Q 5 Two bags each contain n balls In total there are 2n ball

Q 5 Two bags each contain n balls. In total there are 2n balls, n are green and n are red. After each unit of time a ball is selected from each bag and the two are swapped over (i.e the selected balls are put each put in the opposite bag). Let Xn be the number of green balls in the first bag. . What is the transition matrix of Xn. . Find the stationary distribution. . Is the Markov chain reversible?

Solution

There are 2 bags containing n and n.

After I draw from bag I, the ball is put in the second bag.

Hence no of balls will be n-1, n+1

Simultaneously 1 ball is taken from second and swapped thus making

no of balls equal to n, n

If Xn is the no of green balls, prob of drawing green ball =

After I drawn, prob for xn-1 green balls in the first bag =

and prob for xn green bags in the first bag = 1-

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Similar possibility for second bag with say yn no of green balls.

Thus Xn = Xn in the beginning

After first drawE( no of green balls) = (xn-1_+xn(1-xn)= (2xn-1-xn²)

This will go on continuing a no of times.

Yes the Markov chain is reversible.