Derivations for Week 6

The Effort Function

e = \begin{cases} \displaystyle \left(\frac{\,w - x\,}{\,x\,}\right)^{\beta} & \text{if } w > x,\\ 0& \text{otherwise} \end{cases} \tag{1.1}

where x =(1 - bu)w_{a}

e represents the effort exerted by workers.
w is the wage paid by the firm.
w_a is the wage paid by other firms.
x represents the index of alternative job opportunities or outside options.
u is the unemployment rate.
\beta is a parameter that determines how sensitive effort is to relative wages.

The term \frac{w - x}{x} represents how much the wage exceeds the alternative option x as a fraction of x. Workers exert higher effort when their wage is significantly better than their alternative options.

Differentiation of the Effort Function

To determine the elasticity of effort with respect to wages, we differentiate e(w, w_a, u) with respect to w. Using the power rule:

\frac{\partial e}{\partial w} = \beta \left( \frac{w - x}{x} \right)^{\beta - 1} \cdot \frac{1}{x}

Since the elasticity condition requires that:

\frac{w}{e} \frac{\partial e}{\partial w} = 1

we substitute e = \left( \frac{w - x}{x} \right)^\beta:

\frac{w}{\left( \frac{w - x}{x} \right)^\beta} \cdot \beta \left( \frac{w - x}{x} \right)^{\beta - 1} \cdot \frac{1}{x} = 1

Rewriting:

\beta \cdot \frac{w}{x} \cdot \left( \frac{w - x}{x} \right)^{\beta - 1} = 1 \tag{1.2}

This is equation Equation 1.1.

Intuition

The left-hand side represents the elasticity of effort with respect to wages.
The term \beta \cdot \frac{w}{x} represents how effort changes with a small increase in wages.
Since firms optimize wages to maximize effort per dollar, this condition ensures that effort increases just enough to make further wage increases unnecessary.

Solving for the Equilibrium Wage Equation 1.1

We now solve for w. Rewriting Equation 1.1:

\beta \cdot \frac{w}{x} \cdot \left( \frac{w}{x} - 1 \right)^{\beta - 1} = 1

Let:

z = \frac{w}{x}

so that z - 1 = \frac{w - x}{x}. Substituting:

\beta z \cdot (z - 1)^{\beta - 1} = 1

Solving for z, we approximate:

z = \frac{1}{1 - \beta}

Thus:

w = x \cdot \frac{1}{1 - \beta}

Since x is given by:

x = (1 - b u) w_a

we substitute:

w = \frac{(1 - b u) w_a}{1 - \beta} \tag{1.3}

Intuition

w depends on x, which represents job market conditions: If alternative opportunities x increase, firms must raise wages to maintain effort.
The fraction \frac{1}{1 - \beta} acts as a markup factor: A smaller \beta (meaning effort is less sensitive to wages) leads to a larger wage premium.
b u reduces wages: Higher unemployment reduces alternative opportunities, so firms pay less.

Imposing the Equilibrium Condition

At equilibrium, firms must pay the prevailing wage, meaning w = w_a. Substituting this into Equation 1.3:

w_a = \frac{(1 - b u) w_a}{1 - \beta}

Multiplying both sides by (1 - \beta):

(1 - \beta) w_a = (1 - b u) w_a \tag{1.4}

Dividing by w_a (assuming w_a > 0):

1 - \beta = 1 - b u

Rearrange to solve for u:

u = \frac{\beta}{b} \tag{1.5}

Intuition

Higher \beta (higher sensitivity of effort to wages) requires a higher unemployment rate to sustain the equilibrium.
Higher b (weight placed on outside wages) reduces equilibrium unemployment, because workers care more about market conditions.

Equilibrium Unemployment Rate

Equation Equation 1.5 directly gives:

u_{EQ} = \frac{\beta}{b} \tag{1.6}

Thus, equilibrium unemployment is determined by the sensitivity of effort \beta and the importance of external wages b.

Intuition

If \beta is high, effort is very sensitive to wages, requiring higher unemployment to discipline workers.
If b is high, workers care a lot about external wages, making firms increase wages, which reduces unemployment.
This equation provides a structural link between effort, wages, and unemployment.

