Exercise 1.3. From the identityfn(s) =f(fn−1(s)), we obtainfn′(s) =f′(fn−1(s))fn′−1(s). Thus fn′′(1) =
f′′(fn−1(1))fn′−1(1)2+f′(fn−1(1))fn′′−1(1) =f′′(1)fn′−1(1)2+f′(1)fn′′−1(1). Sincef′(1) =mandfn′−1(1) = E(Zn−1) =mn−1, this leads to: fn′′(1) =m2n−1+mfn′′−1(1).
Solving this by induction, we obtain: fn′′(1) =
P2n−1
i=n+1mi+mn−1f1′′(1). Sincef′′(1) =
P∞
i=0i(i−1)pi= E(Z12)−E(Z1) = σ2+m2−m (where σ2 := Var(Z1) < ∞), we arrive at: E(Zn2) = fn′′(1) +E(Zn) =
P2n−1
i=n+1mi+mn−1(σ2+m2−m) +mn. In particular, supnE(Wn2)<∞.
Exercise 1.4. By Exercise 1.3, supnE(Wn2)<∞, which means (Wn) is a martingale bounded in L2 (and
a fortiori uniformly integrable). As a consequence, Wn converges in L2 to a limit, which is W: E(W2) = lim↑E(Wn2)>0, andE(W) = 1.
It remains to check that P(W = 0) = q. Let q′ := P(W = 0); so q′ < 1 because E(W) = 1. By conditioning on the value ofZ1, we obtain: q′ =P∞
i=0piP(W = 0|Z1=i) =Pi∞=0pi(q′)i=f(q′). Since q
is the unique solution off(s) =sfors∈[0,1) in the supercritical case, we haveq′ =q.
Exercise 1.6. Without loss of generality28, we may assume that A1, A2, · · ·, Ak are elements of F
n, for somen. Then b P(N =k, T1∈A1,· · · ,Tk ∈Ak) =E nZn mn 1{N=k,T1∈A1,···,Tk∈Ak} o .
28Assume we can prove the desired identity for allnand allA1,A2,· · ·,A
k ∈Fn. By the monotone class theorem, the
identity holds true for anyA1∈F and allnand allA2,A3,· · ·,Ak∈Fn. By the monotone class theorem again, it holds for
On the event{N=k}, we can writeZn=Pki=1Zn(i−1) , whereZ (i)
n−1denotes the number of individuals in the (n−1)-th generation of the subtree rooted at thei-th individual in the first generation. Accordingly,
b P(N =k, T1∈A1,· · · ,Tk ∈Ak) = 1 mnP(N =k) k X i=1 EnZn(i−1) 1{T1∈A1,···,Tk∈Ak} o . SinceP(N=k) =pk, and E{Zn(i−1) 1{T1∈A1,···,Tk∈Ak}}=E[Zn−11{T∈Ai}] Y j6=i P(Aj) =mn−1P(b Ai)Y j6=i P(Aj)
, the desired identity follows.
Exercise 1.7. Part (i) follows from the law of large numbers, whereas part (ii) from the law of large numbers for i.i.d. non-negative random variables having infinite expectations (which can be easily checked by means
of the Borel–Cantelli lemma).
Exercise 1.8. Write Eb for integration with respect to P. Letb A ∈ Fn. Then E(ξn+11A
n) = P(b An) =
E(ξn1An). Therefore,E(ξn+1|Fn) =ξn, i.e., (ξn) is aP-martingale. Since it is non-negative, it converges
P-almost surely toξ, withξ <∞P-a.s.
Exercise 1.9. Assume firstPb ≪Pand letη:= dPb
dP. Exactly as we proved the martingale property of (ξn),
we see that ξn =E(η|Fn) P-a.s. L´evy’s martingale convergence theorem tells us thatξn →η, P-a.s. In
particular, ξ =η P-a.s., so for anyA ∈F∞, P(b A, ξ <∞) = P(b A, η <∞) = P(b A) =E(ξ1A), whereas
b
P(A∩ {ξ=∞}) = 0: this implies (1.4).
In the general case (i.e., without assuming Pb ≪ P), we use a usual trick. LetQ := 12(P+P); thusb P≪QandPb ≪Q, so that we can apply what we have just proved torn:= dP|Fn
dQ|Fn
andsn:= dPb|Fn
dQ|Fn
. Let
r := lim supn→∞rn and s:= lim supn→∞sn. According to what we have just proved, rn → r = dPdQ and
sn→s= dQdPb,Q-a.s. Sincern+sn = 1Q-a.s., we haveQ(r=s= 0) = 0. Therefore,Q-almost surely, s r = limn→∞sn limn→∞rn = lim n→∞ sn rn = limn→∞ξn=ξ. (In particular,{r= 0}={ξ=∞},{r >0}={ξ <∞},Q-a.s.)
