One of the positions is an unpleasantly hot stove and another is a fly-paper.
Which are they ?
Ex. 3: If the protocol and matrix of Ex. 9/4/1 are regarded as codings of each
Other, which is the direction of coding that loses information?
Ex. 9/4/1 shows how the behaviour of a system specifies its
Matrix. Conversely, the matrix will yield information about the
Tendencies of the system, though not the particular details. Thus
Suppose a scientist, not the original observer, saw the insect’s
Matrix of transition probabilities:
↓
B
W
P
B 1/4 3/4 1/8
W 3/4 0 3/4
P 0 1/4 1/8
166
Equilibrium in a Markov chain. Suppose now that large num-
Bers of such insects live in the same pond, and that each behaves
Independently of the others. As we draw back from the pond the
Individual insects will gradually disappear from view, and all we
Will see are three grey clouds, three populations, one on the bank,
One in the water, and one under the pebbles. These three popula-
Tions now become three quantities that can change with time. If
They are dB, dW, and dP respectively at some moment, then their
values at one interval later, dB' etc., can be found by considering
What their constituent individuals will do. Thus, of the insects in
The water, three-quarters will change over to B, and will add their
Number on to dB, while a quarter will add their number to dP.
Thus, after the change the new population on the bank, dB', will be
1/4 dB + 3/4 dW + 1/8 dP . In general therefore the three populations
Will change in accordance with the transformation (on the vector
With three components)
dB' = 1/4 dB + 3/4 dW + 1/8 dP
dw' = 3/4 dB+ 3/4dP
dP' =1/4 dW + 1/8dP
It must be noticed, as fundamentally important, that the system
167
A N I N T R O D UC T I O N T O C Y B E R NE T I C S
I N CESSA N T TR AN SMI SSIO N
Composed of three populations (if large enough to be free from
Sampling irregularities) is determinate, although the individual
Insects behave only with certain probabilities.
To follow the process in detail let us suppose that we start an
Experiment by forcing 100 of them under the pebbles and then
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Watching what happens. The initial vector of the three populations
(dB , dW, dP) will thus be (0, 0, 100). What the numbers will be at
The next step will be subject to the vagaries of random sampling;
For it is not impossible that each of the hundred might stay under
The pebbles. On the average, however (i.e. the average if the whole
Were tested over and over again) only about 12.5 would
Remain there, the remainder going to the bank (12.5 also) and to
The water (75). Thus, after the first step the population will have
shown the change (0, 0, 100) → (12.5, 75, 12.5).
In this way the average numbers in the three populations may
Be found, step by step, using the process of S.3/6. The next state
Is thus found to be (60.9, 18.8, 20.3), and the trajectory of this sys-
Tem (of three degrees of freedom— not a hundred ) is shown in
Fig. 9/6/1.
The equilibrial values of a Markov chain are readily computed.
At equilibrium the values are unchanging, so dB', say, is equal to
DB. So the first line of the equation becomes
dB = 1/4 dB + 3/4 dW + 1/8 dP
i.e.0 = – 3/4 dB + 3/4 dW + 1/8 dP
The other lines are treated similarly. The lines are not all inde-
Pendent, however, for the three populations must, in this example,
Sum to 100; one line (any one) is therefore struck out and replaced
By
dB + dW + dP = 100
The equations then become, e.g.,
– 3/4 dB + 3/4 dW + 1/8 dP = 0
dW +dP = 100dB +
1/4 dW – 7/8 dP = 0
Which can be solved in the usual way. In this example the equi-
Librial values are (44 9, 42 9, 12 2); as S.9/S predicted, any indi-
Vidual insect does not spend much time under the pebbles.
Ex. 1: Find the populations that would follow the initial state of putting all the
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Insects on the bank.
Ex. 2: Verify the equilibrial values.
Ex. 3: A six-sided die was heavily biased by a weight hidden in face x. When
Placed in a box with face f upwards and given a thorough shaking, the prob-
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