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Consider the $\mathcal{N}(\mu_n,\sigma_n^2)$ distributions, where the $\mu_n$ are real numbers and the $\sigma_n^2$ non-negatives.

A sequence of probability measures $(\xi_n)_{n \in \mathbb{N}}$ on $(\mathbb{R},\mathcal{B})$ is called tight, if $\lim_{M \to \infty} \inf_n \xi_n([-M,M])=1$.

Now, I want to show that the family $\left(\mathcal{N}\left(\mu_n,\sigma_n^2\right)\right)_{n \in \mathbb{N}}$ is tight if and only if $(\mu_n)_{n \in \mathbb{N}}$, $(\sigma_n^2)_{n \in \mathbb{N}}$ are bounded.

We consider first the implication $(\implies)$: we assume that \begin{align} \lim_{M \to \infty} \inf_n\left( \Phi\left(\frac{M-\mu_n}{\sigma_n}\right) - \Phi\left(\frac{-M-\mu_n}{\sigma_n}\right)\right) =1, \end{align} where $\lim_{M \to \infty}\Phi\big(\frac{M-\mu_n}{\sigma_n}\big)=1$ and $\lim_{M \to \infty}\Phi\big(\frac{-M-\mu_n}{\sigma_n}\big)=0$. How to conclude from this information that $(\mu_n)_{n \in \mathbb{N}}$, $(\sigma_n^2)_{n \in \mathbb{N}}$ are bounded?

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    $\xi_n ([-M,M])$ is the probability that a random variable $X_n\sim\mathcal N(\mu_n, \sigma_n^2)$ is in the interval $[-M,M]$, i.e: $$ \xi_n ([-M,M]) = \Phi \biggl( \frac{M-\mu_n}{\sigma_n}\biggr) - \Phi \biggl( \frac{-M-\mu_n}{\sigma_n}\biggr), $$ where $\Phi$ is the cdf of a standard normal distributed random variable. – 2017-01-05
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    Thanks! Assuming that $\lim_{M \to \infty} \inf\{\Phi\big(\frac{M-\mu_n}{\sigma_n}\big) - \Phi\big(\frac{-M-\mu_n}{\sigma_n}\big)\} = 1$. Therefore, $\lim_{M \to \infty}\Phi\big(\frac{M-\mu_n}{\sigma_n}\big) = 1$ and $\lim_{M \to \infty}\Phi\big(\frac{-M-\mu_n}{\sigma_n}\big) = 0$. How to conclude that $(\mu_n)_{n \in \mathbb{N}}, (\sigma_n)_{n \in \mathbb{N}}$ are bounded? – 2017-01-05

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