A short introduction to random trees

Transcription

1 Mongolian Mahemaical Journal (6) 5 A shor inroducion o random rees Chrisina Goldschmid Noes from wo lecures given a he CIMPA-DAAD Research School on Sochasic Processes and heir Applicaions, Naional Universiy of Mongolia, Ulaanbaaar, Mongolia, h July h Augus 5 Absrac Tree srucures are very common in naure, and we ofen wan o model rees which are somehow randomly generaed. In mahemaics, we hink of rees as conneced graphs (or neworks) wih no cycles. In hese lecures, I will discuss some simple models of random rees and wha we can say abou large insances of hem. The firs lecure will focus on Galon Wason branching processes. We hink of an idealised biological populaion (say of cells) in which each individual lives for a uni of ime, gives birh o a random number of children (wih a given disribuion) and hen dies. Wha can we say abou he way he populaion size evolves? Or abou he family ree of ha populaion? In he second lecure, we will ake a more combinaorial perspecive. Wha does a ree wih n verices, chosen uniformly a random from all he possibiliies, look like for large n? I urns ou ha we can use Galon Wason branching processes o help us answer his quesion. Along he way, we will encouner several beauiful bis of combinaorics and probabiliy, including Cayley s formula, random walks and (finally) Brownian moion. Conens. Branching processes Inroducion Generaing funcions Exincion Maringales Galon Wason rees Relaionship o random walks Uniform random rees Inroducion Cayley s formula Leaves The uniform random ree as a Galon Wason ree Taking limis The Brownian coninuum random ree References Mahemaics Subjec Classificaion 6C5, 6J. The auhor is graeful o Lady Margare Hall, Oxford, for funding which enabled her o ravel o Mongolia o deliver hese lecures.

2 54 CHRISTINA GOLDSCHMIDT Figure. The family ree of a branching process. We sar a he boom of he ree, wih a single individual in generaion. Then here are individuals in each of generaions and, 5 individuals in generaion, a single individual in generaion 4 and no individuals in subsequen generaions. (Noice ha in order o produce his picure, we mus have p() >.).. Inroducion. Branching processes Suppose we have a populaion (say of baceria). Each individual in he populaion lives a uni ime and, jus before dying, gives birh o a random number of children in he nex generaion. This number of children has probabiliy mass funcion p(k), k, called he offspring disribuion i.e. p(k) is he probabiliy ha an individual has k children. Differen individuals reproduce independenly in he same manner. Figure shows a possible family ree of such a populaion. Le X n be he size of he populaion in generaion n. The process (X n ) n is ofen called a Galon Wason branching process afer i was popularised by Galon and Wason [], alhough had already been sudied (much earlier) by Bienaymé [5]. As a simple model of he evoluion of a populaion, i has proved o be enormously powerful. I is now considered o be one of he fundamenal processes of probabiliy heory... Generaing funcions Suppose ha X =. Le C (n) i be he number of children of he ih individual in generaion n, so ha we may wrie X n+ = C (n) + C (n) + + C (n) X n. (We inerpre his sum as if X n =.) Noe ha C (n), C (n),... are independen and idenically disribued. Consider he probabiliy generaing funcions G(s) = k= p(k)sk (for s R such ha he righ-hand side is absoluely convergen) and G n (s) = E [ ] s Xn for n.

3 A SHORT INTRODUCTION TO RANDOM TREES 55 Proposiion.. For n, G n+ (s) = G n (G(s)) = G(G(... G(s)...)) = G(G n (s)). }{{} n+ imes Proof. Since X =, we have G (s) = s. Also, we ge X = C () which has probabiliy mass funcion p(k), k. So G (s) = E [ s X ] = G(s). Since we ge G n+ (s) = E [ s Xn+] [ = E X n X n+ = C (n) i, = k= i= s Xn i= C(n) i [ E s Xn ] i= C(n) i ] Xn = k P (X n = k) and, since C (n), C (n),... are independen of X n, his is equal o k= E [s ] k i= C(n) i P (X n = k). Since C (n), C (n),... are independen and idenically disribued, his in urn equals Hence, by inducion, for n, ] k E [s C(n) P (Xn = k) k= = (G(s)) k P (X n = k) k= = G n (G(s)). G n (s) = G(G(... G(s)...)) = G(G n (s)). }{{} n imes Corollary.. Suppose ha he mean number of children of a single individual is µ i.e. k= kp(k) = µ, for some µ <. Then E [X n ] = µ n. Proof. Noe ha µ = G ( ) and ha E [X n ] = G n( ). By he chain rule, Leing s, we ge G n(s) = d ds G(G n (s)) = G n (s)g (G n (s)). E [X n ] = E [X n ] µ = = µ n.