Final Summary of Key Results

Equation	Mathematical Form	Meaning
Equation 1.2	\beta \cdot \frac{w}{x} \cdot \left( \frac{w}{x} - 1 \right)^{\beta - 1} = 1	Elasticity condition for effort
Equation 1.3	w = \frac{(1 - b u) w_a}{1 - \beta}	Wage equation: how wages depend on market conditions
Equation 1.5	u = \frac{\beta}{b}	Equilibrium wage setting condition
Equation 1.6	u_{EQ} = \frac{\beta}{b}	Unemployment rate at equilibrium

Probability of Job Survival

Deriving the Exponential Decay Function

Exponential decay describes a process where the probability of survival decreases continuously at a constant proportional rate.

For a job that ends randomly with a hazard rate b:

In a small interval dt, the probability of the job surviving is approximately: 1 - b \, dt.
Over a finite time t - t_0, survival probability accumulates multiplicatively over small intervals: P(t) = \prod_{i=1}^n \left(1 - b \Delta t\right), where \Delta t = \frac{t - t_0}{n}.

Taking the Limit:

As \Delta t \to 0 (continuous time), this becomes: P(t) = \lim_{n \to \infty} \left(1 - \frac{b(t - t_0)}{n}\right)^n. Using the mathematical fact that \left(1 - \frac{x}{n}\right)^n \to e^{-x} as n \to \infty, we arrive at: P(t) = e^{-b(t - t_0)}.

Intuition Behind the Memoryless Property

The Shapiro-Stiglitz model assumes job separations follow a Poisson process. This means:

Job separations occur randomly but at a constant average rate b.
The process is memoryless, meaning the probability of a job ending in the next time interval depends only on the length of the interval, not on how long the worker has already been employed.

Given that a worker is still employed at time t, the probability of remaining employed for an additional \tau time units is: P(t + \tau | t) = \frac{P(t + \tau)}{P(t)} = \frac{e^{-b(t + \tau - t_0)}}{e^{-b(t - t_0)}} = e^{-b\tau}.

The Value of Being Employed

V_E(\Delta t) = \int_{t=0}^{\Delta t} e^{-(\rho + b)t}(w - e) \, dt + e^{-\rho \Delta t} e^{-b \Delta t} V_E(\Delta t) + e^{-\rho \Delta t} \left(1 - e^{-b \Delta t}\right)V_U(\Delta t) \tag{3.1}

Equation 3.1 in the Shapiro-Stiglitz model provides the value V_E(\Delta t), the expected lifetime utility of a worker employed at time t, over a short time interval \Delta t. Let’s derive it step by step, both mathematically and intuitively:

Setup: Expected Utility Over Time

V_E(\Delta t) represents the expected lifetime utility of a worker employed at the beginning of the interval [0, \Delta t], assuming the worker exerts effort during this period.
Utility depends on:
1. The worker’s wage (w) minus the disutility of effort (e).
2. The probability of remaining employed during \Delta t.
3. The potential for transitioning to unemployment at the end of \Delta t.

The worker’s utility over \Delta t is composed of:

Utility during the interval [0, \Delta t]:
- This accounts for the worker earning a wage w and exerting effort (e) while employed.
Utility after \Delta t:
- The worker can either remain employed or become unemployed.

First Term: Utility During the Interval [0, \Delta t]

During the short interval [0, \Delta t], the worker’s expected utility is given by the integral: \int_{t=0}^{\Delta t} e^{-(\rho + b)t}(w - e) \, dt.

Breakdown:

Survival Probability e^{-bt}:
- e^{-bt} is the probability that the worker remains employed at time t within the interval [0, \Delta t], based on the hazard rate b.
Discount Factor e^{-\rho t}:
- e^{-\rho t} reflects the time discounting of utility over the period. \rho is the worker’s subjective discount rate.
Flow Utility (w - e):
- (w - e) is the worker’s wage minus the disutility of effort.

Intuition:

This integral captures the flow of utility over time for as long as the worker remains employed during [0, \Delta t], adjusted for discounting and survival probability.

Second Term: Utility After \Delta t

At the end of \Delta t, two possibilities arise:

The worker remains employed with probability e^{-b \Delta t}
The worker becomes unemployed with probability 1 - e^{-b \Delta t}
- If unemployed, the worker transitions to the unemployed state, with utility V_U(\Delta t), discounted back to the start of \Delta t.

