Can PDA have empty stack transition?
1- The PDAs in the video seem to have a transition function of the form $\delta: Q\times (\Sigma\cup\{\epsilon\}) \times (\Gamma \cup \{ \epsilon\}) \to 2^{Q\times (\Gamma \cup \{\epsilon\})}$, so you are allowed to take a nondeterministic transition without reading any input letter and without popping a letter from the stack. Note however, that you are allowed to push a single letter (or none) to the stack in a transition that you take.
2- The PDAs from Wikipedia do not allow you to take a transition without popping a letter from the stack (note that as Yuval said:  the stack is initialized with a &quot;bottom of stack&quot; symbol), however note that, unlike the PDAs in item 1, you can push an entire word in the stack in a transition that you take.
So you have two variants of PDAs that have slightly different syntax, however it is not hard to prove that a PDA from one variant can simulate a run of a PDA from the other variant, and thus both variants recognise the same class of languages, CFL. A high-level explanation of how both variants can simulate each other is given next. A PDA $A$ of the 1st variant simulates a run of a PDA B of the 2nd variant by first pushing a &quot;buttom of stack letter&quot;, then $A$ mimics the transitions that $B$ takes: in the case where $B$ takes a transition that pushes a word of length $\geq 2$ into the stack, you can introduce new states that push that word, and then move to the state that $B$ moves to. Conversely, a PDA $B$ of the 2nd variant simulates a run of a PDA $A$ of the 1st variant by first popping the &quot;buttom of stack letter&quot;, then $B$ mimics the transitions that $A$ takes: the non-trivial part here is to mimic popping $\epsilon$ from the stack. This is easy as $B$ is allowed to push an entire word into the stack: popping $\epsilon$, and then pushing a word $w$, is equivalent to popping a letter $\sigma$, and then pushing the word $\sigma\cdot w$.