I have a question concerning this task:
Let $X$ be a random variable and $X_n=X+Y_n$ where $$E[Y_n]=\frac{1}{n}\quad\text{ and }\quad\operatorname{Var}(Y_n)=\frac{\sigma^2}{n}\quad \text{where }\sigma>0$$ Show that $X_n\xrightarrow{P}X$
Now I know that I can prove this by using the Chebyshev inequality, but can I also do it with the Markov inequality, meaning is this correct:
$$\lim\limits_{n\to \infty}P(|X_n-X|\geq \epsilon)=\lim\limits_{n\to \infty}P(|X+Y_n-X|\geq \epsilon)$$ $$=\lim\limits_{n\to \infty}P(|Y_n|\geq \epsilon)\leq \lim\limits_{n\to \infty}\frac{E[|Y_n|]}{\epsilon}= \lim\limits_{n\to \infty} \frac{1}{n\epsilon}=0$$
Here I came across $X-X$ and since both are random variables, I don't know if I can just cancel them? Is this correct?