Proving That A Sequence Does Not Obey The Weak Law Of Large Numbers

Question

I have a sequence $f_n$ of random variables such that $\mu( \{ f_n = n \}) = \mu( \{f_n = -n \}) = \frac{1}{2}$, and need to prove that it doesn't satisfy WLLN.

Clearly, $E(f_n) = 0$, so I have to disprove that $$ \frac{1}{n} \sum_{j=1}^n f_j (\omega) \to 0$$ in measure. In other words, $$\lim_{n \to \infty} \mu \left( \omega: | \frac{1}{n} \sum_{j=1}^n f_j (\omega)| \ge \delta \right) \ne 0$$ $\forall \delta > 0.$

In my textbook, I haven't really learned much about how to do these types of questions. As a hint, I was told to show that if $\frac{1}{n} \sum_{j=1}^n f_j \to 0$ in measure $\mu$, then also $\frac{1}{n} \sum_{j=1}^{n-1} f_j \to 0$ in measure $\mu$.

However, I dont know how to show this and even if I were able to, I don't see why that means $f_n$ doesn't obey the Weak Law of Large Numbers.

I would gladly appreciate any help.

Answer 1

Take the difference, i.e., $$ \frac{1}{n}\sum_{i=1}^n f_i-\frac{1}{n}\sum_{i=1}^{n-1} f_i=\frac{f_n}{n}. $$ The LHS converges to $0$ (in $\mu$-measure). Thus, $f_n/n$ must converge to $0$ as well. However, $$ \mu(|f_n/n|>1/2)=1 $$ for each $n$.

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The observation in the hint follows from the fact that if $\sum_{i=1}^n f_i/n\to 0$, then $$ \frac{1}{n}\sum_{i=1}^{n-1}f_i=\frac{n-1}{n}\times \frac{1}{n-1}\sum_{i=1}^{n-1}f_i\to 0 $$ in $\mu$-measure.

Answer 2

If $$\frac{1}{n} \sum_{j=1}^n f_j \to 0 \quad (1) $$ in measure then (changing indices $n \mapsto n-1$) also $$\frac{1}{n-1} \sum_{j=1}^{n-1} f_j \to 0 $$ in measure. Multiply by $ (n-1)/n$ to get $$\frac{1}{n} \sum_{j=1}^{n-1} f_j \to 0 \,$$ in measure. Subtract this from (1) and obtain $$\frac{1}{n} f_n \to 0 $$ in measure, which contradicts the given law of $f_n$.