Markov Chain Probability (Probability)

Often during interviews, probability questions are stated that contain a Markov chain. A Markov chain is a sequence of random variables with the property that given the present state, the future states and the past states are independent. In other words, in case the current state is known, the past states do not affect the future. Think for example about a cube. An ant is starting at a specific corner of the cube and it tries to reach the exact opposite corner of the cube by only walking straight to neighbor corners. It takes an ant a second to walk a side. A problem statement – containing a Markov chain – could be to compute the expected amount of time that the ant is walking before it reaches the opposite corner. The ant is moving randomly to one of either three neighbor corners. Past choices do not affect the next choice of the ant, which makes this a Markov chain problem. This section will elaborate on the Markov chain.

In case the problem contains a finite-state Markov chain with n-states, then the Markov chain can be described by an n \times ntransition matrix P, given by Equation 1.

(1)   \begin{equation*} P = \begin{bmatrix} p_{11} & \cdots  &p_{1n} \\  \vdots  &\ddots   & \vdots \\  p_{n1} & \cdots  & p_{nn}  \end{bmatrix}\end{equation*}

where p_{ij} is the transition probability from state i to state j. Although a transition matrix P already gives an overview of the possible paths, a transition graph makes it even more clear. All the possible and impossible transitions are easy to recognize in a transition graph. 

Figure 1 – Markov chain transition graph.

The corresponding transition matrix would be:

(2)   \begin{equation*}P = \begin{bmatrix} 0.9 & 0.1  & 0  & 0 \\  0.7 & 0  & 0.2  & 0.1 \\  0.1 & 0.2  & 0  & 0.7 \\  0 & 0  & 0  & 1  \end{bmatrix}\end{equation*}

In the transition graph, we can see which state is accessible from which state(s). For example, state 4 is not directly accessible from state 1, because one has to apply first for the function before the applicant gets the offer. When two state are mutually accessible, then those state communicate. For example, states 2 and 3 do communicate, as do 1 and 2. State 4 is accessible from state 3, but does not communicate with state 3. A state i is recurrent if for every state j that is accessible from i, i is also accessible from j. A state is called transient when it is not recurrent. 

In every problem statement, we are looking for a certain answer. This answer should be the end of the Markov chain, also called the absorbing state. It is impossible to leave this state. In the example with the ant, the absorbing state is when the ant reached the exact opposite corner with respect to the corner from where it started. In the transition graph above, the absorbing state is state number 4. In this case it is assumed that one does not continue with looking for another job after the applicant gets a job offer in trading.

The problem statement could either aim for the probability to reach a specific absorbing state or for the expected number of steps or amount of time before reaching the absorption state. The equation to solve this problem can be derived from the law of total expectation. 

Let’s first elaborate on the probability to reach a specific absorption state s_a. All states are denoted by s_1,s_2,...,s_n for n-states. The absorption state s_a=1, and all other states are equal to zero. For every transient state i we have the equation

(3)   \begin{equation*}s_i= \sum_{i=1}^{N}s_j p_{ij}\end{equation*}

The equation to solve the number of steps X before reaching the absorption state is slightly different. Here, the absorption state s_a=0. For every transient state i we have the equation 

(4)   \begin{equation*}s_i= 1+ \sum_{i=1}^{N}s_j p_{ij}\end{equation*}

These equations can be solved by writing out the equation from the starting point. As an example, let’s work out the example of the ant. The ant starts at a specific corner of a cube and tries to reach the exact opposite corner of the cube. It only walks directly to neighbor corners with an equal probability, 1/3. Every time the ant travels to the next corner, a second passes. What is the expected number of seconds that it takes the ant to reach the exact opposite corner?

Solution: First start with numbering all corners to get a clear overview of all states. The ant starts at s_1 and wants to reach s_a.

Figure 2 – Imagination of the cube.

Now we can start writing out all state equations:

(5)   \begin{equation*}s_1=1+s_2 \end{equation*}

(6)   \begin{equation*}s_2 = 1 + \frac{2}{3}s_3 + \frac{1}{3}s_1\end{equation*}

(7)   \begin{equation*}s_3 = 1 + \frac{2}{3}s_2 + \frac{1}{3}s_a\end{equation*}

s_a is equal to zero as we explained before. We can now substitute s_3=1+\frac{2}{3}s_2 in s_2. This gives us:

(8)   \begin{equation*}s_2=1+\frac{2}{3}(1+\frac{2}{3}s_2)+\frac{1}{3}s_1=\frac{5}{3}+\frac{4}{9}s_2 + \frac{1}{3}s_1 \end{equation*}

(9)   \begin{equation*}\frac{5}{9}s_2 = \frac{5}{3} + \frac{1}{3}s_1\end{equation*}

(10)   \begin{equation*}s_2 = 3 + \frac{3}{5}s_1 \end{equation*}

Now we can substitute s_2 in s_1:

(11)   \begin{equation*}s_1=1+(3+\frac{3}{5}s_1)=4+\frac{3}{5}s_1\end{equation*}

(12)   \begin{equation*}\frac{2}{5}s_1 = 4\end{equation*}

(13)   \begin{equation*}s_1 = 10\end{equation*}

In other words, the expected time that it takes the ant to reach the corner in the exact opposite corner is equal to ten seconds. Other problems and solutions can be found in de exercises section.