On tails of perpetuities

0.09 MB·

by  Pawel Hitczenko

ON TAILS OF PERPETUITIES
PAWEL HITCZENKO†
1 Abstract. We establish an upper bound on the tails of a random variable that arises
1 as a solution of a stochastic difference equation. In the non–negative case our bound is
0 similar to a lower bound obtained by Goldie and Gru¨bel in 1996.
2

n
a
J
1. Introduction
1
1
A random variable R satisfying the distributional identity

]
R (1) R =d MR+Q,
P
. where (M,Q) are independent of R on the right-hand side and =d denotes the equality in
h
t distribution, is referred to as perpetuity and plays an important role in applied probability.
a
m The main reason for this is that it appears as a limit in distribution of a sequence (R )
n
[ given by

d
1 R = M R +Q , n ≥ 1,
n n n−1 n
v
3 provided that limit exists (here, (Mn,Qn) is a sequence of i.i.d. random vectors distributed
8 like (M,Q) and R could be an arbitrary random variable; for convenience we will set
0
1
R = 0). Systematic study of properties of such sequences was initiated by Kesten in [5]
2 0
. and they continue till this day. Once the convergence in distribution of (R ) is established,
1 n
0 at the center ofthe investigation is the tail behavior of R. There aretwo distinctly different
1 cases:
1
: P(|M| > 1) > 0 and P(|M| ≤ 1) = 1.
v
i The first results in R having a heavy tail distribution, that is
X
r P(|R| > x) ∼ Cx−κ,
a
for a suitably chosen constant κ and some constant C (see the original paper of Kesten
[5] or [2]), while in the second case the tails of R are no heavier than exponential. This
was observed by Goldie and Gru¨bel in [3]. Some subsequent work is in [4], but the full
picture in this case is not complete. The purpose of this note is to shed some additional
light on that case by establishing a universal upper bound on the tails of |R|. In a special,
but important, situation when Q and M (and thus also R) are non–negative our bound is
comparable to a lower bound obtained by Goldie and Gru¨bel in [3].
1991 Mathematics Subject Classification. 60E15,60H25.
Key words and phrases. perpetuity, stochastic difference equation, tail behavior.
† Supported in part by the NSA grant #H98230-09-1-0062.
1
2 PAWEL HITCZENKO
2. Bounds on the tails
For a random variable M such that |M| ≤ 1 and 0 < δ < 1 define p := P(1 − δ ≤
δ
|M| ≤ 1). Then, as has been shown in [3] (see also the equation (2.2) in [4]) if 0 ≤ M ≤ 1
and Q ≡ q (q being a positive constant), then for 0 < c < 1 and x > q we have
ln(1−c)
P(R > x) ≥ exp( lnp ).
cq/x
ln(1−cq/x)
Since ln(1− cq/x) ≤ −cq/x, for any particular value of c, say c = 1/2, this immediately
gives
ln(1−c) 2ln2
P(R > x) ≥ exp(− xln(p )) = exp( xlnp ).
cq/x q/(2x)
cq q
Our aim here is to supply an upper bound of a similar form. While our result does not
give the asymptotics of P(R > x) as x → ∞, it shows that it essentially behaves like
exp(c1xlnp ) for some positive constants c ,c . Specifically, we prove
q c2q/x 1 2
Proposition 1. Assume |Q| ≤ q, |M| ≤ 1, and let R be given by (1). Then, for sufficiently
large x
1
P(|R| > x) ≤ exp( xlnp ).
2q/x
4q
Thus, if Q ≡ q > 0 and 0 ≤ M ≤ 1 then
2ln2 1
exp( xlnp ) ≤ P(R > x) ≤ exp( xlnp ).
q/(2x) 2q/x
q 4q
Proof. If P(|M| = 1) > 0 then, as was proved in [3], R has tails bounded by those of an
exponential variable, so we assume that |M| has no atom at 1. Fix 0 < δ < 1 and define a
sequence (T ) as follows
k
T = 0, T = inf{k ≥ 1 : |M | ≤ 1−δ}, m ≥ 1.
0 m Tm−1+k
Then T ’s are i.i.d. random variables, each having a geometric distibution with parameter
k
