Convergence in probability towards a constant

Question

Prove that if a sequence $X_1,X_2..$ of random variables satisfies the following conditions: $$1.lim_{n\to\infty}\mathbb{E}(X_n)=a,\space a\in\mathbb{R}$$ $$2.lim_{n\to\infty}\mathbb{V}(X_n)=0$$ Then this sequence converges in probability towards $a$: $$lim_{n\to\infty}\mathbb{P}(|X_n-a|\geq\epsilon) = 0$$

Answer 1

$E(X_n-a)^{2}=EX_n^{2}+a^{2}-2aEX_n =var(X_n)+(EX_n)^{2}+a^{2}-2aEX_n \to 0+a^{2}+a^{2}-2a^{2}=0$. Hence $P\{|X_n-a| \geq \epsilon\} \leq \frac {E(X_n-a)^{2}} {\epsilon^{2}} \to 0$.

Answer 2

Fix $\varepsilon>0$. Note that $$ |X_n-a|\leq |X_n-EX_n|+|EX_n-a| $$ whence $$ P(|X_n-a|\ge\varepsilon)\leq P(|X_n-EX_n|\ge\varepsilon/2)+P(|EX_n-a|\ge\varepsilon/2) \tag{1} $$ The first term on the RHS of (1) goes to zero by the assumption that $\text{Var}(X_n)\to 0$ combined with Chebeshev's inequality. The second term on the RHS of (1) goes to zero by the assumption that $EX_n\to a$ .