$(X_k)_{k\in\mathbb{N}}$ uncorrelated $|X_k|<52\ \forall k\in\mathbb{N}$. Show $\frac{1}{n}\sum_{k=0}^{n}X_k-\mathbb{E}[X_k]\xrightarrow{\mathbb{P}}0$

Not a homework question but an exercise from an past exam.

Let $$(X_k)_{k\in\mathbb{N}}$$ be uncorrelated real valued random variables on a probability space $$(\Omega, \mathcal{F}, \mathbb{P}$$ and fulfilling $$|X_k|<52$$ for all $$k \in \mathbb{N}$$. Show that $$\frac{1}{n}\sum_{k=0}^{n}X_k-\mathbb{E}[X_k]\xrightarrow[n \to \infty]{\mathbb{P}}0$$ (convergence in probability).

I attempted to use the following theorem from the lecture:

Theorem (Generalisation of the Weak Law of Large Numbers) Let $$(X_k)_{k = 1}^{n}$$ be pairwisely uncorrelated with finite variance. Then $$\frac{1}{n^2} \sum_{k = 1}^{n} \text{Var}[X_k] \xrightarrow{n \to \infty} 0$$ implies $$\frac{1}{n} \sum_{k = 1}^{n} \left( X_k - \mathbb{E}[X_k] \right) \xrightarrow[n \to \infty]{\mathbb{P}} 0.$$

My progress By definition we have $$\frac{1}{n^2} \sum_{k = 1}^{n} \text{Var}[X_k] \overset{\textrm{Def.}}{=} \frac{1}{n^2} \sum_{k = 1}^{n} \mathbb{E}[(X_k - \mathbb{E}[X_k])^2] \le \frac{1}{n^2} \sum_{k = 1}^{n} 2 \cdot 52 \le \frac{1}{n^2} 2n \cdot 52 = \frac{104}{n} \xrightarrow{n \to \infty} 0.$$ Is this correct?

• $|X_k-\mathbb E[X_k]|<2\times 52=104$ so $\mathbb E[(X_k-\mathbb E[X_k])^2] <104^2$. You just forgot about the square.
– Feng
Jul 14 '19 at 3:24

As pointed out by Feng Shao, the only minor mistake is that you forgot the squares in bounding $$\mathbb E[(X_k-\mathbb E[X_k])^2]$$.
In order to save some efforts, we could also assume without loss of generality that $$\left\lvert X_k\right\rvert<1$$ (work with $$X'_i=X_i/52$$).