algorithms – A successful search takes $Theta(1 + alpha)$ time on average when resolving collisions by chaining

I would to discuss a proof found in CLRS book please.

Theorem: In a hash table in which collisions are resolved by chaining, a successful search takes time $Theta(1 + alpha)$, on the average, under the assumption of simple uniform hashing.

Proof: Assume we $n$ elements are equally likely to be searched in the table. The number of elements examined during a successful search for an element $x$ is 1 more than the number of elements that appear before $x$ in $x$’s list. New element will be added at the end of the list.

To find the expected number of elements examined, we take the average, over the n elements $x$ in the table, of 1 plus the expected number of elements added to $x$’s list after x was added to the list. Let $x_i$ denote the ith element inserted into the table, for $i = 1, 2, cdots, n$ and let $k_i = key(x_i)$. For keys $k_i$ and $k_j$, we define the indicator random variable $X_{ij} = I {h(k_i) = h(k_j)}$. Under the assumption of simple uniform hashing, we have $Pr{h(k_i) = h(k_j)} = 1/m$, $E (X_{ij}) = 1/m$. Thus, the expected number of elements examined in a successful search is

$$
begin{gather}
Eleft( frac{1}{n} sum_{i=1}^{n} left( 1 + sum_{i=1}^{n} X_{ij} right)right) tag{1} label{1}\
=frac{1}{n} times left(sum_{i=1}^{n} left( 1 + sum_{i=1}^{n} Eleft(X_{ij}right) right)right) &text{(Based on Linearlity of Expectation)}\
vdots \
=Theta(1+alpha)
end{gather}
$$

Problems:

  1. The number of elements examined during a successful search for an element $x$ is 1 more than the number of elements that appear before $x$ in $x$’s list. So, if our table is empty and we have only 1 element, then the time needed would be $O(1)$. If all elements are hashed to first entry of the table, then we would need $O(Elements)$ to find the item as I understand. Why the time needed is $Theta(1+alpha)$ please?
  2. How equation ref{1} was brought in first place? From what I understood, since all elements to be found are equally likely, then we would have $frac{1}{n}$, so probably this is why its their in ref{1}.