Convergence in probability for a sequence of rv´s where the second moment exists

Question

Let $ X_n $ be sequence of rv´s on $ ( \Omega, F, P ) $. We know that $ E[X_n^2] < \infty, \lim_{n\rightarrow \infty} E[X_n] = \mu \in \mathbb{R} $ and $ \lim_{n \rightarrow \infty} \operatorname{Var}(X_n) = 0 $

Show that $ X_n \overset{P} \rightarrow \mu $

My idea: i have to show that $ \lim_{n \rightarrow \infty} P( \vert X_n - X \vert \geq \epsilon ) = 0 $

in my case i can write : $ \lim_{n \rightarrow \infty} P(\vert X_n - \mu \vert \geq \epsilon) \overset{\mathrm{Tschebychev}} \leq \lim_{n \rightarrow \infty} \frac{\operatorname{Var}(X_n)}{\epsilon^2} =0 $ because of $ \lim_{n \rightarrow \infty} \operatorname{Var}(X_n) = 0 $

So i get the convergence in probability and that $ X_n \overset{P} \rightarrow \mu $

Answer 1

Applying Tschebysheff's inequality yields

$$\mathbb{P}( |X_n-\mu| \geq \epsilon) \leq \frac{1}{\epsilon^2} \mathbb{E}((X_n-\mu)^2);\tag{1}$$

note that the expression on the right-hand side does not equal $\frac{1}{\epsilon^2} \text{var}(X_n)$ since, in general, $\mu \neq \mathbb{E}(X_n)$. This means that your reasoning doesn't work but we can argue as follows: Writing

$$X_n-\mu = (X_n-\mathbb{E}(X_n))+\mathbb{E}(X_n-\mu)$$

we get

$$\mathbb{E}((X_n-\mu)^2) =\underbrace{\mathbb{E}((X_n-\mathbb{E}(X_n))^2)}_{\text{var}(X_n)} +2 \mathbb{E}(X_n-\mu) \underbrace{\mathbb{E}(X_n-\mathbb{E}(X_n))}_{=0} + (\mathbb{E}(X_n-\mu))^2.$$

Hence, by $(1),$

$$\mathbb{P}(|X_n-\mu| \geq \epsilon) \leq \frac{\text{var}(X_n)}{\epsilon^2} + \frac{(\mathbb{E}(X_n)-\mu)^2}{\epsilon^2}.$$

As you already pointed out, the first term on the right-hand side converges to $0$ as $n \to \infty$ since $\text{var}(X_n) \to 0$. On the other hand, it follows from $\lim_{n \to \infty} \mathbb{E}(X_n)=\mu$ that the second term on the right-hand side also converges to $0$ as $n \to \infty$. Thus,

$$\lim_{n \to \infty} \mathbb{P}(|X_n-\mu| \geq \epsilon)=0.$$

Answer 2

The idea is good, but the bound you get after an application of Tschebychev's inequality is $\varepsilon^{-2}\mathbb E\left[\left(X_n-\mu\right)^2\right]$, not $\varepsilon^{-2}\operatorname{Var}\left(X_n\right)$. But $$ \mathbb E\left[\left(X_n-\mathbb E\left[X_n\right]+\mathbb E\left[X_n\right]-\mu\right)^2\right]\leqslant 2\operatorname{Var}\left(X_n\right)+2\left(\mathbb E\left[X_n\right]-\mu\right)^2 $$ which is sufficient to conclude.