Consider the Bernoulli process Xni n123 where P X013 and P X

Consider the Bernoulli process {Xni n=1,2,3..), where P [X,-0-1/3 and P [Xn= 1 ] =2/3 for all n=1,2,3, . Let Yn, n-1,2,3, be the difference in ...t, where P 1-2/3 for all absolute value between the number of 0\'s and 1\'s that have occurred in n trails. For the stochastic process {Yn: n=1,2,3, ), prove or disprove that the stochastic process {Yn: n=1,2,3, ) is a homogeneous Markov chain.

Solution

For n trials, say there are k 0\'s and (n-k) 1\'s .
Then, the absolute value of difference between the number of zeroes and number of ones will be |n-2k| .
Hence, the pobability of occurence of event Yn will be |n-2k|/n . This probability for any n, is independent of any of the previous probabilities i.e. occurence of Y(n-1), Y(n-2) etc. It only depends on the value of n (number of trials) and the number of zeroes (k) .
Therefore, the given stochastic process is NOT a markov chain.