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Math  /  Calculus

Question1. The life expectancy in a demographic model is a random variable with a distribution given by the density g(t)=μ1−e−100μe−μt1[0,100](t)g(t)=\frac{\mu}{1-e^{-100 \mu}} e^{-\mu t} 1_{[0,100]}(t) for some parameter μ>0\mu>0. Determine the median and the mean life expectancy in this model.

Studdy Solution

STEP 1

1. We are given the probability density function (pdf) g(t)=μ1−e−100μe−μt1[0,100](t) g(t) = \frac{\mu}{1-e^{-100 \mu}} e^{-\mu t} 1_{[0,100]}(t) .
2. The parameter μ\mu is positive (μ>0\mu > 0).
3. We need to determine both the median and the mean of the distribution.

STEP 2

1. Understand the structure of the pdf.
2. Determine the median of the distribution.
3. Calculate the mean of the distribution.

STEP 3

Understand the structure of the pdf:
The given pdf is a truncated exponential distribution on the interval [0,100][0, 100]. The function 1[0,100](t)1_{[0,100]}(t) indicates that the pdf is zero outside this interval. The term μ1−e−100μ\frac{\mu}{1-e^{-100 \mu}} is a normalizing constant ensuring the total probability integrates to 1 over [0,100][0, 100].

STEP 4

Determine the median of the distribution:
The median mm is the value such that the cumulative distribution function (CDF) equals 0.5.
∫0mg(t) dt=0.5 \int_0^m g(t) \, dt = 0.5
Substitute the pdf:
∫0mμ1−e−100μe−μt dt=0.5 \int_0^m \frac{\mu}{1-e^{-100 \mu}} e^{-\mu t} \, dt = 0.5

STEP 5

Solve for the median:
Calculate the integral:
∫0me−μt dt=[−1μe−μt]0m=−1μ(e−μm−1) \int_0^m e^{-\mu t} \, dt = \left[ -\frac{1}{\mu} e^{-\mu t} \right]_0^m = -\frac{1}{\mu} (e^{-\mu m} - 1)
Set the equation:
μ1−e−100μ(−1μ(e−μm−1))=0.5 \frac{\mu}{1-e^{-100 \mu}} \left( -\frac{1}{\mu} (e^{-\mu m} - 1) \right) = 0.5
Simplify:
1−e−μm1−e−100μ=0.5 \frac{1 - e^{-\mu m}}{1-e^{-100 \mu}} = 0.5
Solve for mm:
1−e−μm=0.5(1−e−100μ) 1 - e^{-\mu m} = 0.5(1-e^{-100 \mu})
e−μm=1−0.5(1−e−100μ) e^{-\mu m} = 1 - 0.5(1-e^{-100 \mu})
e−μm=0.5+0.5e−100μ e^{-\mu m} = 0.5 + 0.5e^{-100 \mu}
−μm=ln⁡(0.5+0.5e−100μ) -\mu m = \ln(0.5 + 0.5e^{-100 \mu})
m=−1μln⁡(0.5+0.5e−100μ) m = -\frac{1}{\mu} \ln(0.5 + 0.5e^{-100 \mu})

STEP 6

Calculate the mean of the distribution:
The mean E[T]\mathbb{E}[T] is given by:
E[T]=∫0100t⋅g(t) dt \mathbb{E}[T] = \int_0^{100} t \cdot g(t) \, dt
Substitute the pdf:
E[T]=∫0100t⋅μ1−e−100μe−μt dt \mathbb{E}[T] = \int_0^{100} t \cdot \frac{\mu}{1-e^{-100 \mu}} e^{-\mu t} \, dt

STEP 7

Solve for the mean:
Integrate by parts, let u=t u = t and dv=μ1−e−100μe−μt dt dv = \frac{\mu}{1-e^{-100 \mu}} e^{-\mu t} \, dt .
du=dt du = dt v=[−11−e−100μe−μt] v = \left[ -\frac{1}{1-e^{-100 \mu}} e^{-\mu t} \right]
Apply integration by parts:
E[T]=[−t1−e−100μe−μt]0100+∫010011−e−100μe−μt dt \mathbb{E}[T] = \left[ -\frac{t}{1-e^{-100 \mu}} e^{-\mu t} \right]_0^{100} + \int_0^{100} \frac{1}{1-e^{-100 \mu}} e^{-\mu t} \, dt
Calculate:
E[T]=(−1001−e−100μe−100μ+0)+11−e−100μ[−1μe−μt]0100 \mathbb{E}[T] = \left( -\frac{100}{1-e^{-100 \mu}} e^{-100 \mu} + 0 \right) + \frac{1}{1-e^{-100 \mu}} \left[ -\frac{1}{\mu} e^{-\mu t} \right]_0^{100}
E[T]=−100e−100μ1−e−100μ+1μ(1−e−100μ)(1−e−100μ) \mathbb{E}[T] = -\frac{100 e^{-100 \mu}}{1-e^{-100 \mu}} + \frac{1}{\mu(1-e^{-100 \mu})} (1 - e^{-100 \mu})
E[T]=1μ−100e−100μ1−e−100μ \mathbb{E}[T] = \frac{1}{\mu} - \frac{100 e^{-100 \mu}}{1-e^{-100 \mu}}
The median life expectancy is:
m=−1μln⁡(0.5+0.5e−100μ) m = -\frac{1}{\mu} \ln(0.5 + 0.5e^{-100 \mu})
The mean life expectancy is:
E[T]=1μ−100e−100μ1−e−100μ \mathbb{E}[T] = \frac{1}{\mu} - \frac{100 e^{-100 \mu}}{1-e^{-100 \mu}}

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