LetA∈F∞. We have b P(A) = Z A sdQ= Z A s1{r>0}dQ+ Z A s1{r=0}dQ. Since RAs1{r>0}dQ= R
Arξ1{ξ<∞}dQ=E(ξ1{ξ<∞}1A) =E(ξ1A) (because we have already proved in
Exercise 1.8 thatξ <∞P-a.s.) andRAs1{r=0}dQ=RAs1{ξ=∞}dQ=P(b A∩ {ξ=∞}), this yields (1.4).
Exercise 1.10. IfPb ≪P, then ξ <∞ P-a.s. (since ab P-measure zero set is in this case aP-measure zerob set). By (1.4), E(ξ) = 1 (takingA := Ω). Conversely, ifE(ξ) = 1, then by (1.4),P(b ξ =∞) = 0 (taking
A:= Ω again), i.e.,ξ <∞,P-a.s. In particular, (1.4) becomesb P(b A) =E(ξ1A),∀A∈F∞; thusPb ≪P.
It remains to check (1.6). If Pb ⊥ P, then P(b Ac) = 0 = P(A) for some A ∈ F∞, in which case (1.4)
becomesP(b A) =P(b A∩ {ξ=∞}). Since P(b A) = 1, this yieldsP(b ξ=∞) = 1, i.e.,ξ=∞P-a.s. Thus (1.4)b becomes P(b A) =E(ξ1A) +P(b A), ∀A ∈F∞. In particular, takingA = Ω yields E(ξ) = 0. Conversely, if
E(ξ) = 0, then (1.4) becomes P(b A) =P(b A∩ {ξ=∞}),∀A∈F∞; thus ξ=∞P-a.s. (takingb A:= Ω). In
particular,P(b Ac) = 0 =P(A) forA:={ξ=∞} ∈F
∞ ; thusPb ⊥Pas desired. Exercise 1.11. AssumeE(log+Y1) =∞. Then by Exercise 1.7, lim sup
n→∞ logYn
n =∞a.s. SinceZn≥Yn,
it follows that for anyc >0, lim supn→∞Zn
cn =∞, a.s.
Assume nowE(log+Y1)<∞. Exercise 1.7 tells us in this case that for anyc >0,P
k Yk
ck <∞a.s.
LetY be the sigma-algebra generated by (Yn). Clearly,
E(Zn+1|Fn,Y) =mZn+Yn+1≥mZn, thus (Zn
mn) is a sub-martingale (conditionally onY), andE(mZnn|Y) =
Pn k=0 mYkk. In particular, sup n E( Zn mn |Y)<∞ on the setA:={P∞k=0 Yk
mk <∞}. As a consequence, on the setA, limn→∞mZnn exists and is finite29. Since
P(A) = 1, the result follows.
Exercise 2.2. The convexity is a straightforward consequence of H¨older’s inequality. The lower semi-
continuity follows from Fatou’s lemma.
Exercise 2.3. Letψ(·) be as in (2.7), and letψ(t) :=∞fort <0. ThenJ1(a) :=−J(−a) = supt∈R[ta−ψ(t)];
in words,J1is the Legendre transform of the convex functionψ. It is well-known (Rockafellar [49], Theorem 12.2) thatψ(t) = supa∈R[ta−J1(a)] = supa∈R[J(a)−at].
Exercise 2.4. The identity inf{a∈R: J(a) = 0} = sup{a∈R: J(a) = 0} is straightforward: the only difficulty would be ifJ(a) = 0 on an interval but this cannot happen as J is non-decreasing and concave (infimum of linear thus concave functions), henceJ is (strictly) increasing up to its maximum which isψ(0) (Exercise 2.3 witht= 0) and which is (strictly) positive.
The argument also shows thatγis the unique solution ofJ(γ) = 0.
Exercise 2.7. Recall thatJ(a) = inft≥0[at+ψ(t)]. So ψ′(0)<−a.
29We again use the fact that a sub-martingale (X
Ifψ′(β)6=−a,∀β >0, then supt≥0ψ′(t)≤ −a, so for anyb < asufficiently close toasuch thatJ(b)>0, we have supt≥0ψ′(t)< −b, thus 0< J(b) = inft≥0[bt+ψ(t)] = limt→∞[bt+ψ(t)] ≤J(a) < ψ(0), which
yieldsJ(a) = limt→∞[at+ψ(t)] = limt→∞[bt+ψ(t) + (a−b)t] =∞: contradiction.
Acknowledgments. I am grateful to the organizers of the Symposium for the kind invitation, and to my co-authors for sharing the pleasure of random climbs.