# Question about Strong law of large numbers

I am confused by this problem. Let $F$ be a distribution function with $F(0-)=0$ and $F(1)=1$ and let $\mu$ be the associated law.

Let $m_k=\int_{[0,1]}x^k dF(x)$.

Define \begin{array}{c c c} \Omega=[0,1]\times [0,1]^{\mathbb{N}} &\Sigma=\mathcal{B}\times\mathcal{B}^{\mathbb{N}} & P=\mu\times Leb^{\mathbb{N}} \\ \Theta(w)=w_0 &H_k(w)=I_{[0,w_0]}(w_k) & S_n=H_1+...+H_n .\\ \end{array} Here $\mathcal{B}^{\mathbb{N}}$ and $Leb^{\mathbb{N}}$ denote respectively the product Borel sigma algebra and product lebesgue measure on $[0,1]^{\mathbb{N}}$.

The problem says that by the strong law and Fubini's theorem it follows that $$S_n/n\longrightarrow \Theta \quad a.s.$$

I am confused by this claim. The way I interpret the problem is that $w=(w_0,w_1,...)$ and $\Theta$ is a function defined on this sequence acting as the projection on the first component. The strong law of large numbers says that the averages converge to the expected value which is a constant, so how can they converge to a non constant function?

Also if I compute this expected value I get

$$E(H_1)=\int_0^1\int_0^1 I_{[0,\omega_0]}(\omega_1) dF(\omega_0) \: dLeb(\omega_1)=\int_0^1 \omega_0 dF(\omega_0)=m_1$$ because all the other integrals are $1$ since the function doesn't depend on the other variables... Where am I going wrong?

• It took me a few minutes to figure out that $Leb^{\mathbb{N}}$ means "the Lebesgue product measure on $\mathbb{R}^{\mathbb{N}}$", so you might clarify that in your question. – Ian May 3 '15 at 17:39

For each fixed $w_0$, the random variables

$$[0,1]^{\mathbb{N}} \ni \tilde{w} \mapsto H_k(w_0,\tilde{w})$$

are iid. Therefore by, the strong law of large numbers,

$$\frac{S_n(w_0,\cdot)}{n} \to \mathbb{E}H_1(w_0) \qquad \text{\text{Leb}^{\mathbb{N}}-almost surely.}$$

This means that there exists a set $N(w_0)$, $\text{Leb}^{\mathbb{N}}(N(w_0))=0$, such that

$$\frac{S_n(w_0,\tilde{w})}{n} \to \mathbb{E}H_1(w_0) \qquad \text{for all \tilde{w} \in [0,1]^{\mathbb{N}} \backslash N(w_0).}$$

Since

$$\mathbb{E}H_1(w_0,\cdot) = \int_0^{w_0} \, \text{Leb}(dw_k) = w_0$$

we get

$$\frac{S_n(w_0,\tilde{w})}{n} \to w_0 \qquad \text{for all \tilde{w} \in [0,1]^{\mathbb{N}} \backslash N(w_0).}$$

By Tonelli's theorem,

\begin{align*}& P\left(\{(w_0,\tilde{w}): \frac{S_n(w_0,\tilde{\omega})}{n} \, \text{does not converge to \omega_0}\} \right) \\ &= P(\{(w_0,\tilde{w}); \tilde{w} \in N(w_0)\})\\ &= \int_{[0,1]} \underbrace{\int_{[0,1]^{\mathbb{N}}} 1_{N(w_0)}(\tilde{w}) \, d\text{Leb}^{\mathbb{N}}(\tilde{w})}_{0} \, d\mu(w_0)\\ &=0. \end{align*}

Hence,

$$\frac{S_n}{n} \to \theta \qquad \text{P-almost surely}.$$

• Thank you for the detailed explanation – bibo May 3 '15 at 21:53

The $H_k$ are random variables. For each fixed $w_0 \in [0,1]$, they are iid: specifically they are iid Bernoulli($w_0$) variables (since $w_k,k=1,2,\dots$ are iid U(0,1) variables). So by the strong law, for each fixed $w_0$, $S_n/n$ converges a.s. to the expected value of an iid Bernoulli($w_0$) variable, which is $w_0=\Theta(w)$. Now you just need to check that this implies that $S_n/n$ converges a.s. (in your full probability space) to $\Theta(w)$. This is the Fubini/Tonelli step.

What's going on here is only slightly more complicated than a situation where $X_k(t)$ are iid Bernoulli(t) variables for each $t \in [0,1]$, in which case $S_n/n \to t$ by the strong law. The difference is that in this setting $t$ is deterministic, whereas in your setting, $w_0$ is also random, but it is chosen according to a distribution independent of the distribution of the $w_k$ for $k \geq 1$.