So, We are approaching this Model with Linear Algebra. As this is a directed graph, We can just replace this with Adjacent Matrix.

If You don’t what it is, We have already seen its usage in our last discussion with our approach of designing Transition Matrix using Pandas.

So, the Markov Chain above will translate into –

The elements of the matrix just denote the weight of the edge connecting the two corresponding vertices. But, We have already seen the example in our last discussion. So,

- The element of the second row and first column denotes the weight or the Transition Probability from a
`Downside`

state to an`Upside`

state. - If an element is 0, it means there is no edge between two vertices.

This matrix is called as** “Transition Matrix”**. Let’s denote it as `A`

.

Remember that, Our goal is to find the probabilities of each state. So, Conventionally, We take a row vector named `π`

whose element represent the probabilities of the state. Suppose, at the beginning i.e. the first day which is at `t=0`

, We are having a `Downside`

state. So `π`

will look like –

`π_0 = [ 0 1 0 ]`

- The second element which corresponds to the
`Downside`

state becomes`1`

. - And, the other two remains
`0`

.

Now, Let’s see what happens if we multiply our row vector with the transition matrix. So, `π_0.A =`

`Downside`

state.

**The Initial Transition Matrix**

**Future Probabilities corresponding to the Downside state.**

** π_0. **In this case, We get the probabilities of the States of the first day after tomorrow. Then, We keep extending it –

So, `π_1.A =`

Similarly, `π_2.A =`

Now, if there exists a stationary state. After some iterations, the output vector should be the same as the input vector.

Let’s denote this special row vector as ** π**. Now, We should be able to write –

`π.A = π`

If You have ever taken a course of Linear Algebra, You will find this equation is somewhat similar to our famous Eigenvector Equation ** Ax = λx**.

**Consider λ = 1 and just reverse the order of multiplication. You will get out equilibrium state equation!**

Now, coming to context, We can imagine that ** π** is a left eigenvector of the matrix

`A`

with an eigenvalue equal to 1.**Equation 1: π.A = π**

Now, the eigenvalue has to satisfy another condition – The elements of ** π** must add up to

`1`

.**Equation 2: π[1] + π[2] + π[3] = 1**

Because – It denotes a probability distribution. So, after solving the above two equations we will get ** π**. This is the Stationary Distribution calculated from the theories of the Markov Chain with the help of Linear Algebra.

**From π.A = π**

We get three equations from here –

`0.2x + 0.3y + 0.5z = x`

`0.6x = y`

`0.2x + 0.7y + 0.5z = z`

**From: π[1] + π[2] + π[3] = 1**

We get one equation from here –

`x + y + z = 1`

After solving,

π = [25/71, 15/71, 31/71] = [0.35211, 0.21127. 0.43662]

In the last discussion, We calculated the value of π using the Random Walk Theory.

It was –

`π = [0.35191,0.21245,0.43564]`

The value of the eigenvector

`(π)`

calculatd by both methods is almost similar which is what validates Markov Chains even more – `π[0.35211, 0.21127. 0.43662] `

and `[0.35191,0.21245,0.43564]`

.