Equity and Sustainability: An Exposition

Introduction

Perhaps the most frequently quoted ‘definition’ of sustainable development was introduced by the World Commission on Environment and Development (Brundtland Commission, to be abbreviated as BC, 1987, p. 43):

Moreover:

The concept of sustainable development does imply limits—not absolute limits but limitations imposed by the present state of technology and social organization on environmental resources and by the biosphere to absorb the effects of human activities.

The definition has been challenged from various perspectives and, over the years, there has been a move to define, in more quantifiable terms, ‘what’ has to be sustained and ‘how’. But, it is fair to say that it highlights a concern for intergenerational equity: a commitment on the part of the ‘present’ generation to avoid actions/policies that will severely limit or perhaps irrevocably eliminate consumption possibilities of some ‘future’ generations. Such a concern suggests the need for prudent policies on conservation and efficient use of exhaustible and renewable resources, and on the long run challenge posed by environmental degradation primarily due to human activities. Quite rightly, these have been recurrent themes in the huge literature on ‘sustainability.’ In models describing an economy over time, a simple way to capture ‘equity’ is to specify the same consumption in every period (or, in continuous time models a constant rate of consumption). Given the reproduction law of a renewable resource, the first question is to identify conditions under which such equitable programmes can be sustained over a long period, perhaps indefinitely. First, we review a discrete-time model which addresses this issue without a specification of the functional form of the reproduction law. Next, we turn to the widely studied Logistic model and analyze the implications of a constant rate of harvesting. Both of these models are deterministic. The final model is stochastic. It captures the problem of living off a National or Wealth (‘oil’) Fund. Over 20 economies that rely on natural resources as the primary source of revenue from exports have already generated such funds to meet the aspirations of the future. The Norges Bank (Norway) that oversees one of the largest such Funds announced that ‘the Fund is saving for future generations in Norway. One day the oil will run out, but the return on the Fund will continue to benefit the Norwegian population.’ We attempt to characterize the probability of sustaining an equitable plan of consumption when the return from investment is random. It is to be emphasized that, despite the simplicity of the framework, designing a ‘simple’ formula to calculate the probability appears elusive. The paper is entirely expository. The discrete-time model in Section 2 was elaborated in Majumdar and Radner (1992). There are numerous accounts of the Logistic model and its applications. A useful source is the monograph by Brauer and Castillo-Chavez (2001). For a more detailed exploration of the third model and extensions, the reader is referred to Majumdar and Radner (1992), Bhattacharya and Majumdar (2007), Bhattacharya et al. (2015), and Majumdar (2017).

Sustaining an Equitable Programme

Consider an economy which starts with an initial stock y of a renewable resource (e.g., a stock of fish, an animal population, … ). In each period t , the economy plans to consume or harvest a non-negative amount c t of the beginning of the period stock y t . The remaining stock in that period x t = y t − c t is the input that regenerates the (gross) output y t + 1 = f ( x t ) where f is the regeneration function. Assume that:

[F.1] f : R → R + is continuous on R ; f ( x ) = 0 for x ≤ 0 .

[F.2] f ( x ) is twice continuously differentiable at x > 0 , f / ( x ) > 0 , f / / / ( x ) < 0 .

[F.3] lim x → 0 f / ( x ) = 1 + μ 1 , where μ 1 > 0 and lim x → ∞ f / ( x ) = 1 − μ 2 where μ 2 > 0 .

A resource management programme (briefly, a programme) from an initial stock y ̂ > 0 is described by a (non-negative) sequence ( x , y , c ) = ( x t , y t , c t ) satisfying;

y 0 = y ̂ > 0 .

(1)

x t + c t = y t for t ≥ 0 ;

(2)

y t + 1 = f ( x t ) for t ≥ 0 .

(3)

The sequence x = ( x t ) of inputs together with the initial y ̂ is said to support or sustain the non-negative sequence c = ( c t ) of consumptions (or, harvests). Now, given the initial y ̂ > 0, and the regeneration function f , if an arbitrary positive sequence c = ( c t ) of consumption pattern is specified, there may not be any non-negative sequence x = ( x t ) such that the relations (1)–(3) are satisfied. Such a consumption pattern is not sustainable.

