Wright Fisher Chain

Consider an urn that contains d amount of marbles. Some of these marbles are blue and the others are red. Let X0 be the number of blue marbles initially in the urn.

At each step in the Markov chain, perform the following procedure:

First, double the number of marbles of each color that are in the urn (You will know have a total of 2d marbles in the urn).

Second, choose d of these 2d marbles (uniformly, without replacement).

Keep the d marbles you choose in the urn, but throw the others away.

At the end of this procedure, you will have d marbles in the urn. Let Xt be the number of blue marbles in the urn after t steps.

1. Explain why {Xt} as described above is a Markov chain.Xt (the number of blue marbles) is a Markov chain because it's a discrete-time process with a finite or countable (the count of the number of blue marbles) state space.

2. Describe the state space S for this Markov chain.

State Space {0,...d}

3. In class we did an example of an Ehrenfest chain which included an example of how the state of the chain evolved depending on the number of the molecule that was drawn (recall page 5 packet 416-1). Give a similar example showing how the state of the Wright-Fisher chain changes depending on which marbles you choose.

For our example we chose d = 4

DRAW LIST OF OBJECTS X(T) (Number of Blue marbles)

4. Find a formula for P(x,y), the transition function of this Markov chain. We know that we are choosing the marbles uniformly without replacement. (recall back to 414 on Hypergeometic density function)

5. (a) Let A be the set of states of this Markov chain which are absorbing. Find A. (Recall absorbing means once you're in that state A you will never leave)

Our absorbing states are 0 and d, once you get to 0 (no blue marbles) you will never be able to get blue marbles again. Once you get to d, you will always have d number of blue marbles and won't be able to contain any red every again. (b) Suppose d=2. What is P1(TA = ), the probability that you never eventually hit A? Interpret this result in terms of the number of each color of marbles that will eventually be in the urn. Using our transition function (refer to question 4) with d=2

lim (n = 0. (2 or 0 blues, everything will be blue, or nothing) n

(c) Suppose d = 3. For each x S, what is Px(TA = )?

lim (1 - p)n (chance of staying in the box) n lim (1- 1/5)n = 0. Eventually it will always go to the absorbing state n

(d) Do you think results analogous to parts (b) and (c) will hold for general d? Why or why not? Yes will hold for general d because starting at any value of x and moving infinite amount of times you will always eventually move to an absorbing state. (0 or d)

6. As an example, suppose that d=4 and suppose that the distribution of the number of blue marbles in the urn initially is: 0 = (0,

(a) Find the probability that the urn has 3 blue marbles after 1 step.

By using our Transition function:

7. (a) Show that your transition function satisfies the following equation for all x S.

(b) Use the answer to the previous question to find E[X1|X0](x) This can…