4 L I N D Y A S D I S TA N C E F R O M A N ABSORBING BARRIER he Lindy effect (or law) has been investigated through various traditions. In short it corresponds to situations where the conditional expectation of additional life expectancy increases with time, which requires the survival function of survival time to be that of a power lawa . This maps to a declining force of mortality under the standard hazard rate representation. T Here we model it as the exit time of a stochastic process (arithmetic) with drift µ and show how the force of mortality behaves with respect to the distance from absorption. We show that, while a process with a drift µ = 0 produces a Lindy survival function, any amount of negative drift makes it exit that class. (X > x) = L(x)x −α , where α > 0 and L(.) is slowly varying a That is, where X is an r.v. in R, function. 4.1 force of mortality Let τ ∈ R+ be the stopping time random variable and X = (Xt )t∈T be a Brownian motion. Let L ∈ R be a barrier below current or initial level; we define the stopping time "from above" as τ ! inf{t > 0; Xt < L}. Next we look at metrics used in the sciences, insurance industry, and engineering applications. Let us compute the distribution of the force of mortality (a.k.a. hazard rate), which maps, in our representation, to the instantaneous probability of absorption. Consider the probability of conditional stopping time (conditional on survival) in the interval (t, t + ∆t): ! τ ∈ (t, t + ∆ t) | τ > t " Bayes = ! " ! 1 τ > t | τ ∈ (t, t + ∆t) × τ (τ > t) # $% & ∈ (t, t + ∆t) " (4.1) =1 47 48 lindy as distance from an absorbing barrier X 6 4 2 Time 200 400 600 800 1000 -2 Barrier -4 Absorbed path -6 Figure 4.1: Sample paths for an arithmetic Brownian motion (ABM) subjected to an absorbing barrier (from above). An absorbed path is effectively permanently terminated and is allowed no recovery. Now let ϕ(.) be the density function for τ: ! ( t+∆t " 1 ϕ(τ)dτ τ ∈ (t, t + ∆ t) | τ > t = ' +∞ ϕ(τ)dτ t t (4.2) Allora let’s define the force of mortality as the instantaneous rate at τ, " 1 ! τ ∈ (t, t + ∆ t) | τ > t , ∆t→0 ∆t µ(τ) ! lim which maps to µ(τ) = ' +∞ t 1 ϕ(τ). ϕ(τ)dτ (4.3) The insurance letteratura works with survival functions S(τ) or exceedance probabilities, allora: S′ (τ) d µ(τ) = − =− log(S(τ)). (4.4) S(τ) dτ With µ(.) we have a way to integrate over periods. Hence by 4.4 : − ( t+k t µ(τ)dτ = log ! S(t + k) " S(t) , (4.5) and exponentiating we get the behavior of the survival function over the time increment [t, t + k]: ' S(t + k) = e− t+k t µ(τ)dτ S(t). (4.6) 4.1 force of mortality X 2 1 200 400 600 800 1000 Distance from absorption for a given sample path. Time Figure 4.2: -1 Distance from barrier -2 -3 We will return to the mortality business. For now let us examine Lindy and distributions. Let us look at the demarcation properties. Let X be a random variable that lives in either (0, ∞) or (−∞, ∞) and E the expectation operator under "real world" (physical) distribution. By classical results [5], see the exact exposition in The Statistical Consequences of Fat Tails: lim K →∞ 1 E(X | X >K ) = λ K (4.7) • If λ = 1 , X is said to be in the thin tailed class D1 and has a characteristic scale • If λ > 1 , X is said to be in the fat tailed regular variation class D2 and has no characteristic scale. We • If lim E(X | X >K ) − K = λ K →∞ where λ > 0, then X is in the borderline exponential class We can reexpress the above in terms of the force of mortality at a future period t, µ(t). t • If the time random variable belongs to the power law class with pdf → t−α−1 then µ(t) = αt , which declines with time. • If the time random variable is in the borderline