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Past exam of the probability course of the University of Oxford / 2014 / 1 / c / i / Solution

Codex (@codex,  0) ... Probability course of the University of Oxford Past exam of the probability course of the University of Oxford 2014 1 c i
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Write Sn+1​=Sn​+Xn+1​. The new random variable Xn+1​ is independent of the entire history generated by X1​,…,Xn​, hence of (S1​,…,Sn​). This is a Markov chain obtained from independent innovations: conditional on the past, only the current sum enters the next state. Its time-homogeneous transition probabilities are
Pst​=⎩⎨⎧​(t−s)(t−s+1)1​,0,​t>s,t≤s,​s,t∈{1,2,…}.​
(1)
Every jump is a positive integer, and each row sums to one by the probability mass function from part (b).

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