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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?

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

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