exponential class with pdf ϕ(t) = e−λt , t > 0 then µ(t) = λ, which is independent of t. • If the r.v. is in the thin-tailed classes, then µ(.) should increase with time. 49 50 lindy as distance from an absorbing barrier he Sicilian Adventurer and occultist Giuseppe Balsamoa who went by the pseudonym of Count Cagliostro claimed to be immortal. He was frozen in middle age at the height of his physical and intellectual powers (although various representations such as the one above show him plump, soft and sarcopenic, that is, unacquainted with the deadlifting barbell.) In a (fully or perhaps only partially) imagined dinner by Alexandre Dumas in his historical novel Le collier de la reine (the Queen’s Necklace), Cagliostro claimed to be 3,000 years old and having partaken of all manner of historical battles. The Marquis de Condorcet (of the famous paradox) busted his reasoning, showing the difference between immortality and absence of physical aging. The slightest probability of death would cumulate to certainty, and life expectancy must be finite –and (relatively) short. However, Cagliostro appeared to have some counterargument up his sleeve. He explained the notion of experience: 3,000 years allow someone to learn to spot the signs of danger, even predict the trajectory of bullets, which does reduce the probability of absorption and causes the decline of the hazard rate over time. Effectively, people who do not age can be Lindy, hence their survival is power-law distributed. It is quite remarkable that Dumas, almost two centuries ago, captured these probabilistic subtleties while current decision-theory academics still fail to do so. T a Creative Commons. 4.2 the process 4.2 the process Feller’s second volume [7] (page 174) showed that the distribution of the time to first passage for a Brownian motion is "stable with exponent 21 ", using scalability arguments under a Markov property –in fact this would be Lindy, applied to time to hit any point for a one-dimensional Brownian motion (the distribution is Cauchy for two dimensions). We will first rederive then add negative drifts thanks to Girsanov’s theorem to gauge the distribution. 4.2.1 Process without a drift Let Xt be a stochastic process with no drift satisfying dX = √ σdB, where B is a Brownian motion X0 = 0, so by standard result Xt = X0 + σ tW0,1 and W0,1 is a standardized Gaussian r.v.
(
) 0.12 0.10 0.08 Figure 4.3: Stopping time distributions at various absorption levels. L=-2 0.06 L=-1 0.04 0.02 5 10 15 20
Let L < 0 be the level of the absorbing barrier (constant) hitting from above and τ the first passage time τ = inf {t : Xt = L} z We have, using the reflection principle (Fig. 4.2.1), the following distribution of τ: Consider the probability of X below L: ( Xt < L ) = ( X < L|τ <t ) (τ < t) + # ( X < L|τ >t ) (τ > t), $% & (4.8) =0 since by definition one cannot have a conditional probability of hitting L with τ > t, ( X < L|τ >t ) = 0 (in other words, the path is continous, one cannot be below L without having first hit L, even in Saudi Arabia). ( Xt < L ) = ( X < L|τ <t ) (τ < t) (4.9) 51 52 lindy as distance from an absorbing barrier Reflection Principle 2 X 1 2 3 4 6 5 7 -2 L -4 Figure 4.4: The reflection principle, illustrated with a random walk of step size (+1, −1). The number of paths from 0 (the origin) to X via L are equal to those from −2L to X. -6 Now by symmetry, at (or, rather "in the neighborhood of") τ, X is as likely to be above L as to be below L, so ( X < L|τ <t ) = Allora the cumulative ( X > L|τ <t ) = 1 . 