4 56 CHRISTINA GOLDSCHMIDT Exercise.. Show ha if σ := var (X ) < hen var (X n ) = { σ µ n (µ n ) µ if µ nσ if µ =. Corollary. ells us ha we ge exponenial growh on average if µ > and exponenial decrease if µ <. This raises an ineresing quesion: does he populaion die ou?.. Exincion In his secion, we invesigae he exincion probabiliy: q := P (populaion dies ou). If p() = hen every individual has a leas one child and so he populaion clearly grows forever. If p() >, on he oher hand, hen he populaion dies ou wih posiive probabiliy because q P (X = ) = p() >. Noice ha his holds even in he cases where E [X n ] grows as n. Proposiion.4. We have q = lim n P (X n = ). Proof. The even ha he populaion dies ou can be wrien as {X m = }. m= Noice ha he evens in his union are no disjoin: indeed, X m = implies ha X n = for all n > m. I follows ha n {X m = } = {X n = }. m= m= We have ha n m= {X m = } n+ m= {X m = } m= {X m = } and so by monoone convergence, ( ) ( n ) P {X m = } = lim P {X m = } = lim P (X n = ). n n m= Example.5. Suppose ha p(k) = (/) k+, k, so ha each individual has a geomeric number of offspring. We ll find he disribuion of X n. Firs calculae ( ) k+ G(s) = s k = s. k=

5 A SHORT INTRODUCTION TO RANDOM TREES 5 By plugging his ino iself a couple of imes, we ge G (s) = s s, G (s) = s 4 s. A naural guess is ha G n (s) = n (n )s (n+) ns which is, in fac, he case, as can be proved by inducion. If we wan he probabiliy mass funcion of X n, we need o expand his quaniy ou in powers of s. We have (n + ) ns = (n + ) Muliplying by n (n )s, we ge n k+ s k G n (s) = (n + ) k+ k= k= ( ns/(n + )) = k= n k s k (n + ) k+. n k (n )s k (n + ) k = n n + + We can read off he coefficiens now o see ha { n n+ if k = P (X n = k) = n k if k. (n+) k+ k= n k s k (n + ) k+. Noice ha P (X n = ) as n, which indicaes ha he populaion dies ou evenually in his case. Exercise.6. Suppose now ha p(k) = p( p) k, k, for some p (, ). (a) Find he probabiliy generaing funcion G(s) of his disribuion. Wha is is mean, µ? (b) Suppose ha p /. Prove, by inducion, ha he probabiliy generaing funcion of X n is G n (s) = p [( p)n p n ] ( p)s[( p) n p n ] [( p) n+ p n+ ] ( p)s[( p) n p n ], for n. { (c) Find G n () = P (X n = ) and hence show ha q = min p p }., Deduce ha he populaion always dies ou if µ whereas i has posiive probabiliy of surviving forever if µ >. Le us now reurn o he general case. We can find an equaion saisfied by q by condiioning on he number of children of he firs individual. q = P (populaion dies ou X = k) P (X = k) = k= P (populaion dies ou X = k) p(k). k= Remember ha each of he k individuals in he firs generaion behaves exacly like he paren. In paricular, we can hink of each of hem saring is own family, which is an independen copy of he original family. Moreover, he whole populaion dies ou if and only if all of hese sub-populaions die ou. If we had k families, his occurs wih

