DBEs and stationary distributions

The most recent Applied Probability assignment sheet featured various aspects of Detailed Balance Equations for continuous-time Markov chains. We discussed the advantages and disadvantages of using DBEs rather than solving for an equilibrium distribution directly. The equations used in this second case are often called Full Balance Equations.

Briefly, the advantages of DBEs are that they are easy to solve. After all, each one contains only two components of the equilibrium distribution, so generally you can solve one-at-a-time. The disadvantage is that an equilibrium distribution might not satisfy the DBEs. The deductive structure is:

\text{Solves DBEs}\quad \stackrel{\Rightarrow}{\not\Leftarrow}\quad\text{Equilibrium distribution}

Usually, the chain will be irreducible, so the equilibrium distribution is unique. This means that if we can solve the DBEs, the result is the unique equilibrium distribution.

The DBEs are soluble only if the situation is reversible. This is probably the best definition to use in practice, but informally we can say that this means that the behaviour looks qualitatively the same if we reverse time. For example, as in Q1:

Q=\begin{pmatrix}-1 &1&0\\ 0& -1&1\\1&0&-1\end{pmatrix},

gives the Q-matrix which equilibrium distribution (\frac13,\frac13,\frac13), which does not satisfy DBEs. The chain is not reversible because sample paths always go clockwise, so if we reversed time they would go anti-clockwise (or vice-versa depending on how you’ve drawn the diagram).

What I wanted to say in the class, and made a mess of explaining was this, about why it was inappropriate to use DBEs to find stationary distributions in Q3d):

Reversibility is not just a function of the chain. It is a function of the chain AND the initial distribution. This is only in practice a concern when the chain is reducible, but in this case it really can lead you astray. Let’s consider an example, like

Q=\begin{pmatrix}-3&2&0&0&1&0\\ 0&-4&3&1&0&0\\ 0&1&-4&3&0&0\\ 0&3&1&-4&0&0\\ 0&0&0&0&-5&5\\ 0&0&0&0&5&-5\end{pmatrix}.

Then by solving as in the problem sheet, the invariant distributions are given by:

\lambda(0,\frac13,\frac13,\frac13,0,0)+\mu(0,0,0,0,\frac12\frac12),\quad \lambda+\mu=1.

If you attempted to solve the DBEs, you would succeed, but the only solution would be

(0,0,0,0,\frac12,\frac12).

The explanation is fairly simple in the end. Reversibility is a class property, and only one of the communicating classes, \{5,6\} in this example admits a reversible initial distribution, so to solve the DBEs we must assign zero mass on the other class.

Anyway, I hope that clears up any residual confusion from the class.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s