1 − p . Furthermore, |M | ≤ 1 − δ if k = T + ··· + T for some i ≥ 1 and |M | ≤ 1
δ k 1 i k
otherwise. Therefore,
m
Y|Mk| ≤ (1−δ)j for T1 +···+Tj ≤ m < T1 +···+Tj +Tj+1.
k=1
This in turn implies that
k−1 k−1
(cid:12)(cid:12)XYMj(cid:12)(cid:12) ≤ XY|Mj| ≤ T1 +(1−δ)T2 +(1−δ)2T3 +··· = X(1−δ)k−1Tk.
(cid:12)k≥1 j=1 (cid:12) k≥1 j=1 k≥1
Therefore, if |Q| ≤ q we get
k−1
x x
(2) P(|R| > x) ≤ P(XY|Mj| ≥ ) ≤ P(XTk(1−δ)k−1 ≥ ).
q q
k≥1 j=1 k≥1
ON TAILS OF PERPETUITIES 3
To bound the latter probability we use a widely known argument (our calculations follow
[1, proof of Proposition 2]). First, if T is a geometric variable with parameter 1−p then
∞ ∞
eλ(1−p) eλ
EeλT = XeλjP(T = j) = Xeλjpj−1(1−p) = = ,
1−eλp 1− p (eλ −1)
j=1 j=1 1−p
provided eλp < 1. Thus, writing t in place of x/q in the right-hand side of (2), for λ > 0
we have
P(X(1−δ)k−1Tk ≥ t) = P(exp(λX(1−δ)k−1Tk) ≥ eλt) ≤ e−λtEeλPk≥1Tk(1−δ)k−1.
k≥1 k≥1
If λ satisfies eλp < 1 then peλ(1−δ)k−1 < 1 for every k ≥ 1 as well, and by independence of
(T ), the expectation on the right is
k
∞ eλ(1−δ)k−1 ∞ 1
(3) Y = eλ/δY .
1− p (eλ(1−δ)k−1 −1) 1− p (eλ(1−δ)k−1 −1)
k=1 1−p k=1 1−p
Now, choose λ > 0 so that p (eλ−1) ≤ 1. Then, as 1/(1−u) ≤ e2u for 0 ≤ u ≤ 1/2, for
1−p 2
every k ≥ 1 we get
1 p
≤ exp(2 (eλ(1−δ)k−1 −1)).
1− p (eλ(1−δ)k−1 −1) 1−p
1−p
Therefore, the rightmost product in (3) is bounded by
p
exp(2 X(eλ(1−δ)k−1) −1)).
1−p
k≥1
We bound the sum in the exponent as follows
λj(1−δ)(k−1)j λj
XX = X X(1−δ)j(k−1)
j! j!
k≥1 j≥1 j≥1 k≥1
λj 1 1 λj eλ −1
= X ≤ X = .
j! 1−(1−δ)j δ j! δ
j≥1 j≥1
Combining the above estimates we get that
λ 2p eλ −1
(4) P(|R| > qt) ≤ exp(−tλ+ + ).
δ 1−p δ
provided that λ satisfies the required conditions, that is:
p 1
eλp < 1 and (eλ −1) ≤ .
1−p 2
Clearly both are satisfied when eλp ≤ 1/2.
We finish the proof by making a suitable choice of λ. Since we are assuming that |M|
has no atom at 1 and we are interested in large x, we may assume that δ is small enough
4 PAWEL HITCZENKO
so that p < 1/3. This condition implies that 2p /(1−p ) < 3p so that the last term in
δ δ δ δ
the exponent of (4) is bounded by 3p (eλ −1)/δ. Now let t = 2/δ. Then (4) becomes
δ
2 λ 2p eλ −1 1
P(|R| > qt) ≤ exp(−λ + + δ ) ≤ exp(− (λ−3p (eλ −1))).
δ
δ δ 1−p δ δ
δ
Set λ = ln( 1 ) so that eλp = 1. This choice of λ is within the constraints and maximizes
3pδ δ 3
the value of λ−3p (eλ −1), this maximal value being
δ
1 1 1 1
ln( )−3p ( −1) = ln( )−(1+ln3)+3p ≥ ln(1/p ),
δ δ δ
3p 3p p 2
δ δ δ
with the inequality valid for sufficiently small p (less than e−2/9 for example). Thus, using
δ
t = 2/δ we finally obtain
1 t
P(|R| > qt) ≤ exp(− ln(1/p )) = exp( lnp ),
δ 2/t
2δ 4
or, in terms of x,
x
P(|R| > x) ≤ exp( lnp ).
2q/x
4q
(cid:3)
References
[1] Goh, W. M. Y. and Hitczenko, P. (2008).Randompartitions with restrictedpartsizes. Random
Structures Algorithms. 32, 440–462.
[2] Goldie, C. M.(1991).Implicitrenewaltheoryandtailsofsolutionsofrandomequations.Ann.Appl.
Probab. 1, 126–166.
[3] Goldie, C. M. and Gru¨bel, R. (1996). Perpetuities with thin tails. Adv. in Appl. Probab. 28,
463–480.
[4] Hitczenko, P. and Weso lowski, J. (2009). Perpetuities with thin tails, revisited. Ann. Appl.
Probab. 19, 2080 – 2101. (Corrected version available at http://arxiv.org/abs/0912.1694.)
[5] Kesten,H.(1973).Randomdifferenceequationsandrenewaltheoryforproductsofrandommatrices.
Acta Math. 131, 207–248.
Pawe l Hitczenko, Departments of Mathematics and Computer Science, Drexel Univer-
sity, Philadelphia, PA 19104, U.S.A
E-mail address: [email protected]
URL: http://www.math.drexel.edu/∼phitczen