6 5 CHRISTINA GOLDSCHMIDT probabiliy q k. So q = q k p(k) = G(q). k= Exercise.. Check ha for he offspring disribuion in Exercise.6 he exincion probabiliy q you obained does, indeed, solve s = G(s). The equaion q = G(q) doesn quie enable us o deermine q: noice ha is always a soluion, bu i s no necessarily he only soluion in [, ]. I urns ou ha when here are muliple soluions, he one we wan is he smalles one in [, ]. Theorem.. Suppose ha p() > and p() + p() <. (a) The equaion s = G(s) has a mos wo soluions in [, ]. The exincion probabiliy q is he smalles non-negaive roo of he equaion G(s) = s. (b) Suppose ha he offspring disribuion has mean µ. Then if µ hen q = if µ > hen q <. Proof. have We observe ha G() = p() >, G() = and ha µ = G ( ). We also G (s) = j(j )p(j)s j for s <. j= Since p() + p() <, we have G (s) > for s [, ). Hence, G is sricly convex. Consider he case µ. Then G() > and G (s) G ( ) = µ for all s [, ]. I follows ha G(s) > s for all s [, ) and so he unique roo in [, ] is s =, which mus necessarily be q. See Figure. Consider now he case µ >. Then since G() >, G() =, G ( ) > and G is sricly convex, here exiss precisely one roo α of s = G(s) in [, ). See Figure. Now G() G(α) = α. So G(G()) G(α) = α and, ieraing, G n () α for all n. I follows ha q = lim n G n () α. Bu since q mus be a roo of G(s) = s, we deduce ha q = α. We refer o he case µ < as subcriical, he case µ = as criical and he case µ > as supercriical..4. Maringales We have jus seen ha, in he subcriical and criical cases, we have X n as n. In he supercriical case, on he oher hand, here is posiive probabiliy ha X n. We also have ha he populaion grows exponenially on average. Wha can we say almos surely? For n, define W n = µ n X n and le F n = σ(x m, m n).

7 A SHORT INTRODUCTION TO RANDOM TREES 5 G(s) G(s) s s G(s) s Figure. Solving s = G(s) for µ <, µ = and µ > respecively. Proposiion.. (W n, n ) is a non-negaive maringale (wih respec o he filraion (F n ) n ) and so exiss almos surely. W := lim n W n Proof. We have already shown ha E [X n ] = µ n so ha E [W n ] = for all n. Hence, W n is inegrable. Moreover, [ Xn ] E [W n+ F n ] = µ (n+) E [X n+ F n ] = µ (n+) E F n [ = µ W n E i= C (n) C (n) i ],

8 6 CHRISTINA GOLDSCHMIDT since X n is F n -measurable, [ ] C (n) i, i are independen of F n and are idenically disribued. Since E = µ, we ge C (n) E [W n+ F n ] = W n, as required. The convergence resul now follows from he almos sure maringale convergence heorem. Proposiion. ells us ha, in a rough sense, W n µ n W. Using Faou s lemma, we obain ha E [W ] lim inf n E [W n ] =. I is, however, sill possible ha W. (Indeed, Proposiion. holds even if µ, in which case we mus have W.) Proposiion.. or o. Suppose ha < µ <. Then P (W = ) is eiher equal o q Proof. We have X X n+ = X n (i), where X n (), X n (),... are i.i.d. copies of X n, independen of X. So, similarly, i= X W n+ = µ W n (i). Leing n and using Proposiion., we see ha W has he same disribuion as µ X i= W (i). (.) i= ] In paricular, P (W = ) = E [P (W = ) Z = G(P (W = )). So P (W = ) is a roo of s = G(s). Bu, by Theorem. he possible roos are q and and he resul follows. Exercise.. Consider again he Geomeric offspring disribuion of Exercise.6. Assume ha p < /, so ha µ = p p >. Suppose ha W has he disribuion specified by P (W = ) = p p ( p)w, P (W > w) = p p e p for w >. Show ha W solves he recursive disribuional equaion (.). [Hin: Firs show ha he number of erms in he sum (.) which are non-zero has a Geomeric disribuion, wih parameer p/( p). Then use momen generaing funcions.] A necessary and sufficien condiion for E [W ] = is given in he following heorem, which we won prove.

