Examples of uniformly bounded random variables In my textbook there is the following definition:
A sequence of random variables $\{X_n\}$ is uniformly bounded if there exists $K>0$ such that $P(|X_n|\ge K)=0$ for all $n$.
There are no examples given, so am I right saying that a sequence of normal random variables, $X_n\sim N(\mu_n, \sigma_n)$, will not be uniformly bounded, but a sequence of Bernoulli random variables, $X_n\sim B(p_n)$, or Binomial random variables, $X_m\sim Bin(n_m, p_m)$ will be if the parameters of the distributions are fixed? 
 A: You are correct, a normal random variable has density $e^{-(x-\mu)^2/2\sigma^2}$ which does not have compact support, hence $P(X\le x)>0$ for all $x\in\Bbb R$. On the other hand, if you have a sequence of Bernoulli random variables, parameter fixed or not does not matter, you know they take values in the set $\{0,1\}$, so let $K=2$ no matter what $p$ is $P(X<-2)=P(X>2)=0$. Binomial random variables are sums of Bernoulli random variables, so if you have $n$ bounded, then the answer is yes as it will be supported on $[-n,n]$ so $P(X<-n-1)=P(X>n+1)=0,\;$ independent of $p$. The converse is also true.
In the case of any discrete random variable on $\Bbb N$ you can easily see that it is bounded as an individual random variable iff its generating function is a polynomial, so a sequence of discrete, random variables is uniformly bounded iff the degrees of their generating functions are finite and uniformly bounded.
This all comes from the more general fact that a random variable (resp. sequence of random variables) is bounded (resp. uniformly bounded) iff it has compact support (resp. are all supported on a common compact set). In the case it is discrete, you use that discrete + compact implies finite to get the second paragraph I mention.