A programme ( x , y , c ) y ̂ = ( x t , y t , c t ) y ̂ is equitable if, in addition to satisfying (1)–(3) one has:

c t = c > 0 , x t = x , y t = y ̂ = f ( x ) for all t ≥ 0

(4)

The problem of identifying equitable consumption patterns that are sustainable leads to the study of:

x t + 1 = f ( x t ) − c , x 0 = y ̂ − c .

(5)

Let T be the first period t , if any, such that the process described by (5) leads to x t < 0 ; if there is no such t , then T = ∞ . If T is finite we say that the target c is sustainable up to (but not including) period T . We say that the target c is attainable (forever) if T = ∞ (i.e., if x t ≥ 0 for all t ). Actually, we are only interested in following the process (5) up to the ‘failure’ or ‘extinction’ time T .

Proposition 2.1. (i) There is some k ̅ > 0 such that f ( x ) > x for all x satisfying 0 < x < k ̅ and f ( x ) < x for all x > k ̅ . Hence, f ( k ̅ ) = k ̅ > 0 .

(ii) f / ( x ) is decreasing on R + + .

(iii) f / ( k ̅ ) < 1 .

(iv) There is a unique k * satisfying 0 < k * < k ̅ such that f / ( k * ) = 1 . For all 0 < x < k * , f / ( x ) > 1 , and for all x > k * , f / ( x ) < 1 .

Proof. To establish (i) we want to show that:

To show (A), take x 1 > x 2 > 0 . Then, for some t ∈ ( 0 , 1 ) , x 2 = t · x 1 + ( 1 − t ) · 0 . Since f is strictly concave (from F.2, refer to Nikaido, 1968, Chapter 1, Section 3.5), we have

f ( x 2 ) = f ( t · x 1 + ( 1 − t ) · 0 ) > t · f ( x 1 ) + ( 1 − t ) · f ( 0 ) = t · f ( x 1 )

Hence,

f ( x 2 ) / x 2 > [ t · f ( x 1 ) ] / x 2 = [ t · f ( x 1 ) ] / t · x 1 = f ( x 1 ) / x 1 .

Thus, we have established (A).

To establish (B), observe that there is some x ̅ > 0 such that f ′ ( x ) > 1 for all x ∈ ( 0 , x ̅ ] . By the mean value theorem, there exists z such that:

f x ¯ = f 0 + x ¯ f ′ z , 0 < z < x ¯ , = x ¯ f ′ z > x ¯ f ′ x ¯ , o r , f x ¯ / x ¯ > A x ¯ > f ′ x ¯ > 1.

Thus, we have established (B).

To establish (C) note that if f ( x ) is bounded, that is, if there is some N > 0 such that f ( x ) ≤ N for all x > 0 , then [ f ( x ) / x ] ≤ [ N / x ] for all x > 0 . Hence, there is some x ′ > 0 (‘sufficiently large’)

such that [ f ( x ′ ) / x ′ ] ≤ [ N / x ′ ] < 1 . Otherwise, f ( x ) → ∞ as x → ∞ , and, by L’ Hospital's rule, lim x → ∞ [ f ( x ) / x ] = lim x → ∞ f / ( x ) = 1 − θ 2 < 1 . Hence, there is some x ′ > 0 such that [ f ( x ′ ) / x ′ ] = A ( x ′ ) < 1 .

By the intermediate value theorem, there is some k ̅ satisfying x ′ > k ̅ > x ̅ > 0 such that A ( k ̅ ) = 1 , that is , f ( k ̅ ) = k ̅ . From (A) we get:

(a) A ( x ) > 1 for 0 < x < k ̅ , or, f ( x ) > x for 0 < x < k ̅ ; and (b) A ( x ) < 1 for x > k ̅ , or, f ( x ) < x for x > k ̅ . . This completes the verification of ( i ) .

(ii) follows from the fact that f ′ ′ ( x ) < 0 (refer to Nikaido 1968, p. 49).