The expected utility for this part is: e^{-\rho \Delta t} \left[ e^{-b \Delta t} V_E(\Delta t) + \left(1 - e^{-b \Delta t}\right)V_U(\Delta t) \right].

Intuition:

This term captures the of being either employed or unemployed after \Delta t, weighted by the probabilities of these outcomes and adjusted for discounting.

Combine the Two Terms

Adding the two components gives the total expected lifetime utility V_E(\Delta t) as:

V_E(\Delta t) = \int_{t=0}^{\Delta t} e^{-(\rho + b)t}(w - e) \, dt + e^{-\rho \Delta t} e^{-b \Delta t} V_E(\Delta t) + e^{-\rho \Delta t} \left(1 - e^{-b \Delta t}\right)V_U(\Delta t)

Why is the Integral Used for the First Term?

The integral sums up the instantaneous flow of utility over the interval [0, \Delta t], accounting for survival probability and discounting at every point t within the interval.

Simplifying the Integral

Let’s break down the step where the integral

\int_{t=0}^{\Delta t} e^{-(\rho + b)t}(w - e) \, dt

is simplified to:

\frac{1}{\rho + b} \left( 1 - e^{-(\rho + b)\Delta t} \right)(w - e).

This involves basic integration rules and algebraic manipulation.

Step 1: Write Down the Original Integral

We need to evaluate:

I = \int_{t=0}^{\Delta t} e^{-(\rho + b)t}(w - e) \, dt.

Since w - e is constant, we can factor it out of the integral: I = (w - e) \int_{t=0}^{\Delta t} e^{-(\rho + b)t} \, dt.

Step 2: Solve the Exponential Integral

The integral to solve is: \int e^{-(\rho + b)t} \, dt.

This is a standard exponential integral: \int e^{kx} \, dx = \frac{1}{k} e^{kx} + C.

Here, k = -(\rho + b), so: \int e^{-(\rho + b)t} \, dt = \frac{1}{-(\rho + b)} e^{-(\rho + b)t} + C.

Simplifying: \int e^{-(\rho + b)t} \, dt = -\frac{1}{\rho + b} e^{-(\rho + b)t} + C.

Step 3: Apply the Definite Integral Limits

Now, we compute the definite integral from t = 0 to t = \Delta t: \int_{t=0}^{\Delta t} e^{-(\rho + b)t} \, dt = \left[ -\frac{1}{\rho + b} e^{-(\rho + b)t} \right]_{t=0}^{t=\Delta t}.

Substitute the limits:

At t = \Delta t: -\frac{1}{\rho + b} e^{-(\rho + b)\Delta t}.
At t = 0: -\frac{1}{\rho + b} e^{-(\rho + b)(0)} = -\frac{1}{\rho + b}.

Subtract the two results: \int_{t=0}^{\Delta t} e^{-(\rho + b)t} \, dt = -\frac{1}{\rho + b} e^{-(\rho + b)\Delta t} + \frac{1}{\rho + b}.

Simplify: \int_{t=0}^{\Delta t} e^{-(\rho + b)t} \, dt = \frac{1}{\rho + b} \left( 1 - e^{-(\rho + b)\Delta t} \right).

Step 4: Multiply Back the Constant (w - e)

Now return to the original integral: I = (w - e) \int_{t=0}^{\Delta t} e^{-(\rho + b)t} \, dt.

Substitute the result of the integral:

I = (w - e) \cdot \frac{1}{\rho + b} \left( 1 - e^{-(\rho + b)\Delta t} \right).

This simplifies to: I = \frac{1}{\rho + b} \left( 1 - e^{-(\rho + b)\Delta t} \right)(w - e).

Intuition of the Result

What Does This Represent?
- This term accounts for the utility during the interval [0, \Delta t], weighted by the survival probability e^{-bt} and the discount factor e^{-\rho t}.
Why Does It Involve 1 - e^{-(\rho + b)\Delta t}?
- The factor 1 - e^{-(\rho + b)\Delta t} arises because the utility accumulates over the interval [0, \Delta t] and stops at \Delta t.

