FIVE- M IN U T E A N A LYST
There are several approaches we
could take to solving this problem.
One approach would be to simulate it,
another would be to consider the
discrete-time, discrete-space Markov
chain and compute the limiting distributions the usual way. We’re going to take
a different approach, which requires a
bit of machinery but ultimately is more
straightforward. We’re going to use
the generating function of the continuous time Markov Chain, which we (and
others!) denote as the G matrix [2].
This matrix differs from the more
commonly used P matrix in discrete
time chains because it is specified in
terms of rates, not probabilities. Have
no fear, knowledge of one fundamental
matrix implies the other. The practical
difference is that the rows of the P matrix sum to 1, while the rows of the G
sum to zero, with the on-diagonal elements being negative. For instance, we
might specify:
Where λ represent the sleeping times
of the children, and μ represents their
wakeful times. To make use of our model,
we need to use P 't = tG , which is readily
solved in matrix form as
P(t ) = etG
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A N A LY T I C S - M A G A Z I N E . O R G
To evaluate this expression, we need
to consider what it means to take e to a
matrix power. This turns out not to be any
harder than taking e to a scalar power –
you just have to use Taylor’s Theorem:
Depending on your programming
environment, efficient algorithms exist
to compute this numerically.
As an example, if we take
λm=.1λn=.125, μm=.2, μn= .2, and recalling that the rate parameters of exponential random variables are the
reciprocals of their expectations, we
see that Mary is a child who sleeps
for 10 hours at a stretch (without interruption from her brother), and Neil is a
child who sleeps an average of eight
hours at a time. Both infants are up
for an average of five hours at a time
between sleep stretches. We can use
these values to populate the G matrix
mentioned earlier.
To find the long-run behavior of the
system, we choose a value of t which
is large enough so that the system is
stable but small enough to avoid problems with numerical stability. We select, somewhat arbitrarily, t = 30 to
make sure that transients are out of
the system. Now, it’s simply a matter
of evaluating the expression. We find
W W W. I N F O R M S . O R G