Necessary and sufficient conditions for tightness of a sequence of Normal variables. Let $\mu_n$ be the Normal$(m_n,\sigma_n^2)$ probability distribution, $n=1,2,\dots$. Prove that the sequence $\{\mu_n\}_n$ is tight if and only if the number sequences $m_n$ and $\sigma_n^2$ are both bounded.
Intuitively, when I think about the shape of a normal distribution curve, this makes sense. However I am really stuck when it comes to trying to actually prove this. I will demonstrate where I get stuck by showing you my workings so far in trying to prove that $m_n$ being unbounded implies that $\mu_n$ is not tight:
First, assume that $m_n$ is not bounded, and that $\sigma_n$ is bounded. Let the compact set $K_\varepsilon$ be $[-C_\varepsilon,C_\varepsilon]$. Then we want to show that for every $K_\varepsilon$, we can choose an $n$ such that $\mathbb{P}(\mu_n\in K_\varepsilon)\geq1-\varepsilon$. This is equivalent to showing that we can choose $n$ such that
$$
\Phi\left(\frac{C_\varepsilon-m_n}{\sigma_n}\right)-\Phi\left(\frac{-C_\varepsilon-m_n}{\sigma_n}\right)\geq1-\varepsilon,
$$
or equivalently that we can choose $n$ such that
$$
\Phi\left(\frac{C_\varepsilon-m_n}{\sigma_n}\right)\geq\frac{2-\varepsilon}2,
$$
where $\Phi$ is the standard normal distribution function. The issue here is that I have no idea how to prove this, since $\Phi$ is not a simple function that I can manipulate and do things to. I am wondering if the approach I am taking is wrong, and I need to use some other theorems to answer this. Am I on the right track at all? Any help would be appreciated!
 A: Firstly, let's prove that tighness implies boundedness. Start with assuming $m_n$ is not bounded, but that $\sigma_n$ is bounded. Then we want to show that for all $r$, there exists $m_n$ such that
$$
\mathbb{P}(|\mu_n|>r)\geq\frac12.
$$
Since $m_n$ is unbounded, there exists an $n$ such that $m_n>r$. Then since $m_n$ is the median, we would have that $\mathbb{P}(|\mu_n|>r)\geq\mathbb{P}(\mu_n>r)\geq\frac12$.
Now assume that $m_n$ is bounded, but that $\sigma_n$ is unbounded. Then we have that the density of the normal distribution is bounded above by $\frac1{\sigma_n\sqrt{2\pi}}$. So since $\sigma_n$ is unbounded, there exists $n$ such that  $\frac{2r}{\sigma_n\sqrt{2\pi}}\leq\frac12$. Then we would have that
$$
\begin{align*}
&\mathbb{P}(|\mu_n|<r)\leq\frac{2r}{\sigma\sqrt{2\pi}}\leq\frac12\\
\implies&\mathbb{P}(|\mu_n|>r)\geq\frac12.
\end{align*}
$$
So I have now proven the forwards direction.
Now let's prove that boundedness implies tightness. Without loss of generality, assume that $m_n\geq0$ for all $n$. Since $m_n$ and $\sigma_n$ are bounded, we can say that for some $M,D$ are the bounds. That is, $|m_n|\leq M, |\sigma_n|\leq D$. Let $f_{m,\sigma^2}(x)$ be the density of the Normal$(m,\sigma^2)$. Then for every $\varepsilon$, we choose $r_\varepsilon$ such that
$$
\frac{D^2}{(r_\varepsilon-M)}\leq\varepsilon.
$$
Now we use Chebyshev's inequality. Note that since $m_n\geq0$, we can say that
$$
\{|\mu_n|>r_\varepsilon\}\implies\{|\mu_n-M|>r_\varepsilon-M\}.
$$
So we have that
$$
\mathbb{P}(|\mu_n|>r_\varepsilon)\leq\mathbb{P}(|\mu_n-M|>r_\varepsilon-M)\leq\frac{D^2}{r_\varepsilon-M}\leq\varepsilon.
$$
And we are done. $\square$
I think that this solution works, but I am not 100% confident in all my ideas and workings. I would also appreciate feedback on the clarity of the solution, so any comments and questions or tips would be appreciated still.
