Question:
Given two sequences of random variables {$X;n=1,2,...$} and {$Y;n=1,2,...$} and a random variable $X$,suppose that with probability one $|X_n-X| \le Y_n$ all $n$ and that $E[Y_n] \to 0$ as $n \to \infty$. Prove that $E[X_n] \to E[X]$ and that $X_n$ converges to $X$ in probability as $n \to \infty$
Hint:
$E[X_n] \to E[X]$ means $|E[X_n]-E[X]| \to 0$,try to prove $Pr(|E[X_n]-E[X]| \gt 0)$ with Chebyshve’s inequality.
Solution:$|E[X_n]-E[X]| \le E[|X_n-X|] \le E[Y_n] =0$,that means,$E[X_n] \to E[X]$
By Chebyshve’s inequality $Pr(|E[X_n]-E[X]| \gt \epsilon) \le \frac{E[|X_n-X|^2]}{\epsilon} \to 0$
$\mathbf {Here}$ $\mathbf {are}$ $\mathbf {my}$ $\mathbf {are}$ $\mathbf {question}$ $\mathbf {about}$ $\mathbf {the}$ $\mathbf {hint}$ $\mathbf {and}$ $\mathbf {solution}$
1.The question wants me to prove $E[X_n] \to E[X]$ and that $X_n$ converges to $X$ in probability as $n \to \infty$,but i can't understand why do i have to prove $Pr(|E[X_n]-E[X]| \gt 0)$ with Chebyshve’s inequality?Do they have some relation?
2.I can't understand why $|E[X_n]-E[X]| \le E[|X_n-X|]$,which inequality does the author use?
3.According to the information about Chebyshve’s inequality in wiki https://en.wikipedia.org/wiki/Chebyshev%27s_inequality ,why is it $\le \frac{E[|X_n-X|^2]}{\epsilon}$,but not $\le \frac{E[|X_n-X|^2]}{\epsilon ^2}$
If the problem/"solution" it is from a book, I don't think it's a good book.
The convergence in probability is $\operatorname{Pr}(|X_n-X|>\epsilon)\to 0$
$\operatorname{Pr}(|E[X_n]-E[X]| \gt 0)$ does not make any sense.