Suppose pk k 0 is the offspring distribution so that

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Suppose {pk, k ::: 0} is the offspring distribution so that Pk represents the probability of k offspring per in- dividual. Let m = Lk kpk be the mean number of offspring per individual. Let tz<n>(i), n ::: 0, i ::: 1} be an iid sequence whose common mass function is the offspring distribution {Pk} and define recursively Zo = 1 and Zn+I = l z<n>(1) + · · · + z<n>(Zn). 0, if Zn > 0, if Zn = 0, which represents the number in the (n + 1)- generation. Then {Zn} is a Markov chain and p;,· := P(Zn+I = j!Zn = i] = 01 .' J if i = 0, if i ::: 1, where for i ::: 1, pji is the jth component of the i-fold convolution of the se- quence {Pn }. Note fori ::: 1 while for i = 0, 00 00 "\' "\' >t<i. = J = tm, j=O }=I 00 L PiJ j = Poo · 0 + 0 = 0 = mi. j=O With I (j) = j we have PI = mI. This means that the process {(Zn/mn, a(Zo, ... , Zn)), n ::: 0} is a martingale. (10.24) (7) Likelihood ratios. Suppose {Yn. n ::: 0} are iid random variables and sup- pose the true density of Y1 is lo· (The word "density" can be understood with respect to some fixed reference measure JJ..) Let ft be some other probability density. For simplicity suppose lo(y) > 0, for ally. Then for n ::: 0 is a martingale since (( n7=o /t(Y;)) /I(Yn+I) ) E(Xn+IIYo, ... , Yn) = E D?=o lo(Y;) lo(Yn+I) !Yo,···, Yn ( ft(Yn+I) ) =XnE I. IYo, ... ,Yn . ,o(Yn+I)
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360 10. Martingales By independence this becomes = XnE (/I (Yn+I)) = Xn I /I (y) fo(y)JJ.(dy) fo(Yn+I) fo(y) = Xn I ftdf.J. = Xn · 1 = Xn since /I is a density. 10.6 Connections between Martingales and Submartingales This section describes some connections between martingales and submartingales by means of what is called the Doob decomposition and also some simple results arising from Jensen's inequality. 10.6.1 Doob's Decomposition The Doob decomposition expresses a submartingale as the sum of a martingale and an increasing process. This latter phrase has a precise meaning. Definition. Given a process {Un, n ::: 0} and a-fields {Bn, n ::: 0}. We call {Un, n ::: 0} predictable if Uo e Bo, and for n ::: 0, we have Call a process {An, n ::: 0} an increasing process if {An} is predictable and almost surely Theorem 10.6.1 (Doob Decomposition) Any submartingale can be written in a unique way as the sum of a martingale and an increasing process {An, n ::: 0}; that is Xn = Mn +An, n::: 0.
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10.6 Connections between Martingales and Submartingales 361 Proof. (a) Existence of such a decomposition: Define dg=Xo, dJ=Xj-E(XjiBj-t), n Mn:="'LdJ. j=O Then {Mn} is a martingale since {dJl is a fair sequence. Set An= Xn- Mn. Then Ao = Xo- Mo = Xo- Xo = 0, and An+t- An= Xn+l- Mn+l- Xn + Mn = Xn+t - Xn - (Mn+l - Mn) = Xn+t - Xn - d!+l = Xn+l- Xn- Xn+l + E(Xn+til3n) = E(Xn+til3n)- Xn 0 by the submartingale property. Since n n An+t = L(Aj+t-Aj) = L(E(Xj+til3j) -Xj) e Bn, j=O j=O this shows {An} is predictable and hence increasing. (b) Uniqueness of the decomposition: Suppose Xn = Mn +An, and that there is also another decomposition Xn = where is a martingale and is an increasing process. Then and = Xn+t - Xn - - Because is predictable and is a martingale, = iBn)= E(Xn+til3n)- Xn- 0 and An+t -An= E(An+t-Anil3n) = £(Xn+tll3n) -Xn.
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