Only for the questions with * marks
Worksheet # 5 Statistics 150, Pitman/Calvert, Spring 2019 Reading: Text Section 1.11. Homework: For each of the five exercises, turn in solutions of the part marked *. 1. Branching processes. (a) Durrett Exercise 1.76 on page 94 (b) Durrett Exercise 1.77 on page 94 (c) * Consider a supercritical branching process with Poisson(µ) offspring distribution for some µ > 1. Given a number p with 0 < p="">< 1,="" find="" an="" explicit="" formula="" for="" µ="µ(p)" such="" that="" the="" branching="" process="" started="" with="" one="" individual="" has="" extinction="" probability="" p.="" plot="" the="" graph="" of="" µ(p),="" and="" evaluate="" µ(1/2)="" as="" a="" decimal.="" 2.="" a="" death="" and="" immigration="" process.="" starting="" with="" x0="" individuals="" at="" time="" 0,="" for="" n="1," 2,="" .="" .="" .="" let="" xn+1="Xn∑" i="1" bn,i="" +="" yn+1="" where="" the="" bn,i="" are="" independent="" bernoulli(p)="" and="" the="" y="" ′s="" are="" independent,="" all="" poisson(λ)="" independent="" of="" the="" b′s.="" in="" words,="" from="" one="" generation="" to="" the="" next,="" each="" individual="" dies="" with="" probability="" 1−="" p,="" and="" there="" are="" a="" poisson="" (λ)="" number="" of="" immigrants.="" (a)="" give="" an="" explicit="" summation="" formula="" for="" the="" transition="" matrix.="" (warning:="" the="" sum="" does="" not="" simplify!)="" (b)="" find="" the="" exact="" distribution="" of="" x1="" if="" x0="" ∼="" poisson="" (µ0)="" (use="" the="" thinning="" property="" of="" the="" poisson="" distribution,="" which="" you="" should="" check="" by="" generating="" functions).="" (c)="" *="" find="" the="" exact="" distribution="" of="" xn="" if="" x0="" ∼="" poisson="" (µ0).="" you="" may="" figure="" this="" out="" first="" by="" recursion.="" but="" once="" you="" find="" the="" form="" of="" the="" answer,="" you="" should="" see="" there="" is="" a="" nice="" explanation:="" generation="" n="" is="" formed="" by="" immigrants,="" plus="" children="" of="" immigrants="" in="" the="" previous="" generation,="" plus="" children="" of="" children="" of="" immigrants,="" and="" so="" on.="" detail="" this="" argument="" with="" an="" adequate="" notation.="" (d)="" find="" the="" limit="" distribution="" of="" xn="" as="" n="" →="" ∞,="" no="" matter="" what="" the="" distribution="" of="" x0.="" 3.="" kac="" identities.="" let="" x0,="" x1,="" .="" .="" .="" be="" a="" stationary="" sequence="" of="" random="" variables="" (not="" necessarily="" markov),="" a="" a="" subset="" of="" the="" set="" of="" possible="" values="" of="" the="" xn,="" and="" ta="" :="min{n" ≥="" 1="" :="" xn="" ∈="" a}.="" show="" the="" following;="" (a)="" p(ta="n)" =="" p(x0="" ∈="" a,="" ta="" ≥="" n)="" (n="1," 2,="" .="" .="" .)="" (b)="" *="" deduce="" that="" if="" x="" is="" an="" irreducible="" markov="" chain="" with="" countable="" state="" space="" s,="" with="" stationary="" initial="" distribution="" π,="" and="" x="" ∈="" s="" and="" tx="" :="T{x}" then="" 1="πxExTx" (1)="" and="" eπtx="πx" ex="" 12tx(tx="" +="" 1)="" (2)="" 1="" (c)="" continuing="" for="" such="" a="" markov="" chain,="" let="" mxy="" :="ExTy" and="" σ2xy="" :="VarxTy." find="" a="" formula="" for="" σ2xx="" in="" terms="" the="" mean="" first="" passage="" times="" myz="" for="" various="" y,="" z="" ∈="" s.="" 4.="" tail="" generating="" functions.="" (adapted="" from="" feller="" vol="" i)="" for="" a="" suitable="" sequence="" f0,="" f1,="" .="" .="" .