Below question is from the book 'Probability course.com'. The book provides a solution using Chebyshev's inequality. Before reading that solution, I used Markov's inequality. Is my solution correct?
Question: Let $X$ be a random variable, and $X_n = X + Y_n$, where $E[Y_n] = \frac{1}{n}, V[Y_n] = \frac{\sigma^2}{n}$, where $\sigma > 0$ is a constant. Show that $X_n$ converges in probability to $X$.
Answer: By defn of convergence in probability, I must show that $\lim_{n \rightarrow \infty} P(|X_n - X| \ge \epsilon) = 0, \forall \epsilon > 0$.
$P(|X_n - X| \ge \epsilon) = P(|X + Y_n - X| \ge \epsilon) = P(|Y_n | \ge \epsilon) = P(Y_n \le -\epsilon) + P(Y_n \ge \epsilon) $.
My strategy was to use Markov's inequality on each of the summands of the last equality.
$ P(Y_n \le -\epsilon) = P(- Y_n \ge \epsilon) \le \frac{E[-Y_n]}{\epsilon} = \frac{-E[Y_n]}{\epsilon} = \frac{-1/n}{\epsilon} = \frac{-1}{n\epsilon} \rightarrow 0 \text{ as } n \rightarrow \infty.$
$ P(Y_n \ge \epsilon) \le \frac{E[Y_n]}{\epsilon} = \frac{1/n}{\epsilon} = \frac{1}{n\epsilon} \rightarrow 0 \text{ as } n \rightarrow \infty.$
Since both summands go to 0, we get the final result that $P(Y_n \le -\epsilon) + P(Y_n \ge \epsilon) \rightarrow 0 \text{ as } n \rightarrow 0$ proving that $X_n$ converges in prob to $X$ as required.
Is this solution correct? If not, where are the error please?
To use Markov's inequality, you need to have a positive random variable. As far as we know, $Y_n$ can take positive and negative values. Even if it is a.s positive (or a.s negative), you can just use one time Markov's inequality so the two inequalities
cannot be simultaneously true in general.