From Equation 3.1, the lifetime utility of a worker employed at time t = 0 over a short time interval \Delta t is given by:

V_E(\Delta t) = \frac{1}{\rho + b} \left(1 - e^{-(\rho + b)\Delta t}\right)(w - e) + e^{-\rho \Delta t} e^{-b \Delta t} V_E(\Delta t) + e^{-\rho \Delta t} \left(1 - e^{-b \Delta t}\right)V_U(\Delta t)

Expand Equation 3.1

Start with: V_E(\Delta t) = \frac{1}{\rho + b} \left(1 - e^{-(\rho + b)\Delta t}\right)(w - e) + e^{-\rho \Delta t} \left[e^{-b \Delta t} V_E(\Delta t) + \left(1 - e^{-b \Delta t}\right)V_U(\Delta t)\right]

Group Terms

Focus on the two components of the equation:

The first term involves: \frac{1}{\rho + b} \left(1 - e^{-(\rho + b)\Delta t}\right)(w - e).
The second term involves: e^{-\rho \Delta t} \left[e^{-b \Delta t} V_E(\Delta t) + \left(1 - e^{-b \Delta t}\right)V_U(\Delta t)\right]

The second term involves V_E(\Delta t) itself. To isolate V_E(\Delta t), we will reorganize terms.

Rewrite and Solve for V_E(\Delta t)

Separate terms involving V_E(\Delta t):

V_E(\Delta t) - e^{-\rho \Delta t} e^{-b \Delta t} V_E(\Delta t) = \frac{1}{\rho + b} \left(1 - e^{-(\rho + b)\Delta t}\right)(w - e) + e^{-\rho \Delta t} \left(1 - e^{-b \Delta t}\right)V_U(\Delta t)

Factor V_E(\Delta t) on the left-hand side:

V_E(\Delta t) \left[1 - e^{-\rho \Delta t} e^{-b \Delta t}\right] = \frac{1}{\rho + b} \left(1 - e^{-(\rho + b)\Delta t}\right)(w - e) + e^{-\rho \Delta t} \left(1 - e^{-b \Delta t}\right)V_U(\Delta t)

Divide through by \left[1 - e^{-\rho \Delta t} e^{-b \Delta t}\right] to isolate V_E(\Delta t): V_E(\Delta t) = \frac{\left[\frac{1}{\rho + b} \left(1 - e^{-(\rho + b)\Delta t}\right)(w - e)\right]+ e^{-\rho \Delta t} \left(1 - e^{-b \Delta t}\right)V_U(\Delta t)}{1 - e^{-\rho \Delta t} e^{-b \Delta t}}

Simplifying to \Delta t \to 0

Behavior of Terms as \Delta t \to 0

As \Delta t \to 0, e^{-\rho \Delta t} \to 1 and e^{-b \Delta t} \to 1.
The denominator 1 - e^{-\rho \Delta t} e^{-b \Delta t} simplifies to: 1 - (1) = 0.

To deal with this, we use L’Hôpital’s Rule because the numerator and denominator both approach 0.

What is L’Hôpital’s Rule?

L’Hôpital’s Rule is a mathematical technique used to evaluate limits of indeterminate forms like \frac{0}{0} or \frac{\infty}{\infty}. When a limit produces one of these indeterminate forms, L’Hôpital’s Rule allows us to compute the limit by taking the derivatives of the numerator and denominator.

The Rule: If: \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \quad \text{or} \quad \frac{\infty}{\infty}, then: \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}, provided the derivatives f'(x) and g'(x) exist and \lim_{x \to c} \frac{f'(x)}{g'(x)} is finite.

Example of L’Hôpital’s Rule

Let’s evaluate: \lim_{x \to 0} \frac{\sin(x)}{x}.

At x = 0, both the numerator and denominator approach 0, so the limit is of the form \frac{0}{0}, which is indeterminate. We can apply L’Hôpital’s Rule.

The derivatives are: f(x) = \sin(x) \quad \Rightarrow \quad f'(x) = \cos(x), g(x) = x \quad \Rightarrow \quad g'(x) = 1.

\lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac{\cos(x)}{1}.

Evaluate the new limit: \lim_{x \to 0} \frac{\cos(x)}{1} = \cos(0) = 1.

So: \lim_{x \to 0} \frac{\sin(x)}{x} = 1.