9 A SHORT INTRODUCTION TO RANDOM TREES 6 Theorem. (Kesen and Sigum []). Suppose ha < µ <, and wrie log + x = max{log x, }. Then he following are equivalen: E [W ] = P (W > non-exincion) = k= kp(k) log+ k <..5. Galon Wason rees We have so far jus considered he populaion sizes X, X, X,... in each of he successive generaions. Bu i s also naural o wan o hink abou a richer srucure: he family ree (also called he genealogical ree). In order o do his, i will urn ou o be useful o consider a canonical labelling for rees (called he Ulam Harris labelling). Le N = {,,...} and le U = { } n= Nn. In general, an elemen u U is a sequence of naural numbers represening a poin in an infiniary ree. We will hink of u as a sring and wrie u u... u n. The concaenaion of srings u = u u... u n and v = v v... v m is wrien uv = u u... u n v v... v m. The label of a verex u U indicaes is genealogy: u has paren p(u) = u u... u n. u has children u, u,.... We wrie u = n for he generaion of u. Definiion.. A rooed ordered ree is a subse of U such ha for all u such ha u, p(u) for all u, here exiss c(u) {,,,...} such ha for j N, uj if and only if j c(u). The ree is rooed a. The number of children of u in is given by c(u). We wrie #() for he size (number of verices) of and noe ha #() = + u c(u). We can hink of he family ree of a branching process as a random rooed ordered ree which is obained by sampling independen and idenically disribued random variables C(u), u U, each having he offspring disribuion. The corresponding Galon Wason ree T is hen obained by finding he larges subse of U such ha he condiions of Definiion. are fulfilled. Noe ha he ree is finie if and only if he branching process becomes exinc..6. Relaionship o random walks Trees are slighly awkward objecs o manipulae and so we ofen find i useful o encode hem in erms of funcions. This will be useful in he nex lecure. There are several differen ways o do his we will concenrae on one which works well wih our labelling. See Secions. and. of Le Gall [] for more deails. For a rooed ordered ree, le v = and, for i, le v i be he ih verex of when he verices are lised in lexicographic order of label.

10 6 CHRISTINA GOLDSCHMIDT Definiion.4. The Lukasiewicz pah associaed wih a ree of size #() = n is he funcion l : {,,..., n} {,,,...} defined by l() = and, for i n, l(i) = l(i ) + c(v i ) i.e. l(i) = i j= (c(v j) ). Noe ha by (.), we mus have l(n) =. Moreover, l(i) for i n. Exercise.5. unique ree. Show ha a given Lukasiewicz pah (l(i), i n) encodes a In oher words, here is a bijecion beween rooed ordered rees and Lukasiewicz pahs. In he case of Galon Wason rees, his correspondence gives us somehing paricularly nice. Wrie (L(i), i #(T)) for he Lukasiewicz pah associaed wih a Galon Wason ree T. Le (S(i), i ) be a random walk wih sep-sizes having disribuion p(k + ), k and le τ = inf{i : S(i) = }. Proposiion.6. he same disribuion. The processes (L(i), i #(T)) and (S(i), i τ) have Proof (skech). We have L(i) = i j= (C(v j) ), so ha L(i) for i #(T) and L(#(T)) =. Since he random variables C(u), u U are independen, ake values in {,,,...} and have probabiliy mass funcion p(k + ), k, he resul seems inuiively clear. The problem is ha he labels v, v,... are random and depend on (C(u), u U), so i is no, in fac, obvious ha C(v ), C(v ),... are independen and idenically disribued. The poin is ha, a sep n + of he lexicographic exploraion of he ree, he labels v,..., v n and he numbers C(v ),..., C(v n ) enable us o deermine (a) if we have reached he end of he ree and (b) if no, which is he label v n+. Bu, in case (b), he informaion we already possess can influence he value of C(v n+ ). See Proposiion.5 of Le Gall [] for he deails. Noe ha we can (and will) hink of he random walk as coninuing on beyond he firs ime i his. Remark.. Proposiion.6 enails ha we can hink of he exincion probabiliy of a branching process as he probabiliy ha he random walk ever his. We observe ha he sep sizes of he random walk have mean µ. Assuming addiionally ha p() > and p() + p() <, we see ha i is, indeed, he case ha he walk his wih probabiliy if µ (corresponding o exincion) and has posiive probabiliy of never hiing if µ >.

11 A SHORT INTRODUCTION TO RANDOM TREES 6.. Inroducion. Uniform random rees In his lecure, we re going o ake a more combinaorial perspecive on random rees, bu i will evenually connec back o wha we discussed in he firs lecure. Here we will concenrae on labelled rees. We le T n be he se of (unordered) labelled rees on n verices (labelled by [n] := {,,..., n}. For example, T consiss of he rees Unordered means ha hese rees are all he same: bu his one is differen: 4 5 Suppose we ake all of he rees on n labelled verices and pick one uniformly a random call i T n. Wha does T n look like? Phrased anoher way: wha does a ypical ree on n verices look like?.. Cayley s formula A good saring place is o ask how many elemens T n has. Theorem. (Cayley s formula). For n, T n = n n. This heorem was firs proved by Borchard [6] and laer exended by Cayley []. Proof. We give a beauiful proof due o Piman [4]. Call he quaniy we re afer a(n). I s useful for he purposes of his proof o hink of rooed rees wih direced edges, where he edges are all direced away from he roo. (We can simply forge he roo and he direcions of he arrow a he end.) Sar from he empy graph on our n

