Help with showing that $\lim_{n \to \infty} E(|\frac{\sum_i^{n-1} I_i}{n}-\frac{1}{3}|)$ = 0

Question

Say we have an indicator function with probability of being realized equal to $1/3.$ I want to show that:

$$\lim_{n \to \infty} E\left(\left|\frac{\sum_{i=1}^{n-1}{I_i}}{n}-1/3\right|\right)=0$$

How could I do this? This is one step in a multi part proof. I would appreciate any ideas.

Answer 1

I suppose your $I_n$'s form an ii.d sequence of indicators with assgined mean. In that case a.s. convergence follows by SLLN's and $L^{1}$ convergence follows by DCT since $0 \leq \frac 1 n \sum\limits_{i=1}^{n} I_i\leq 1$.

Answer 2

I am assuming that the $X_k$ are independent.

Let $X_k=I_k -{1 \over 3}$, we have $E X_k = 0$, $E X_k^2 = {6 \over 27}$.

Then Cauchy Schwartz gives $E [ | {1 \over n} \sum_{k=1}^n X_k | ]^2 \le E[1^2] E[ ({1 \over n} \sum_{k=1}^n X_k )^2] = {1 \over n^2}\sum_k E[X_k^2] = {1 \over n} {6 \over 27}$.