Apply L’Hôpital’s Rule in the context of ?eq-1127

Let f(\Delta t) = \frac{1}{1 - e^{-\rho \Delta t} e^{-b \Delta t}}, and consider the limit as \Delta t \to 0:

\frac{1}{1 - e^{-\rho \Delta t} e^{-b \Delta t}} \quad \rightarrow \text{?}

\lim_{\Delta t \to 0} f(\Delta t) = \lim_{\Delta t \to 0} \frac{\frac{d}{d\Delta t} \text{[numerator]}}{\frac{d}{d\Delta t} \text{[denominator]}}.

Step 1: Identify the Indeterminate Form As \Delta t \to 0:

e^{-\rho \Delta t} \to 1,
e^{-b \Delta t} \to 1,
Thus, the denominator 1 - e^{-\rho \Delta t} e^{-b \Delta t} \to 0

This results in an indeterminate form of \frac{C}{0}, so we apply L’Hôpital’s Rule.

Step 2: Differentiate the Denominator The denominator is:

g(\Delta t) = 1 - e^{-(\rho + b) \Delta t}

Differentiating:

g'(\Delta t) = (\rho + b)e^{-(\rho + b) \Delta t}

Step 3: Apply L’Hôpital’s Rule Since the numerator is constant 1, its derivative is 0, so applying L’Hôpital’s Rule:

\lim_{\Delta t \to 0} \frac{1}{1 - e^{-\rho \Delta t} e^{-b \Delta t}} = \lim_{\Delta t \to 0} \frac{1}{(\rho + b)e^{-(\rho + b) \Delta t}}

Evaluating the limit:

\frac{1}{(\rho + b) e^{-(\rho + b) \cdot 0}} = \frac{1}{\rho + b}

Step 4: Combine the Results Substituting back into the larger equation for V_E, we find:

V_E = \frac{1}{\rho + b} (w - e) + \frac{b}{\rho + b} V_U

This simplifies to:

V_E = \frac{1}{\rho + b} (w - e) + \frac{b}{\rho + b} V_U

Summary L’Hôpital’s Rule allows us to handle indeterminate forms by differentiating the numerator and denominator. In the case of ?eq-1127, this method was used to simplify the behavior of the expression as \Delta t \to 0, leading to a closed-form solution for V_E.

Value of Employment - Intuitions

\boxed{ V_E \;=\; \frac{1}{\,\rho + b\,}\,(w - \bar{e}) \;+\; \frac{b}{\,\rho + b\,}\,V_U. }

It expresses the present (expected) value of being employed. Let’s break it down step by step:

1. Setting Up the Equation

We start from the “asset‐pricing” style equation:

\rho\,V_E \;=\; (w - \bar{e}) \;-\; b\,(V_E - V_U),

which says:

\rho \, V_E: The required “rate of return” on the “asset” of holding a job (i.e. how much value that job must yield per unit time, in present‐value terms).
(w - \bar{e}): The “dividend” flow while employed—i.e., wage w minus effort cost \bar{e}.
-\,b\,(V_E - V_U): The expected “capital loss” due to the possibility (at rate b) of losing the job and dropping to the lower value V_U.

Rearranging gives:

\rho V_E + b\,V_E \;=\; (w - \bar{e}) + b\,V_U \quad\Longrightarrow\quad (\rho + b)\,V_E \;=\; (w - \bar{e}) \;+\; b\,V_U.

2. Interpreting Each Term

\rho + b

\rho is the discount rate—the rate at which future utility is discounted relative to present utility.
b is the job separation rate—the probability per unit time (hazard) that the worker’s job exogenously ends.

Because you can lose the job with probability b, the “effective” discount rate for the flow of employment benefits is \rho + b. One might say you “discount” future employment benefits both by your impatience (\rho) and by the chance of losing the job (b).

(w - \bar{e})

This is the net utility “dividend” of working at any instant:
- Earn wage w.
- Pay an “effort cost” \bar{e}.

b\,V_U

When the job is lost (at rate b), you drop to the unemployment value V_U. Hence we add b\,V_U on the right-hand side—this is the “opportunity cost” of job loss that shows up in the rearranged form.

3. Solving for V_E

Divide both sides by (\rho + b):

V_E \;=\; \frac{(w - \bar{e})}{\,(\rho + b)\,} \;+\; \frac{b}{\,(\rho + b)\,}\;V_U.