12 64 CHRISTINA GOLDSCHMIDT verices and coun he number of differen sequences of direced edges ha we can add o form a rooed ree. Firs way o coun. Sar wih one of he a(n) unrooed rees on n verices. There are n ways o pick a roo. Direc all edges away from he roo. There are (n )! differen orders in which we migh add he n direced edges. So, we have a(n) n (n )! = a(n)n! differen sequences of direced edges. Second way o coun. Sar by hinking of each of he n verices as a roo. A each sep, we have a fores of rooed rees, wih edges direced away from he roos. Add an edge from an arbirary verex v o he roo of a differen ree, and roo he resuling ree a he roo of v s old ree. If we have added k edges already, hen we have a fores conaining n k rees (we reduce he number of rees by a each sep). So here are n(n k ) choices for he edge we add: he saring verex can be any one of he n verices and here are n k roos of oher rees o which i can connec. So he oal number of choices is n k= n(n k ) = n n (n )! = n n n!. So a(n)n! = n n n! and cancelling gives a(n) = n n. There are many quesions we migh ask abou our uniform random ree T n. For example, How many leaves (i.e. verices wih only one neighbour) are here? How many verices have wo neighbours? Or hree? Or more? Wha is he diameer of he ree (i.e. he lengh of he longes pah beween wo poins in he ree)? Pick a verex uniformly a random o be he roo. How many verices are here a disance d from he roo? Because he ree is random, all of hese quaniies are random variables... Leaves We firs invesigae he proporion of leaves in a uniform random ree. Theorem.. as n. Le N n denoe he number of leaves in T n. Then N n n p e Proof. where Noe ha I i = N n = n i= { if he verex labelled i is a leaf oherwise. I i

13 A SHORT INTRODUCTION TO RANDOM TREES 65 We have E [I i ] = P (i is a leaf). In order o calculae his probabiliy, we need o deermine how many rees on [n] have i as a leaf. Each such ree can be hough of as a ree on [n] \ {i} plus an edge from one of he oher verices o i. Since here are (n ) n rees on n labelled verices, i follows ha here are (n ) n rees which have i as a leaf. So P (i is a leaf) = (n )n n n = ( n) n e, as n. If he indicaor random variables I, I,..., I n were independen, we would now be able o apply he weak law of large numbers. Bu his is no quie he case. Le i j. Then P (i is a leaf and j is a leaf) = (n ) (n ) n 4 ( n n = n) n e, so ha I i and I j are asympoically independen. In paricular, ( cov (I i, I j ) = n) n ( n) (n ), as n. We also have ( var (I i ) = Hence, using symmery, var ( Nn n ( ) ) n ( ) n ( e )e. n n ) = n n var (I i ) + n cov (I i, I j ) i= i<j = n var (I ) + n n cov (I, I ), as n. Finally, for any ɛ >, we can apply Chebyshev s inequaliy o obain ( ) N n P n e > ɛ var (N n/n) ɛ, as n, as desired..4. The uniform random ree as a Galon Wason ree We now make he link beween our sudy of he uniform random ree and he las lecure on branching processes. Proposiion.. The uniform random ree T n has he same disribuion as a ree generaed as follows: Take a Galon Wason ree wih Poisson() offspring disribuion Condiion i o have oal progeny precisely n Assign he roo he label and he oher verices random labels chosen from {,..., n}. Forge he roo and he original ordering.

