# 10.4. Examples¶

Here are two examples to illustrate how to find the stationary distribution and how to use it.

## 10.4.1. A Diffusion Model by Ehrenfest¶

Paul Ehrenfest proposed a number of models for the diffusion of gas particles, one of which we will study here.

The model says that there are two containers containing a total of \(N\) particles. At each instant, a container is selected at random and a particle is selected at random independently of the container. Then the selected particle is placed in the selected container; if it was already in that container, it stays there.

Let \(X_n\) be the number of particles in Container 1 at time \(n\). Then \(X_0, X_1, \ldots\) is a Markov chain with transition probabilities given by:

The chain is clearly irreducible. It is aperiodic because \(P(i, i) > 0\).

**Question.** What is the stationary distribution of the chain?

**Answer.** We have computers. So let’s first find the stationary distribution for \(N=100\) particles, and then see if we can identify it for general \(N\).

```
N = 100
states = np.arange(N+1)
def transition_probs(i, j):
if j == i:
return 1/2
elif j == i+1:
return (N-i)/(2*N)
elif j == i-1:
return i/(2*N)
else:
return 0
ehrenfest = MarkovChain.from_transition_function(states, transition_probs)
Plot(ehrenfest.steady_state(), edges=True)
```

That looks suspiciously like the binomial (100, 1/2) distribution. In fact it *is* the binomial (100, 1/2) distribution. Since you’ve guessed it, all you have to do is plug it into the balance equations and check that they work out.

The balance equations are:

You have already guessed the solution by looking at the answer calculated for \(N=100\). But if you want to start from scratch, you’ll have to simplify the balance equations.

To do this, **it’s a great idea to write all the elements of \(\pi\) in terms of one of the elements**.

Try writing all the elements of \(\pi\) in terms of \(\pi(0)\). You will get:

and so on by induction:

In other words, the stationary distribution is proportional to the binomial coefficients. So \(\pi(0) = 1/2^N\) to make all the elements sum to 1, and the distribution is binomial \((N, 1/2)\).

## 10.4.2. Expected Reward¶

Suppose I run the sticky reflecting random walk from the previous section for a long time. As a reminder, here is its stationary distribution.

```
stationary = reflecting_walk.steady_state()
stationary
```

Value | Probability |
---|---|

1 | 0.125 |

2 | 0.25 |

3 | 0.25 |

4 | 0.25 |

5 | 0.125 |

**Question 1.** Suppose that every time the chain is in state 4, I win 4 dollars; every time it’s in state 5, I win 5 dollars; otherwise I win nothing. What is my expected long run average reward?

**Answer 1.** In the long run, the chain is in steady state. So I expect that on 62.5% of the moves I will win nothing; on 25% of the moves I will win 4 dollars; and on 12.5% of the moves I will win 5 dollars. My expected long run average reward per move is 1.65 dollars.

```
0*0.625 + 4*0.25 + 5*.125
```

```
1.625
```

**Question 2.** Suppose that every time the chain is in state \(i\), I toss \(i\) coins and record the number of heads. In the long run, how many heads do I expect to get on average per move?

**Answer 2.** Each time the chain is in state \(i\), I expect to get \(i/2\) heads. When the chain is in steady state, the expected number of coins I toss at any given move is 3. So, by iterated expectations, the long run average number of heads I expect to get is 1.5.

```
stationary.ev()/2
```

```
1.5000000000000013
```

If that seems artificial, consider this: Suppose I play the game above, and on every move I tell you the number of heads that I get *but I don’t tell you which state the chain is in.* I *hide* the underlying Markov Chain. If you try to recreate the sequence of steps that the Markov Chain took, you are working with a Hidden Markov Model. These are much used in pattern recognition, bioinformatics, and other fields.