(iii) Using the Mean value Theorem, we see that f ( k ̅ ) − f ( 0 ) = k ̅ · f / ( z ) , where 0 < z < k ̅ . This leads to:

k ̅ = f ( k ̅ ) = k ̅ · f / ( z ) > k ̅ · f / ( k ̅ ) , or , 1 > f / ( k ̅ )

To establish (iv), use (F3) and the Intermediate Value Theorem to assert the existence of k * such that f / ( k * ) = 1 . Uniqueness and the other properties follow from (ii).  ▪

Define the net return function h ( x ) = f ( x ) − x . It follows that h satisfies

h ( x ) ≥ < 0 for [ c ] c 0 < x < k ̅ ; x = 0 , k ̅ ; x > k ̅ .

Since f ( x ) < 0 for all x ≤ 0 , all statements about f and h will be understood to be for non-negative arguments unless something explicit is said to the contrary.

The maximum sustainable harvest or consumption is

H = max [ 0 , k ̅ ] h ( x ) .

(6)

Observe that h ( x ) > 0 on ( 0 , k ̅ ) . Hence, H > 0 , and the maximum is attained at some point(s) in ( 0 , k ̅ ) , as h ( 0 ) = h ( k ̅ ) = 0 . Let ξ be some point in ( 0 , k ̅ ) such that h ( ξ ) = H . Then

h / ( ξ ) = 0 , or , , f / ( ξ ) = 1

(7)

From Proposition 2.1 (iv), we can conclude that ξ = k * and h ( k * ) > h ( x ) for all x ∈ [ 0 . k * ] such that x ≠ k * . Also, h / ( x ) > 0 for 0 < x < k * , and h / ( x ) < 0 for k * < x < k ̅ .

We write

x t + 1 = x t + h ( x t ) − c .

If c > H , x t + 1 − x t = h ( x t ) − c < H − c < 0 . Hence, x t will fall below (extinction) 0 after a finite number of periods.

On the other hand, if 0 < c < H , we show that there will be two roots ξ ′ and ξ ′ ′ of the equation

c = h ( x )

which have the properties

0 < ξ ′ < k * < ξ ′ ′ < k ̅ .

First, observe that h ( 0 ) = 0 < c < H = h ( k * ) . By the Intermediate Value Theorem, there is some ξ ′ , 0 < ξ ′ < k * , satisfying c = h ( ξ ′ ). Since h ′ ( x ) > 0 , for 0 < x < k * , there cannot be two distinct points ξ ′ , ξ ′ 1 in the interval ( 0 , k * ) satisfying c = h ( ξ ′ ) = h ( ξ ′ 1 ) . If there are, and, say, ξ ′ > ξ ′ 1 , we have an immediate contradiction, by using the Mean Value Theorem,

0 = h ( ξ ′ ) − h ( ξ ′ 1 ) = ( ξ ′ − ξ ′ 1 ) h ′ ( z ) > 0 , where ξ ′ > z > ξ ′ 1

By a similar argument there is one and only one point ξ ′ ′ satisfying k * < ξ ′ ′ < k ̅ and c = h ( ξ ′ ′ ).

We write y ′ = c + ξ ′ . If y 0 = y ′ , set x 0 = ξ ′ . Since c = h ( ξ ′ ), we get x 1 = x 0 + h ( x 0 ) − c = ξ ′ + h ( ξ ′ ) − c = x 0 = ξ ′ . Repeating the argument we see that x t = ξ ′ , y t = y ′ = c + ξ ′ . Hence, the target c is sustainable forever. It follows that whenever y 0 ≥ c + ξ ′ , the target c is sustainable forever. Indeed, note that c = h ( ξ ′ ) is equivalent to c + ξ ′ = f ( ξ ′ ) . For any y 0 > c + ξ ′ , set x 0 = y 0 − c > ξ ′ . Hence, y 1 = f ( x 0 ) > f ( ξ ′ ) = c + ξ ′ > c . Again, x 1 = f ( x 0 ) − c > ξ ′ . This leads to y 2 = f ( x 1 ) > f ( ξ ′ ) = c + ξ ′ > c . Repeating the steps we see that y t > c for all t ≥ 1 , confirming that the target c is sustainable forever. We can, in fact, show that x t converges monotonically to ξ ′ ′ .