="" and="" suitable="" s="" define="" generating="" functions="" f="" (s)="" :="∞∑" n="0" fns="" n="" (3)="" q(s)="" :="∞∑" n="0" qns="" n="" where="" qn="" :="fn+1" +="" fn+2="" +="" ·="" ·="" ·="" (4)="" r(s)="" :="∞∑" n="0" rns="" n="" where="" rn="" :="qn+1" +="" qn+2="" +="" ·="" ·="" ·="" (5)="" (a)="" show="" that="" under="" suitable="" assumptions="" on="" f0,="" f1,="" .="" .="" .="" and="" s="" (to="" be="" specified)="" q(s)="F" (1)−="" f="" (s)="" 1−="" s="" and="" r(s)="Q(1)−Q(s)" 1−="" s="" (6)="" (b)="" *="" suppose="" that="" fn="P(T" =="" n)="" for="" a="" non-negative="" integer="" valued="" random="" variable="" t="" .="" show="" that="" qn="P(T"> n) and rn = E(T − 1− n)+ Q(1) = ET and R(1) = E1 2 T (T − 1) 5. Renewal generating functions. (Adapted from Feller Vol I) Suppose now that T0 = 0 and Tk is the sum of k independent copies of T as above, with P(T > 0) = 1, so f0 = 0, and let In be the sequence of renewal indicators In := 1(Tk = n for some k ≥ 0), so that by definition {n : In = 1} = {0, T1, T2, . . .} Let un := P(In = 1) be the probability of a renewal at time n. Note that if Xn is a recurrent Markov chain with transition matrix P started in state x, and Tk is the time of the kth return of X to its initial state, then In = 1(Xn = x), a renewal occurs each time the chain returns to x, and un = P n(x, x). See also text Example 1.37 on page 41, and Section 3.3.1 on page 137, where T0 ≥ 0 is allowed to have any distribution, and Tk is T0 plus the sum of k independent copies of T . With F (s) := EsT as above, let µ = ET and σ2 = VarT. (a) * Show that for suitable s (to be determined) ∞∑ n=0 uns n = 1 1− F (s) (7) ∞∑ n=0 ( un − 1 µ ) sn = R(s) µQ(s) (8) ∞∑ n=0 ( un − 1 µ ) = σ2 + µ2 − µ 2µ2 (9) 2 Beware that the sum (9) is divergent if the distribution of T is periodic, but the formula is still correct assuming that ET 2 < ∞="" and="" the="" sum="" is="" interpreted="" by="" abel’s="" method="" of="" summation,="" i.e.="" taking="" the="" limit="" as="" s="" ↑="" 1="" of="" the="" generating="" function.="" for="" an="" aperiodic="" distribution="" of="" t="" with="" et="" 2=""><∞, it="" can="" be="" shown="" (and="" you="" can="" assume)="" that="" the="" series="" is="" in="" fact="" convergent.="" then="" (9)="" is="" correct="" by="" abel’s="" theorem="" https://en.wikipedia.org/wiki/abel%27s_theorem.="" (b)="" the="" formulas="" (8)="" and="" (9)="" are="" corrected="" versions="" of="" formulas="" (12.1)="" and="" (12.2)="" in="" feller’s="" book.="" check="" that,="" unlike="" feller’s="" version="" of="" these="" formulas,="" the="" above="" formulas="" are="" correct="" in="" the="" case="" when="" the="" in="" for="" n="" ≥="" 1="" are="" independent="" bernoulli(p)="" variables.="" what="" are="" the="" evaluations="" of="" µ="" and="" σ2="" as="" functions="" of="" p="" in="" this="" case?="" (c)="" use="" the="" bound="" on="" page="" 53="" in="" step="" 4="" of="" the="" proof="" of="" theorem="" 1.23="" of="" the="" text="" to="" show="" that="" if="" un="P" n(x,="" x)="" for="" an="" irreducible="" and="" aperiodic="" p="" with="" finite="" state="" space,="" then="" ∑∞="" n="0" |un="" −="" πx|="">< ∞,="" and="" hence="" the="" above="" formula="" is="" justified="" by="" abel’s="" theorem.="" deduce="" the="" same,="" under="" the="" weaker="" assumption="" that="" the="" radius="" of="" convergence="" of="" f="" (s)="" is=""> 1. In fact, all that is needed is ET 2 < ∞, but this is more technical to prove. 3 https://en.wikipedia.org/wiki/abel%27s_theorem ∞,="" but="" this="" is="" more="" technical="" to="" prove.="" 3="">