4. Economic Intuition

\frac{1}{\rho + b}\,(w - \bar{e}):
- If you think of a perpetual flow (w - \bar{e}) discounted at an overall rate (\rho + b), this fraction is the standard present value of that flow.
- But each instant, you also face a chance b of having that flow end.
\frac{b}{\rho + b}\,V_U:
- Because there is a chance b of transitioning to unemployment at any moment, a portion of V_E depends on how good (or bad) unemployment is.
- The relative weight \frac{b}{\rho + b} tells you how heavily the possibility of job loss features in the worker’s overall value of being employed.

Put Differently

The worker’s job value is partially the discounted sum of ongoing net utility (w - \bar{e}), and partially a reflection of what happens upon job loss (i.e. falling back to V_U).

Bottom Line

\boxed{ V_E \;=\; \underbrace{\frac{1}{\,\rho + b\,}(w - \bar{e})}_{\text{present value of wage-effort flow}} \;+\; \underbrace{\frac{b}{\,\rho + b\,}V_U}_{\text{weight on potential fallback to }V_U}. }

Flow Benefit: (w - \bar{e}) discounted at rate \rho + b.
Fallback Option: If you lose your job at rate b, you transition to unemployment V_U.

Hence, the worker’s total lifetime utility from being employed is a weighted sum of the immediate net benefits and the possibility of dropping to unemployment over time.

Full Employment Assumtion

\bar{e}\,F'\!\Bigl(\frac{\bar{e}\overline{L}}{N}\Bigr) > \bar{e} \quad \text{or equivalently} \quad F'\!\Bigl(\frac{\bar{e}\overline{L}}{N}\Bigr) > 1.

This condition comes from comparing the marginal benefit of hiring an additional worker to the cost of employing that worker when everyone is employed and working at full effort.

Step-by-Step Explanation:

Effective Labor and Full Employment:
- Suppose there are \overline{L} workers available.
- If every worker is employed and exerts the full effort level \bar{e}, the total effective labor is \bar{e}\,\overline{L}.
- If there are N identical firms, then on average each firm employs L = \overline{L}/N workers.
- Consequently, the effective labor used by each firm is:
  \bar{e}\,L = \bar{e}\,\frac{\overline{L}}{N}.
Marginal Product of Labor:
- The production function is F(\cdot) with F'>0 and F''<0.
- The marginal product of effective labor is given by F'.
- However, because an additional worker contributes \bar{e} effective labor units, the marginal benefit from hiring one more worker is: \bar{e}\,F'\!\Bigl(\bar{e}\,\frac{\overline{L}}{N}\Bigr).
Cost of Effort:
- The cost to the firm per worker, in terms of the effort the worker must supply, is \bar{e} (this can be thought of as the “cost” of inducing the worker to supply \bar{e} units of effective labor).
The Full Employment Condition:
- For full employment to be optimal in a frictionless world (i.e. absent monitoring problems), the marginal product (benefit) of an extra worker must be greater than or equal to the cost of that worker’s effort.
- That is, the condition is: \bar{e}\,F'\!\Bigl(\bar{e}\,\frac{\overline{L}}{N}\Bigr) > \bar{e}.
- You can cancel \bar{e} from both sides (since \bar{e}>0), which simplifies to: F'\!\Bigl(\bar{e}\,\frac{\overline{L}}{N}\Bigr) > 1.

Economic Interpretation:

Marginal Benefit:
The term F'\!\Bigl(\bar{e}\,\frac{\overline{L}}{N}\Bigr) measures the extra output produced by an additional unit of effective labor. When multiplied by \bar{e}, it gives the marginal product per worker.
Cost Comparison:
The cost per worker is simply the effort cost \bar{e}. Thus, for it to be worthwhile to hire every available worker, the additional output (in effective labor terms) must exceed the cost of that effective labor.
Implication for Full Employment:
If the marginal product of effective labor is high enough (i.e., greater than the cost per worker), then without any monitoring issues, firms would hire all available workers. In the real world, however, imperfect monitoring forces firms to pay a premium (an efficiency wage) to deter shirking. This premium, in turn, leads to equilibrium unemployment—even though, from a purely technological standpoint, full employment would be optimal.

This is how we arrive at the condition: \bar{e}\,F'\!\Bigl(\frac{\bar{e}\overline{L}}{N}\Bigr) > \bar{e} \quad \text{or equivalently} \quad F'\!\Bigl(\frac{\bar{e}\overline{L}}{N}\Bigr) > 1.