On the other hand, consider y 0 < c + ξ ′ . Then, x 0 = y 0 − c < ξ ′ , If x 0 > 0 , writing c − h ( x 0 ) = β > 0

x 1 = x 0 + h ( x 0 ) − c = x 0 − β ,

Either x 1 ≤ 0 , or, x 1 > 0 . In the latter case, h ( x 1 ) < h ( x 0 ) , and we get

x 2 = x 1 + h ( x 1 ) − c < x 1 + h ( x 0 ) − c = x 1 − β = x 0 − 2 β

Repeating the argument we see that x t ≤ 0 after a finite number of periods.

The implications of the foregoing discussion for sustainability and extinction are summarized as follows.

Theorem 2.1. Let c > 0 be the planned consumption for every period and H the maximum sustainable consumption.

Example 2.1. Consider the following family:

For x ≥ 0 , f x = θ 1 x x + θ 2 where the parameters θ 1 , θ 2 satisfy θ 1 > θ 2 > 0. For x ≤ 0 , f x = 0.

Here , it is easy to verify for all values of θ 1 , θ 2 the family satisfies (F2) and (F3). f / ( x ) = θ 1 · θ 2 / [ x + θ 2 ] 2 and f / / ( x ) = − 2 θ 1 θ 2 / ( x + θ 2 ∖ 2 ) 3 .

Then, h ( x ) = f ( x ) − x = x [ ( θ 1 − θ 2 ) − x ] / ( x + θ 2 ) . So, h ( 0 ) = h ( θ 1 − θ 2 ) = 0 . Clearly , h ( x ) . > 0 on ( 0 , θ 1 − θ 2 ) .

Choosing θ 1 = 4 , θ 2 = 1 , we get h ( x ) = ( 3 x − x 2 ) / ( x + 1 ) . The k ̅ appearing in Proposition 2.1 is k ̅ = θ 1 − θ 2 = 3 . On [ 0 , 3 ] , h ( x ) attains a maximum value at k * = 1 . This value H = 1 . Hence, no target c > 1 is attainable. Computation of the critical stock ξ / corresponding to different values of c in ( 0 , 1 ) is left as an exercise.

The Logistic Population Model

Let x ( t ) be the size of the population at time t , and its rate of change is denoted by x / (or, x ˙ , or, dx / dt ) . We look at an early, influential effort to link the growth rate to the size of a population. The logistic model is a formal description of the dynamic process of population change in terms of a differential equation:

dx / dt ≡ x ˙ = rx ( t ) [ 1 − ( x ( t ) / K ) ]

(8)

Here the parameters r and K are positive and will be interpreted shortly. Informally, when x is ‘rather small’ relative to K , one has x ˙ ≈ rx , whereas when x is ‘close to’ K , one has x ˙ ≈ 0 . In other words, when population size x is small, it experiences exponential growth, whereas when x is near K , its change is negligible. Of course, from the point of view of interpretation, we need x ( t ) ≥ 0 . Let us denote the initial population size (i.e., size at time t = 0 ) x ( 0 ) by x 0 ≥ 0 . The Equation (8) can be solved by using the method of ‘separation of variables’. For us, it is enough to point out that, by direct substitution in (8), we can verify that the solution to (8), given x 0 ≥ 0 , is

x ( t ) = Kx 0 e rt / [ K − x 0 + x 0 e rt ] = Kx 0 / [ x 0 + ( K − x 0 ) e − rt ]

(9)

Examining the solution (9), we see that (a) 0 and K are two steady states (b) the population size x ( t ) approaches the limit K as t → ∞ if the initial x 0 > 0 . The value K is often called the carrying capacity of the population: it represents the population size that the environment (available resources) can carry or sustain. The value r is called the intrinsic growth rate because it represents the per capita growth rate achieved when the population size is small enough to ensure negligible resource constraints. Note finally, that if x 0 > 0 , x ( t ) > 0 for all t . Thus, left on its own, there is no possibility of extinction: in the long run, the population stays close to its carrying capacity.

