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My statistics book defines the concept of "bounded in probability" in the following way: 
... But doesn't this mean that any sequence of R.V.'s that does not include any R.V.'s with a pdf with infinite support (i.e. $ -\infty < x < \infty $) is bounded in probability as you can always pick a higher $B_{\epsilon}$?
But doesn't this mean that any sequence of R.V.'s that does not include any R.V.'s with a pdf with infinite support (i.e. $−∞<x<∞$) is bounded in probability as you can always pick a higher $B_ϵ$?
The mistake you are making lies in the logic order. For a given $\epsilon$ you have to be able to choose a $B_{\epsilon}$ that works for every $\boldsymbol{X_n}$ for $\boldsymbol{\ n\geq N_{\epsilon}}$ simultaneously. You are thinking the other way around: Once you have your $X_n$ you choose your $B_{\epsilon}$. But that's not what the definition says. So, even if all your random variables have bounded support, since your are forced to first choose $B_{\epsilon}$, an $X_n$ might come along in the sequence such that, while still having bounded support, it's support is not bounded by $B_{\epsilon}$. Note that this alone doesn't necessarily stops the sequence from being bounded in probability, but it shows why your argument doen't work. As shown by Kjetil, this can indeed happen in a way that makes the sequence not bounded in probability.
No, let $X_n$ be distributed with the uniform distribution on $(n, n+1)$ for $n=1,2,3,\dotsc$ Then $\{X_n\}$ is a sequence of bounded random variables which is not bounded in probability.