14 66 CHRISTINA GOLDSCHMIDT Proof. Le T be a Poisson() Galon Wason ree, and le be a paricular ree wih #() = n. Then P (T = ) = e c(v)! = e n c(v)!. v v Now observe ha P (#(T) = n) is a funcion only of n. Hence, P (T = #(T) = n) = f(n) v c(v)! for some funcion f. Moreover, here are (n )! differen ways o label wih he roo labelled and he oher labels chosen from {,..., n}. Now consider a paricular unordered labelled ree T n, and hink of i as rooed a. Then we can say which are he children of a verex v (i.e. hose which are furher away from he roo han he verex iself). Then here are v c(v)! differen possible orderings of he labelled ree. In paricular, here are v c(v)! differen labelled ordered rees which yield he same labelled ree when we forge he ordering. So he probabiliy ha our condiioned Galon Wason ree gives a paricular ree T n is c(v)! v f(n) (n )! v c(v)! = f(n) (n )!. Since his depends only on n, and no on any oher feaure of he ree, i mus be he case ha he ree is uniformly disribued on T n. Remark.4. Oher classes of combinaorial rees may also be obained as Galon Wason rees condiioned on heir oal progeny. For example, aking p(k) = ( ) k+, k, gives a uniform plane ree aking p() = / and p() = / gives a uniform complee binary ree (as long as he oal progeny is odd)..5. Taking limis We ve already seen ha if we look a he proporion of he verices in T n which are leaves hen we obain a meaningful limi as n. Wha can we say more generally abou T n as n ges large? We will make crucial use of he link o Galon Wason rees oulined in he las secion. We won prove he resuls in his secion or he nex see Aldous [] and Le Gall [] for more deails. Recall ha if T is a Galon Wason ree wih offspring disribuion p(k), k, hen is Lukasiewicz pah (L(i), i #(T)) has he same disribuion as a random walk (S(k), k ) wih sep-size disribuion p(k + ), k, sopped a τ, he firs ime S his. We will concenrae on he criical case, where µ =. The fundamenal idea ha we will make use of in his secion comes from he following heorem. Theorem.5 (Donsker s heorem). Suppose ha (S(k), k ) is a random walk wih sep-sizes of mean and finie variance σ. Then, as n, ( ) σ n S( n ), d (B(), ),

15 A SHORT INTRODUCTION TO RANDOM TREES 6 Figure. A simulaion of a sandard Brownian excursion. where (B(), ) is a sandard Brownian moion. In oher words, in he criical case, here is a good way o rescale he random walk in such a way ha we obain a non-rivial limiing process. If we have a Poisson() offspring disribuion, we ge ha he sep-sizes precisely have mean and variance. Raher han a criical Galon Wason of random size, we would like o consider a criical Galon Wason ree condiioned o have size n. The corresponding Lukasiewicz pah is a random walk condiioned o hi for he firs ime a ime n. Le (S (n) (k), k n) have he same disribuion as (S(k), k n) condiioned on he even {τ = n}. Theorem.6 (Kaigh []). As n, ( ) σ n S(n) d ( n ), (e(), ), where (e(), ) is a sandard Brownian excursion. Informally, e is a Brownian moion condiioned o say posiive in he inerval (, ) and o come back o a ime (see Figure ). This heorem suggess ha here should be a limi for he ree iself, and ha limi should somehow be encoded by e. This is, indeed, he case, as was proved by Aldous in a series of papers in he early s [,, ]. The limi is known as he Brownian coninuum random ree..6. The Brownian coninuum random ree In order o see how he limi ree arises, we will go hrough he consrucion in he special case where he offspring disribuion is given by p(k) = ( ) k+, k i.e. when

16 6 CHRISTINA GOLDSCHMIDT we have a uniform plane ree. We will use an encoding for our ree which is somewha easier o visualise han he Lukasiewicz pah. The conour funcion is obained by simply racing he conour of he ree from lef o righ a speed, so ha we pass along each edge wice. Record he disance from he roo a each ime o ge (C(k), k n )

17 A SHORT INTRODUCTION TO RANDOM TREES

18 CHRISTINA GOLDSCHMIDT The conour funcion is a sor of expanded version of he ree. For convenience, we will define a slighly shifed version: le C() =, C(n) = and, for k n, C(k) = + C(k ). Exercise.. Show ha ( C(k), k n) has he same disribuion as a simple symmeric random walk condiioned o reurn o he origin for he firs ime a ime n. I s easy o linearly inerpolae o give a coninuous funcion C : [, n] [, ), and similarly for C. Since he variance of he offspring disribuion is, i follows from Theorem.6 ha n ( C(n), ) d (e(), ) as n and, hence, n (C(n), ) d (e(), ). I s also relaively sraighforward o see how o ge back a ree from a conour funcion. Inuiively speaking, we pu glue on he underside of he funcion and hen push he wo sides ogeher unil hey mee (see Figure 4). The Brownian coninuum random ree is he objec we obain by doing his gluing procedure o he funcion (e(), ) see Figure 5.