We now study the effect of harvesting (removing, consuming … ) on the population. The simple case is constant yield harvesting, described formally by the differential equation:

x ˙ = rx ( t ) [ 1 − ( x ( t ) / K ) ] − Θ

(10)

where Θ is the positive constant rate of harvesting. It is to be stressed that the rate does not depend on t . The steady state or equilibrium values of (10) are given by given by the condition x ˙ = 0 , and these may be found by solving the quadratic equation rx [ 1 − ( x / K ) ] − Θ = 0 , or , x 2 − Kx + ( K Θ ) / r = 0 . There are two equilibria:

x L = [ K − ( K 2 − ( 4 Θ K / r ) ) 1 / 2 ] / 2 and x U = [ K + ( K 2 − ( 4 Θ K / r ) ) 1 / 2 ] / 2

(11)

provided K 2 − ( 4 Θ K / r ) ≥ 0 or Θ ≤ rK / 4 . If Θ > rK / 4 , x ˙ < 0 for all initial x 0 ≥ 0 , and every solution ‘hits’ 0 in finite time. If a solution ‘hits’ zero at a (finite) time T , we say that the population is extinct at time T . If 0 ≤ Θ ≤ rK / 4 , the upper equilibrium, x U decreases from K to K / 2 as Θ increases from 0 to rK / 4 , and the lower equilibrium x L increases from 0 to K / 2 as Θ increases from 0 to rK / 4 . Writing Θ ̂ = rK / 4 , a critical value, we can summarize the qualitative properties of the trajectories and the possibilities of extinction as follows. For any harvest rate Θ < Θ ̂ , the population size tends to the equilibrium size x U , provided x 0 ≥ x L . For all x 0 < x L , population size hits zero in finite time. For Θ > Θ ̂ , the population size hits zero in finite time for all initial population sizes (i.e., ruin is inevitable due to overharvesting). Despite the simplicity of the example, two implications stand out for the policymaker/social planner concerned about extinction: first, if the population size is already small (namely below x L ) , an immediate intervention, a ban on harvesting, is called for. Second, while a modest harvesting rate is consistent with growth (provided the initial size is above the threshold stock x L ) , monitoring is essential if the harvest rate keeps increasing. When Θ is close to Θ ̂ , a small change in the rate may have a disastrous impact on the population.

In many expositions, the logistic model is specified as:

x ˙ = x ( t ) [ λ − ax ( t ) ]

(12)

where λ and a are parameters (positive). Comparing (10) and (12) we get:

λ = r , a = r / K

(13)

The critical value Θ ̂ = λ 2 / 4 a . For any Θ > Θ ̂ , the extinction time T from an initial x 0 is given by (refer to Brauer & Sanchez, 1975, p. 16):

T = ∫ 0 x 0 du / [ au 2 − λ u + Θ ] = 2 / [ 4 a Θ − λ 2 ] 1 / 2 [ arctan { ( 2 ax 0 − λ ) / ( 4 a Θ − λ 2 ) 1 / 2 } + arctan { ( λ / ( 4 a Θ − λ 2 ) 1 / 2 } .

(14)

Despite the simplicity of the model, it has been used to compare the projections on population size and predicted time of extinction due to overharvesting. Brauer and Sanchez reported some contrasting numerical estimates from a study by Miller and Botkin (1974). The Miller–Botkin model, depending on 10 parameters described the dynamics of the population of sandhill cranes. They had performed computer simulation to estimate, among other things, the extinction times for harvests greater than the critical Θ ̂ .

‘Observations suggest that with no harvesting the equilibrium population ( λ / a ) is approximately 194,6000, and the critical harvest Θ ̂ = 4 , 800 per year’. Measuring in units 1,000. we have:

λ / a = 194.6 , Θ ^ = λ 2 / 4 a , l e a d i n g t o λ = 0.0987 , a = 5.07 × 10 − 4 .

