# A special case of the law of large numbers for (possibly) dependent, identically distributed random variables

I want to show that $\dfrac{1}{N}\sum\limits_i^N x_iw_i \to 0$ as $N\to\infty$, where $w_i\sim \mathcal{N}(0,\sigma^2)$, $\mathbb{E}[x_i] = 0$ and $\mathbb{E}[x_i^2] = \rho$. The $x_i$s are independent of the $w_i$s. The $x_i$s are identically distributed but not necessarily independent. The convergence is in probability

I have tried showing that the terms in the sum are mutually independent (so I could invoke the law of large numbers) but I have not succeeded. My next attempt was to find something like the law of large numbers for dependent variables, but I couldn't get that to work either.

I'm not sure if it can be shown that the above holds under the given assumptions, but by gut feeling says it should.

Can anyone point me in the right direction?

Edit

After spending some more time, I think I managed to show what I wanted. See the answer below. Thanks everyone!

• Please explain the meaning of "$\to$" when applied to a linear combination of random variables: would this be absolute convergence, convergence in probability, convergence in distribution, or something else? For some of these meanings your claim might be true but for others it is false. – whuber Aug 12 '18 at 13:55
• Surely you must need some condition on the $x_i$ so that the $x_iw_i$ still meet the conditions of the LLN. – BruceET Aug 12 '18 at 17:11
• @BruceET My bad, I also know that the $x_i$s have finite variance. To be precise, they have finite variance and are zero mean. This should settle it, right? – Vanligvodka Aug 12 '18 at 19:47
• That sounds like that ought to do it. Rather than abandoning the question, though, it would be of more benefit to the community if you post an answer. Answering your own question is perfectly ok. – eric_kernfeld Aug 12 '18 at 20:15
• OK, nice. I'll post an answer tomorrow to make sure everything I ok. Thanks everyone. – Vanligvodka Aug 12 '18 at 20:22

Let's define the random variable $a_i=x_iw_i$. We don't know the distribution of $a_i$, but we know that $\mathbb{E}[a_i] = 0$ and $\mathbb{E}[a_i^2] = \rho\sigma^2$. With this, we know that $\mathbb{E}[a_ia_j] = 0$, for $i\neq j$, so the terms in the sum are uncorrelated. Now, we can follow the steps here to show that the sum converges to zero in probability. To make the answer self contained, the steps are as follows:
Let's define $A = \frac{1}{N}\sum\limits_{i=1}^N a_i$. Since the terms in the sum are uncorrelated, the variance of the sum is the sum of the variances, i.e., $\text{Var}(A) = \frac{\rho\sigma^2}{N}$. From Chebyshev's inequality, we have $\mathbb{P}[|A|\geq \epsilon] \leq \frac{\rho\sigma^2}{N\epsilon^2}$. Thus, $\mathbb{P}[|A|< \epsilon] = 1 - \mathbb{P}[|A|\geq \epsilon] \geq 1 - \frac{\rho\sigma^2}{N\epsilon^2}$, which proves that $A\to 0$ in probability, for large N.