19 A SHORT INTRODUCTION TO RANDOM TREES Figure 4. Trees from excursions Figure 5. A simulaion of a Brownian coninuum random ree For a general criical Galon Wason ree wih finie offspring variance σ condiioned o have oal progeny n, i urns ou ha he convergence of he (suiably rescaled) Lukasiewicz pah o a sandard Brownian excursion acually implies he convergence of he conour funcion of he ree (alhough his is far from rivial o see). Hence, any such condiioned ree converges in disribuion as n o he Brownian coninuum random ree. See Aldous [] and Le Gall [] for more deails, in paricular he specificaion of he opology in which convergence occurs. Furher reading In hese lecures, I have jus scraped he surface of a large and fascinaing opic. For an accoun of he hisory of branching processes, see Kendall []. For much more on he heory of branching processes, see he classic accoun of Ahreya and Ney [4] or he recen lecure course of Shi [5]. To learn abou maringales, see he beauiful book of

20 A SHORT INTRODUCTION TO RANDOM TREES Williams [6]. The accoun I have presened here of he relaionship beween Galon Wason rees and random walks owes much o he excellen survey paper of Le Gall []. Tha paper also gives he rigorous deails of he convergence of a criical Galon Wason ree wih finie offspring variance o he Brownian coninuum random ree. For overview of he Brownian coninuum random ree, see Aldous aricle []. Finally, for an accoun of a variey of models of random rees, see he book of Drmoa []. Acknowledgemens The simulaion of a large Galon Wason ree (approximaing he Brownian coninuum random ree) in Figure 5 was done using Mahemaica code wrien by Igor Korchemski. I am very graeful o him for leing me use his program. I would like o express my graiude o he organisers of he research school in Ulaanbaaar a which hese lecures were given for heir ireless effors and wonderful hospialiy. Finally, I would like o hank he audience for heir warm welcome and infecious enhusiasm, which made he research school an unforgeable experience. References. D. Aldous, The coninuum random ree. I, Ann. Probab. (), no.,.., The coninuum random ree. II. An overview, Sochasic analysis (Durham, ), London Mah. Soc. Lecure Noe Ser., vol. 6, Cambridge Universiy Press, Cambridge,, pp..., The coninuum random ree. III, Ann. Probab. (), no., K. Ahreya and P. Ney, Branching processes, Springer-Verlag, New York,, Die Grundlehren der mahemaischen Wissenschafen, Band I. J. Bienaymé, De la loi de muliplicaion e de la durée des familles, Soc. Philomah. Paris Exrais, Sér. 5 (45),. 6. C. W. Borchard, Über eine der Inerpolaion ensprechende Darsellung der Eliminaions- Resulane, J. Reine Angew. Mah. 5 (6),.. A. Cayley, A heorem on rees, Q. J. Pure Appl. Mah. (), 6.. M. Drmoa, Random rees: An inerplay beween combinaorics and probabiliy, Springer-Verlag, Vienna,.. F. Galon and H.W. Wason, On he probabiliy of exincion of families, J. Anhropol. Ins. 4 (4), 44.. W. D. Kaigh, An invariance principle for random walk condiioned by a lae reurn o zero, Ann. Probabiliy 4 (6), no., 5.. D.G. Kendall, Branching processes since, J. London Mah. Soc. 4 (66), H. Kesen and B. Sigum, A limi heorem for mulidimensional Galon Wason processes, Ann. Mah. Sais. (66),.. J.-F. Le Gall, Random rees and applicaions, Probab. Surv. (5), J. Piman, Coalescen random foress, J. Combin. Theory Ser. A 5 (), no., Zhan Shi, Random walks and rees, X Symposium on Probabiliy and Sochasic Processes and he Firs Join Meeing France-Mexico of Probabiliy, ESAIM Proc., vol., EDP Sci., Les Ulis,, pp.. 6. David Williams, Probabiliy wih maringales, Cambridge Mahemaical Texbooks, Cambridge Universiy Press, Cambridge,. Chrisina Goldschmid Deparmen of Saisics and Lady Margare Hall Universiy of Oxford Unied Kingdom goldschm@sas.ox.ac.uk