(15)

Brauer–Sanchez compared the extinction times predicted by the logistic model and the Miller–Botkin model corresponding three possible rates above the critical Θ ̂ :

Θ T L o g i s t i c T M i l l e r B o t k i n 6 90 71 8 44 38 12.77 21 19

(16)

Living off a Wealth Fund

First, let us pose the sustainability problem in the deterministic case. An economy starts with a ‘Wealth Fund’ of size y > 0 . From this, a positive quantity c is subtracted. The parameter c is a datum: it is a target consumption level that the economy wishes to sustain in each period. If the remainder i = y − c is zero or negative, the Fund is ‘exhausted’ (bankrupt). If the remainder is strictly positive, it is interpreted as an an ‘investment’. The return from the investment is then the size of the Fund at the beginning of the next period and is given by y 1 = r · i = r · ( y − c ) , where r > 0 is also a parameter (return factor). Again, in period one, the parameter c is subtracted from y 1 , and the story is repeated. Let N be the first period, if any, such that y N < 0 . If N is finite, we say that the Fund can support or sustain c up to (but not including) the period N from the initial y > 0 . If N is infinite (i.e., y n > 0 for all n ), we say that the consumption target c is sustainable from y (or, the Fund can support c forever).

We have the following

Proposition 3.1. If r ≤ 1 , no target c > 0 is sustainable from any y . If r > 1 , the economy can sustain c > 0 forever if and only if y ≥ [ r / ( r − 1 ) ] c .

Proof. Suppose the Fund can sustain c in the first two periods. Then:

c + i 0 = y , where i 0 > 0 ; c + i 1 = r · i 0 where i 1 > 0

(17)

It follows that c ( 1 + 1 / r ) + i 1 / r = y , leading to:

c ( 1 + 1 / r ) < y .

Hence, if the Fund can sustain c up to period N , we must have

∑ n = 0 N − 1 ( 1 / r n ) < y / c .

(18)

From (4.2) the conclusion follows.▄

To extend the scope of our analysis, suppose that the returns from the investment are uncertain rather than deterministic. We model this by introducing an i.i.d. sequence ϵ n of positive random variables. An investment i n in period n generates the return Y n + 1 according to the rule Y n + 1 = ( ϵ n + 1 ) · i n . As in the deterministic case, the Fund starts with an initial reserve y , and has a target consumption c . It is exhausted if ( y − c ) ≤ 0 . If y − c > 0 , then the size of the Fund in period one is Y 1 = ϵ 1 ( y − c ) . Again, if Y 1 − c ≤ 0 , it is exhausted. Otherwise, after consumption, Y 1 − c is invested to generate Y 2 = ϵ 2 · ( Y 1 − c ) . In general, one studies the process;

Y 0 = y , Y n + 1 = ( ϵ n + 1 ) ( Y n − c ) + , where a + = max ( a , 0 ) .

(19)

If y > c , the probability of sustaining c is defined as

ρ ( y ) = P ( Y n > c for all n ≥ 0 | Y 0 = y ) .

(20)

From (19), successive iteration yields

Y n + 1 > c iff Y n > c + c ϵ n + 1 iff Y n − 1 > c + c + c / ϵ n + 1 ϵ n … iff Y 0 ≡ y > c + c ϵ 1 + c ϵ 1 ϵ 2 + … + c ϵ 1 ϵ 2 … ϵ n + 1 .

Hence,

{ Y n > c for all n } = { y > c + c ∑ j = 1 n 1 ϵ 1 ϵ 2 … ϵ j for all n } = { y > c + c ∑ n = 1 ∞ 1 ϵ 1 ϵ 2 … ϵ n } = { ∑ n = 1 ∞ 1 ϵ 1 ϵ 2 … ϵ n < y c − 1 }

In other words,

ρ ( y ) = P ( ∑ n = 1 ∞ 1 ϵ 1 ϵ 2 … ϵ n < y c − 1 ) .

(21)

First, we review some conditions on the common distribution of ϵ n under which one has ρ ( x ) = 0 for all y > 0 . The Strong Law of Large Numbers gives

1 n ∑ r = 1 n log ϵ r ⟶ a . s . E log ϵ 1 .

Thus, if E log ϵ 1 < 0 , ϵ 1 ϵ 2 … ϵ n → 0 a.s. This implies that the infinite series in (21) diverges a.s., that is,

ρ ( y ) = 0 for all y if E log ϵ 1 < 0 .

(22)

. By Jensen's Inequality, E log ϵ 1 ≤ log E ϵ 1 , with strict inequality if ϵ 1 is non-degenerate, which we assume. Therefore, E ϵ 1 ≤ 1 implies E log ϵ 1 < 0 , so that ρ ( y ) = 0 for all y > 0 .

