Law of large numbers to get the limit of $\frac{1}{n}\sum\limits_{i=1}^n\frac{1}{n}\sum\limits_{j=1}^nf(X_i)g(X_j)$

Question

Suppose I have $X_i,i=1,\dots,n$, i.i.d. finite mean random variables. Can I use the law of large numbers to get $$\frac{1}{n}\sum_{i=1}^n\frac{1}{n}\sum_{j=1}^nf(X_i)g(X_j) \longrightarrow\mathbb{E}[f(X_i)g(X_i)]\; a.s.$$ I know that in general we have $$\frac{1}{n}\sum_{i=1}^nf(X_i)\longrightarrow \mathbb{E}[f(X_i)]\;a.s.$$ and same for $g(X_i)$. But I can't see what do we do for the product.

Answer 1

Note that $$\frac{1}{n} \sum_{i=1}^n \frac{1}{n} \sum_{j=1}^n f(X_i) g(X_j) = \left( \frac{1}{n} \sum_{i=1}^n f(X_i) \right) \left( \sum_{j=1}^n g(X_j) \right).$$ By the law of large numbers, $$\frac{1}{n} \sum_{i=1}^n f(X_i) \stackrel{a.s.}{\to} \mathbb{E}f(X_1) \qquad \qquad \frac{1}{n} \sum_{j=1}^n g(X_j) \stackrel{a.s.}{\to} \mathbb{E}g(X_1).$$

Hence,

$$\frac{1}{n} \sum_{i=1}^n \frac{1}{n} \sum_{j=1}^n f(X_i) g(X_j) \stackrel{a.s.}{\to} \mathbb{E}f(X_1) \cdot \mathbb{E}g(X_1).$$

Answer 2

$$\frac{1}{n}\sum_{i=1}^n\frac{1}{n}\sum_{j=1}^nf(X_i)g(X_j)=\left(\frac{1}{n}\sum_{i=1}^nf(X_i)\right)\cdot\left(\frac{1}{n}\sum_{j=1}^ng(X_j)\right)$$