I'm taking a course in asymptotic statistics, and I'm having trouble with how to start proving the following statement. I learn best from examples, rather than just reading the theorems, however I can't find an example that seems similar. Any and all help is appreciated!
$$ \left. \begin{matrix} k_n \left( Y_n -c \right) \xrightarrow{D} H \\ k_n \to \infty \end{matrix} \right\} \implies Y_n \xrightarrow{P} c $$ There is the additional hint: You may assume without loss of generality that $H$ is the cumulative distribution function of a random variable $V$ which is tight meaning that, given any $\epsilon > 0$, there exists a finite number $M$ such that $P(|V | \leq M) \geq 1 − \epsilon/2$.
When I look at this, I instinctively think of the Central Limit Theorem with $k_n = \sqrt{n}$, $Y_n = \bar{X}_n$, $c=\mu$ and $H$ the CDF some normally distributed r.v. $V$. I understand the concept of tightness which implies that none of the PMF escapes to infinity, so that makes me think I'm incorrect in assuming $V$ is normal.
Let $\epsilon >0$. Let $M$ be as given in the hint. Without loss of generality we may assume that $H$ is continuous at $\pm M$. If $k_n >\frac M {\epsilon}$ then $P\{|Y_n-c| >\epsilon\} =P\{|k_n(Y_n-c)| >k_n \epsilon\}\leq P\{|k_n(Y_n-c)| >M\} \to P\{|V| >M\} <\epsilon /2$.