Next, let us consider the case E log ϵ 1 > 0 . This case turns out to be surprisingly challenging. Define

m : = inf { z ≥ 0 : P ( ϵ 1 ≤ z ) > 0 } .

(23)

We will show that

ρ ( y ) < 1 for all y , if m ≤ 1 .

(24)

Fix A > 0 , however large. One can find n 0 such that n 0 > A ∏ r = 1 ∞ ( 1 + r − 2 ) as ∏ ( 1 + r − 2 ) < exp { ∑ r − 2 } < ∞ . If m ≤ 1 then P ( ϵ 1 ≤ 1 + r − 2 ) > 0 for all r ≥ 1 . Consequently,

0 < P ( ϵ r ≤ 1 + r − 2 for 1 ≤ r ≤ n 0 ) ≤ P ( ∑ r = 1 n 0 1 ϵ 1 ϵ 2 … ϵ r ≥ ∑ r = 1 n 0 1 ∏ j = 1 r ( 1 + 1 / j 2 ) ) ≤ P ( ∑ r = 1 n 0 1 ϵ 1 ϵ 2 … ϵ r ≥ n 0 ∏ j = 1 ∞ ( 1 + 1 / j 2 ) ) ≤ P ( ∑ r = 1 n 0 ( ϵ 1 ϵ 2 … ϵ n ) − 1 > A ) ≤ P ( ∑ r = 1 ∞ ( ϵ 1 ϵ 2 … ϵ r ) − 1 > A ) .

Since A is arbitrary, the right side of (21) is less than 1 for all y , proving (24). One may also prove that, for m > 1 ,

ρ y < 1 if y < c ( m m − 1 ) , = 1 if y ≥ c ( m m − 1 ) . ( m > 1 ) .

(25)

Observe that ∑ n = 1 ∞ ( ϵ 1 ϵ 2 … ϵ n ) − 1 ≤ ∑ n = 1 ∞ m − n = 1 / ( m − 1 ) , with probability 1 if m > 1 . The second relation in (25) is implied by (21). In order to prove the first relation in (25), let y < cm / ( m − 1 ) − c δ for some δ > 0 . Then, ( y / c ) − 1 < 1 / ( m − 1 ) − δ . One can choose n ( δ ) such that ∑ r = n ( δ ) ∞ m − r < δ / 2 and then choose δ r > 0 ( 1 ≤ r ≤ n ( δ ) − 1 ) such that

∑ r = 1 n ( δ ) − 1 1 ( m + δ 1 ) … ( m + δ r ) > ∑ r = 1 n ( δ ) − 1 1 m r − δ 2 .

Then

0 < P ( ϵ r < m + δ r for 1 ≤ r ≤ n ( δ ) − 1 ) ≤ P ( ∑ r = 1 n ( δ ) − 1 1 ϵ 1 … ϵ r > ∑ r = 1 n ( δ ) − 1 1 m r − δ 2 ) ≤ P ( ∑ r = 1 ∞ 1 ϵ 1 … ϵ r > ∑ r = 1 ∞ 1 m r − δ ) = P ( ∑ r = 1 ∞ 1 ϵ 1 … ϵ r > 1 m − 1 − δ ) .

For δ > 0 small enough, the last probability is smaller than P ( ∑ ( ϵ 1 … ϵ r ) − 1 > y / c − 1 ) provided y / c − 1 < 1 / ( m − 1 ) , that is, if y < cm / ( m − 1 ) . For such y , 1 − ρ ( y ) > 0 . This establishes the first relation in (25). The following proposition summarizes the above results:

Proposition 4.1. (A) If E log ϵ 1 ≤ 0 , then ρ ( y ) = 0 for all y and c .

(B) If E log ϵ 1 > 0 , then

ρ ( y ) = < 1 for all y , if m ≤ 1 , if y < c ( m m − 1 ) if m > 1 = 1 if y ≥ c ( m m − 1 ) if m > 1 .

For a more detailed exploration of ρ ( y ) , the reader is referred Bhattacharya et al. (2015).