# Based on accumulated data concerning the heights of male children relative to their fathers, it has been determined that the probabilities that a tall man will have a tall medium-height or short...

Based on accumulated data concerning the heights of male children relative to their fathers, it has been determined that the probabilities that a tall man will have a tall medium-height or short child are 0.6, 0.2 and 0.2 respectively. The probabilities that a man of medium height will have a tall, medium-height or short child are 0.1, 0.7 and 0.2 respectively, and the probabilities that a short man will have a tall, medium-height or short child are 0.2, 0.4 and 0.4 respectively.

(a) Write down the transition matrix for this Markov chain.

(b) What is the probability that a short man will have a tall grandson?

(c) If 20% of the current male population is tall, 50% is of medium height and 30% is short, what will the distribution be in three generations?

(d) According to this model, what will the distribution be in the long run?

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### 2 Answers

(a) The transition matrix is the matrix whose entries `a_(i,j) ` are the probabilities that an object in state i moves to state j. The transition matrix is:

`T=[[.6,.2,.2],[.1,.7,.2],[.2,.4,.4]] ` (Note that the .7 in entry 2,2 is the probability that a medium height man will have a medium height son.) Here the first row is for tall men and the column entries are tall, medium, then short.

(b) To find the probability that a short man will have a tall grandson we look at the square of the transition matrix. (Multiplying a state vector by the transition matrix once moves through one time step -- multiplying by the square is multiplying twice and moves through 2 time steps.)

`T^2=[[.42,.34,.24],[.17,.59,.24],[.24,.48,.28]] `

Thus the probability is .24

(c) To find the distribution in three generations, we take the initial state vector times the cube of the transition matrix.

`[[.2,.5,.3]]*T^3~~[[.2457,.5039,.2504]] `

(d) The distribution in the long run is the steady state vector. Since all entries in T are nonzero there is a unique steady state vector.

We could solve the following matrix equation:

`[[x,y,z]]*T=[[x,y,z]] ` where `[[x,y,z]] ` is the steady state vector. This is equivalent to solving the following linear system:

.6x+.1y+.2z=x

.2x+.7y+.4z=y

.2x+.4y+.4z=z

Or we could evaluate `T^n ` for "large" values of n. In either case we find the steady state vector to be `[[.25,.5,.25]] `

Thus the distribution in the long run would be 25% tall, 50% medium and 25% short.

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Is there any way you could explain d) in more detail? I understand that the long-range distribution is the steady state vector but I'm not sure how you concluded that T^n was equal to [.25 .5 .25]

The algebraic way is to solve the system.

If you evaluate T^n for "large" values of n, you will see that the result gets closer and closer to [.25 .5 .25]. Try T^25 and T^30 to see what happens. You can do this with any transition matrix -- occasionally you will have to use fairly large values for n (say in the hundreds or even larger) but you will see that the entries eventually stay the same.