Riemann surface

matlab

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In the Dark Ages, Harvard, Dartmouth, and Yale admitted only male students. Assume that, at that time, 80 percent of the sons of Harvard men went to Harvard and the rest went to Yale, 40 percent of the sons of Yale men went to Yale, and the rest split evenly between Harvard and Dartmouth; and of the sons of Dartmouth men, 70 percent went to Dartmouth, 20 percent to Harvard, and 10 percent to Yale. (i) Find the probability that the grandson of a man from Harvard went to Harvard. (ii) Modify the above by assuming that the son of a Harvard man always went to Harvard. Again, find the probability that the grandson of a man from Harvard went to Harvard..

Solution. We first form a Markov chain with state space $S={H, D, Y}$ and the following transition probability matrix :

$$

\mathrm{P}=\left(\begin{array}{ccc}

.8 & 0 & .2 \

.2 & .7 & .1 \

.3 & .3 & .4

\end{array}\right)

$$

Note that the columns and rows are ordered: first $H$, then $D$, then $Y$. Recall: the $i j^{\text {th }}$ entry of the matrix $\mathrm{P}^n$ gives the probability that the Markov chain starting in state $i$ will be in state $j$ after $n$ steps. Thus, the probability that the grandson of a man from Harvard went to Harvard is the upper-left element of the matrix

$$

\mathrm{P}^2=\left(\begin{array}{ccc}

.7 & .06 & .24 \

.33 & .52 & .15 \

.42 & .33 & .25

\end{array}\right)

$$

It is equal to $.7=.8^2+.2 \times .3$ and, of course, one does not need to calculate all elements of $\mathrm{P}^2$ to answer this question.

If all sons of men from Harvard went to Harvard, this would give the following matrix for the new Markov chain with the same set of states:

$$

P=\left(\begin{array}{ccc}

1 & 0 & 0 \

.2 & .7 & .1 \

.3 & .3 & .4

\end{array}\right)

$$

The upper-left element of $\mathrm{P}^2$ is 1 , which is not surprising, because the offspring of Harvard men enter this very institution only.

A certain calculating machine uses only the digits 0 and 1 . It is supposed to transmit one of these digits through several stages. However, at every stage, there is a probability p that the digit that enters this stage will be changed when it leaves and a probability $q=1-p$ that it won’t. Form a Markov chain to represent the process of transmission by taking as states the digits 0 and 1 . What is the matrix of transition probabilities?

Now draw a tree and assign probabilities assuming that the process begins in state 0 and moves through two stages of transmission. What is the probability that the machine, after two stages, produces the digit 0 (i.e., the correct digit)?

Solution. Taking as states the digits 0 and 1 we identify the following Markov chain (by specifying states and transition probabilities):

$$

\begin{array}{lll}

0 & 1 \

1 & q & p \

1 & p & q

\end{array}

$$

where $p+q=1$. Thus, the transition matrix is as follows:

$$

\mathrm{P}=\left(\begin{array}{ll}

q & p \

p & q

\end{array}\right)=\left(\begin{array}{cc}

1-p & p \

p & 1-p

\end{array}\right)=\left(\begin{array}{cc}

q & 1-q \

1-q & q

\end{array}\right) .

$$

It is clear that the probability that that the machine will produce 0 if it starts with 0 is $p^2+q^2$.

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