2 ! " (τ < t)) = 2 ( P ( Xt < L)) = 2 CDF(L) = erf √ L√ + 1 2σ τ and the distribution of τ becomes (assuming X = 0, so it is expressed as arithmetic distance from X) − L2 ∂ (τ < t) L e 2σ2 τ ϕ(τ) = = −√ ∂τ 2π στ 3/2 " ! Noting the CDF for the stopping time, Φ(τ) = erf √ L√ + 1, we get: (4.10) 2σ τ lim τ →∞ ! " log 1 − Φ(τ) log(τ) 1 =− , 2 1 meaning that the survival function → Kt− 2 , where K is a constant.1 In fact we can see that ϕ(τ) maps to a Lévy distribution with parameters (0, 1 As it has no moments, we can get an idea of its median −1 2 2 erfc ( 21 ) σ2 1 ). σ2 We note the hazard function: − L2 ∂ log(1 − Φ(τ)) Le 2σ2 τ τ 1 ! " → µ(τ) = − = √ . L ∂τ 2τ 2πστ 3/2 erf √ √ 2σ τ 1 Feller’s reasoning using dimensionless analysis is as follows. The residual time to reach L + L′ after having hit L, which we write as τL+L′ - τL is independent of τL and has the same distribution as τL′ . As we can see in the graph in Fig. 4.2.1, the transition probability, given that is is a standard random walk, is proportional to L2 t (assuming a time increment of t). 4.2 the process Force of Mortality for a Driftless Brownian Motion 1.0 Figure 4.5: The force of mortality for a driftless Brownian Motion is not declining monotonically as the "strong Pareto property" is not reached until later in the game. 0.8 Pre-Lindy 0.6 2
1
(t) 0.4 0.2 t 0.5 1.0 1.5 2.0 2.5 3.0 Remark 7 The force of mortality is not "Lindy" over the pre-asymptotic zone in the power law log(S(x)) where the distribution log(x) is not a constant. From the standard definition of an r.v. in the power law class, one must have cleared the "Karamata point", that is the area where the slowly varying function wanes. The standard representation is (X > x) = L(x)x −α , where α > 0 and L : [ xmin , +∞) → (0, +∞) is a slowly varying function, defined as lim x →+∞ L(kx) =1 L(x) for any k > 0. In other words, we must have L(x) a constant. If L(x) is not a constant, then the force of mortality The arguments used in this section do not adapt to an arithmetic Brownian motion with drift (it would cancel the symmetry of the reflection principle), nor, besides, do they adapt to a geometric Brownian motion (see Dynamic Hedging, [22]). For that, one needs a change of measure as we will do next. 4.2.2 Adding a drift via Girsanov’s theorem Let µ < 0 be a negative drift. We have the Radon-Nikodym derivative for X̂ = µt + X, where X andX̂ are martingales under the P and Q measures, respectively: µ2 t dQ = e 2 −µXt dP (4.11) 53 54 lindy as distance from an absorbing barrier P(t ∈ dt) = EP ( t∈dt ) = EQ Now t∈dt = X̂=L , )! dQ "−1 t∈dt dP * = EQ ) e µXt − µ2 t 2 t∈dt * (4.12) allora (adjusting for σ2 since we did not normalize 4.10: P(t ∈ dt) = eµL− µ2 σ 2 t 2 Q(t ∈ dt) (4.13) ) * µ2 t L2 Le− 2t eµL− 2 √ 2πt3/2 Since, from the earlier result 4.10 we have P(t ∈ dt) = − , and since L is negative, we focus on µ negative as they must be of the same sign for P(t ∈ dt) to represent a density "from above": − ( L−µσ2 τ ) 2 Le , µ < 0. ϕµ (τ) = − √ 2π στ 3/2 2σ2 τ (4.14) Remark 8 Any amount of negative drift causes the stopping time distribution to exit the powerlaw class, hence lose the "Lindy" attribute. Proof. The moments of order p are finite for all values of µ < 0: + ! " 1 1 2 −eµL L p+ 2 µ 2 − p σ−2p K p− 1 (Lµ) (p) = 2 π (4.15) and the Bessel function K p− 1 (0) = −∞. We note the finite mean 2 (1) = L , µσ2 which is positive for all µ < 0. Remark 9 Why do we need µ to be ≤ 0 to get ϕ(.) to be a density? Simply: (t < +∞) = ( ∞ 0 − Le − √ ( L−µσ2 τ ) 2 2σ2 τ 2πστ 3/2 dτ = e !√ " µ2 +µ L . Should µ be positive, it is worse than when µ = 0 where no moment > 0 exists. Here the 0th moment would not exist as there is e2µL − 1 probability of the hitting time never taking place. This is, literally, immortality.