Probabilistic user behavior markov models on web

Question

I am considering the following probabilistic Markov model of actions of a user on the results page of a search engine.

The user examines the first result, with a probability $A$ he is satisfied with it (finds what he was looking for) and stops searching; otherwise he considers the second result, with probability $A$ he is satisfied with it and stops searching; otherwise he looks through the third result, and so on. For the sake of simplicity, we assume that the output contains an infinite number of results, so theoretically this random process can continue as long as needed, or indefinitely.

I want to formulate the mathematical expectation and the variance of the number of user-studied results?

My expectation of how to solve this problem: The probability $A$ doesn't vary and the user continues searching as long as needed. This means that we have:

• Independent trails,
• Identical trails,
• The outcome of a trail must be a success or a fail.

Thanks in advance.

Answer 1

For a geometric distribution with probability $A$ of success, the probability that exactly $k$ failures occur before the first success is: $A(1-A)^{k-1} \hspace{10pt} \forall k \geq 1$. This is written as a probability mass function $P(X=k)$.

• The mean or expected value of distribution gives useful information about what average one would expect from a large number of repeated trials.

  $\mathbb{E} (X) = \sum_{k\geq1}kP(X=k) = \sum_{k\geq1}kA(1-A)^{k-1} = A\sum_{k\geq1}k(1-A)^{k-1} = \frac{A}{(1-A)^2} = \frac{1}{A}$

• The variance of a distribution measures how "spread out" the data is.

  $\mathbb{V}(X) = \mathbb{E}(X^2) - \mathbb{E}^2(X)$

$\mathbb{E}(X^2) = \sum_{k\geq1} k^2P(X=k) = \sum_{k\geq1} k^2A(1-A)^{k-1} = \sum_{k\geq1} k(k+1)A(1-A)^{k-1} - \sum_{k\geq1} kA(1-A)^{k-1} = \frac{2A}{(1-A)^3)} - \frac{A}{(1-A)^2} = \frac{2}{A^2} - \frac{1}{A}$

$\mathbb{V}(X) = \frac{2}{A^2} - \frac{1}{A} - \frac{1}{A^2} = \frac{1-A}{A^2}$