Show that $\mathbb{P}(|\frac{1}{n}\sum_{i=1}^nX_i-\frac{1}{2}|>\frac{1}{2})\leq\frac{1}{3n}$

Question

Suppose $X_1, X_2, ..., X_n$ are independent uniformly distributed random variables on [0,1].

Show that $\mathbb{P}(|\frac{1}{n}\sum_{i=1}^nX_i-\frac{1}{2}|>\frac{1}{2})\leq\frac{1}{3n}$

I've started rewriting this equation \begin{align} \mathbb{P}(|\frac{1}{n}\sum_{i=1}^nX_i-\frac{1}{2}|>\frac{1}{2}) & = \mathbb{P}(\frac{1}{n}\sum_{i=1}^nX_i-\frac{1}{2}>\frac{1}{2}) + \mathbb{P}(\frac{1}{n}\sum_{i=1}^nX_i-\frac{1}{2}<-\frac{1}{2})\\ &=\mathbb{P}(\frac{1}{n}\sum_{i=1}^nX_i>1) + \mathbb{P}(\frac{1}{n}\sum_{i=1}^nX_i<0) \end{align} How can I go continue from here? Or is this not the way to go?

Answer 1

We can use Chebyshev's inequality on the random variable $\frac{1}{n}\sum_{i=1}^nX_i$, which states the following: $$\mathbb{P}(|Y-\mathbb{E}(Y)|\geq a)\leq\frac{1}{a^2} \text{Var}(Y)$$ Thus $$\mathbb{P}(|\frac{1}{n}\sum_{i=1}^nX_i-\frac{1}{2}|>\frac{1}{2})\leq4 \text{Var}\left(\frac{1}{n}\sum_{i=1}^nX_i\right)=\frac{4}{n^2}n\cdot\frac{1}{12}=\frac{1}{3n}$$