By Soeren Asmussen

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1 Let {Xt } be explosive and modify the process so as to restart in some ﬁxed state i after each explosion. Show that we obtain a Markov jump process. 2 Let E = Z\ {0} and λ(k) = k 2 , qn(−n−1) = q(−n)(n+1) = 1/n2 , qn(n+1) = q(−n)(−n−1) = 1 − 1/n2 , n > 0. Show that the process is explosive and that 0 < PF+ < 1, PF+ + PF− = 1 where F± = limt↑ω(∆) Xt = ±∞ . Show that we get a Markov process by letting Xω(∆) = 1 on F+ , = −1 on F− (and similarly for the explosions after ω(∆)). 3 Let E = Z ∪ {∆} and λ(k) = (k + 1)2 , qk(k+1) = 1 for all k ∈ Z.

1) then easily yield that Xn = 0 eventually and that taking h0 = 0 makes λ−n hXn a martingale. If |λ| ≥ 1, boundedness would imply L1 – convergence (necessarily to h0 ) so that taking X0 = i yields hi = h0 = 0 which contradicts h = 0. Hence |λ| < 1 and spr(Q) < 1. 2 for nonnegative matrices A. We shall adopt the deﬁnitions of irreducibility and the period d from transition matrices to nonnegative matrices by noting that they depend only on the pattern of entries i, j with aij > 0. 4 If A is an irreducible nonnegative matrix, then the greatest common divisor d of the m with am ii > 0 does not depend on i.

Proof. Let λ be an eigenvalue of absolute value spr(Q) and let h ∈ Eλ . Consider a Markov chain {Xn } on {0, 1, . . , p} such that 0 is absorbing, qi and the probability of a transition i → j is qij for i, j ≥ 1 and 1 − for j = 0. The assumptions on Q and a geometrical trials argument (cf. 1) then easily yield that Xn = 0 eventually and that taking h0 = 0 makes λ−n hXn a martingale. If |λ| ≥ 1, boundedness would imply L1 – convergence (necessarily to h0 ) so that taking X0 = i yields hi = h0 = 0 which contradicts h = 0.