what you had in mind?
– dawo
Jun 4 '19 at 5:10
• You don't need to have $[0,1]$. Any compact will do. Jun 4 '19 at 14:45
• Any luck proving this? Jun 8 '19 at 7:52

Assuming uniform integrability of $$X^m$$, we have that $$\mathrm E[X^m]\to \mathrm E[X]$$, $$m\to\infty$$. Therefore, it is enough to show that $$\left| (\overline{X^m})_m - \mathrm E[X^m]\right|\to 0, m\to \infty,\tag{1}$$ almost surely, where $$(\overline{X^m})_m = \frac1m \sum_{i=1}^m X_i^m$$.

One possibility is to go through concentration inequalities. For example, if the variables are bounded, as in your question, then by the Hoeffding inequality, for any $$\varepsilon>0$$, $$\mathrm P\left(\left| (\overline{X^m})_m - \mathrm E[X^m]\right|>\varepsilon\right)\le e^{-C \varepsilon^2 m}$$ with some $$C>0$$. Using the Borel-Cantelli lemma, we easily get $$(1)$$.

Another possibility is, as I commented, to deduce the uniform convergence $$\sup_m \left| (\overline{X^m})_n - \mathrm E[X^m]\right|\to 0, n\to \infty,\tag{2}$$ from the uniform law of large numbers. However, it seems unlikely that the almost sure convergence can be shown this way; I will only outline the convergence in probability.

Let $$F^m$$ be the cdf of $$X^m$$ and $$Q^m(t) = \sup\{x\in \mathbb R: F^m(x), be its quasi-inverse (quantile function). Then, as it is well known, $$X^m \overset{d}{=} Q^m(U)$$, where $$U$$ is a uniform $$[0,1]$$ variable. Therefore, $$(\overline{X^m})_n \overset{d}{=} \frac1n \sum_{k=1}^n Q^m(U_k),$$ where $$U_1,U_2,\dots$$ are iid uniform $$[0,1]$$ variables. Also it follows from the weak convergence of $$X^m\to X^0$$ that $$Q^m\to Q^0$$ pointwise in the continuity points of $$Q^0$$, hence, almost everywhere on $$(0,1)$$.

Now let $$\Theta = \{m^{-1}, m\ge 1\}\cup \{0\}$$ and set $$f(t,m^{-1}) = Q^m(t)$$, $$m\ge 1$$, $$f(t,0) = Q^0(t)$$. Then, as is explained above, $$f(t,\theta)$$ is continuous in $$\theta$$ for almost all $$t$$ (modulo the distribution of $$U$$). Therefore, assuming existence of integrable majorant of $$f(U,m^{-1})=Q^m(U)$$ (which is easily seen to be equivalent to uniform integrability of $$X^m$$), we get that $$\sup_{\theta\in \Theta}\left| \frac1n \sum_{k=1}^n f(U_k,\theta) - \mathrm{E}[f(U,\theta)]\right| \to 0, n\to \infty,$$ almost surely, whence we get the convergence $$(2)$$ in probability (remember that we replaced $$(\overline{X^m})_n$$ by its distributional copy).

The convergence in probability might sound bad, but there are at least two advantages:

1. Only uniform integrability is required.

2. The approach works for any $$(n_m,m\ge 1)$$ such that $$n_m\to\infty$$, $$m\to\infty$$, i.e. we have $$\left| (\overline{X^m})_{n_m} - \mathrm E[X^m]\right|\to 0, m\to \infty,$$ in probability. The first approach fails (to establish the almost sure convergence) for "small" $$n_m$$.

• Thank you for this extensive answer! If I understand your point correctly, both approaches 'only' yield convergence in probability. You might be right that in the general setting outlined above, almost sure convergence will be hard to establish. I also largely agree with your assessment of convergence in probability. However, my problem has a little more structure and the original post was written to abstract from my concrete situation. In the process, I might have lost what actually might help to establish almost sure convergence (see post below).
– dawo
Jun 19 '19 at 5:28
• @dawo, no, the first approach gives an a.s. convergence. Jun 20 '19 at 3:28
• Thank you - I must have missed the Borel-Cantelli Lemma on the first reading.
– dawo
Jun 26 '19 at 13:39

In the case I am considering the random variable $$X^M$$ can actually be written as function of $$X^0$$ and $$M$$, i.e., $$X^M = f(X^0, M)$$ for all $$M \in \mathbb{N}$$.

A priori, this function is only defined on the integers. However, I could extend it to a function $$f: \mathbb{R} \times [0,1] \to \mathbb{R}$$ by defining $$f(x, \theta) = \begin{cases} f(x,M), & \text{if } \theta = M^{-1}, \text{ for } M \in \mathbb{N} \\ \frac{\lambda}{M^{-1}-(M+1)^{-1}} f(x,M) + \frac{1-\lambda}{M^{-1}-(M+1)^{-1}} f(x,M+1), & \text{for } \theta = \frac{\lambda M^{-1}}{M^{-1}-(M+1)^{-1}} + \frac{(1-\lambda)(M+1)^{-1}}{M^{-1}-(M+1)^{-1}} \end{cases}.$$

By my assumption on the almost sure convergence $$X^M \to X^0$$, it follows that $$f(x, \theta) \xrightarrow{\theta \to 0} f(x, 0)$$ and the continuity of $$f$$ at all other points $$\theta \in (0,1]$$ follows from the way $$f$$ is defined.

Hence, in this case the conditions for the uniform LLN are fulfilled and it would follow that $$\sup_{\theta \in [0,1]} \left| n^{-1} \sum_{i=1}^n f(X_i^0, \theta) - \mathbb{E} (f(X^0, \theta)) \right| \xrightarrow{n \to \infty} 0 \quad \text{almost surely.}$$ This would imply $$\sup_{M \in \mathbb{N}} \left| n^{-1} \sum_{i=1}^n f(X_i^0, M^{-1}) - \mathbb{E} (f(X^0, M^{-1})) \right| = \sup_{M \in \mathbb{N}} \left| n^{-1} \sum_{i=1}^n X_i^M - \mathbb{E} (X_i^M) \right| \xrightarrow{n \to \infty} 0 \quad \text{almost surely.}$$

Hence, for every $$\epsilon > 0$$, there is a $$N(\epsilon)$$ such that $$\left| n^{-1} \sum_{i=1}^n X_i^M - \mathbb{E} (X_i^M) \right| < \epsilon, \quad \forall n \geq N(\epsilon),$$ which means that for $$M \geq N(\epsilon)$$, we have $$\left| M^{-1} \sum_{i=1}^M X_i^M - \mathbb{E} (X_i^M) \right| < \epsilon \quad \text{almost surely.}$$

As mentioned in the post by zhoraster above, $$|\mathbb{E}(X^M) - \mathbb{E}(X^0)| \xrightarrow{M \to \infty} 0 \quad \text{almost surely}$$ using the boundedness assumption on the $$X^M$$ (or indeed the weaker condition of uniform integrability).

Putting these two things together establishes (via an application of the triangle inequality) that $$|M^{-1} \sum_{m=1}^M X_m^M - \mathbb{E}(X^0)| \xrightarrow{M \to \infty} 0 \; \text{ almost surely.}$$

• As I said, considering piecewise linear functions is overcomplicating things. If indeed $F(X^0,M)\to X_0$, $M\to\infty$, a.s., then you have a strong (i.e. almost sure) LLN provided that $\sup_{M\ge 1} F(X^0,M)$ is integrable (uniform integrability of $F(X^0,M)$ might be enough). Just take $\Theta$ as in my answer. Jun 20 '19 at 3:35