# Probabilistic user behavior markov models on web

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.

• Welcome to our site. This doesn't sound Markov and it doesn't have enough information to determine a model. Perhaps you could formulate a specific question in which a model is more clearly specified?
– whuber
Commented Jan 15, 2019 at 22:47
• @whuber I know that the correct answers should be explicit analytic expressions in the form of rational fractions of the variable $A$.
– Joe
Commented Jan 15, 2019 at 22:53
• Unfortunately, this question doesn't have enough information for a correct answer to be determined, even given such a hint. Indeed, how do you know this would be the form of an answer?
– whuber
Commented Jan 15, 2019 at 22:54
• Is the probability $A$ after every result or does it vary? Is it possible that the user stops searching without finding or do we assume the searching continues as long as needed? It might also be helpful to add references to the aforementioned articles and explain why you couldn't use those results Commented Jan 16, 2019 at 7:07
• @whuber I think this problem is well defined. I was thinking to use geometric distribution it might be useful for this specific problem.
– Joe
Commented Jan 19, 2019 at 10:34

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.

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

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

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