Math Snap

STEP 1

1. We are given the probability density function (pdf) $g(t) = \frac{\mu}{1-e^{-100 \mu}} e^{-\mu t} 1_{[0,100]}(t)$.
2. The parameter $\mu$ is positive ($\mu > 0$).
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 $[0, 100]$. The function $1_{[0,100]}(t)$ indicates that the pdf is zero outside this interval. The term $\frac{\mu}{1-e^{-100 \mu}}$ is a normalizing constant ensuring the total probability integrates to 1 over $[0, 100]$.

STEP 4

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

STEP 5

Solve for the median:
Calculate the integral:
$$\int_0^m e^{-\mu t} \, dt = \left[ -\frac{1}{\mu} e^{-\mu t} \right]_0^m = -\frac{1}{\mu} (e^{-\mu m} - 1)$$ Set the equation:
$$\frac{\mu}{1-e^{-100 \mu}} \left( -\frac{1}{\mu} (e^{-\mu m} - 1) \right) = 0.5$$ Simplify:
$$\frac{1 - e^{-\mu m}}{1-e^{-100 \mu}} = 0.5$$ Solve for $m$:
$$1 - e^{-\mu m} = 0.5(1-e^{-100 \mu})$$ $$e^{-\mu m} = 1 - 0.5(1-e^{-100 \mu})$$ $$e^{-\mu m} = 0.5 + 0.5e^{-100 \mu}$$ $$-\mu m = \ln(0.5 + 0.5e^{-100 \mu})$$ $$m = -\frac{1}{\mu} \ln(0.5 + 0.5e^{-100 \mu})$$

STEP 6

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

Solve for the mean:
Integrate by parts, let $u = t$ and $dv = \frac{\mu}{1-e^{-100 \mu}} e^{-\mu t} \, dt$.
$$du = dt$$ $$v = \left[ -\frac{1}{1-e^{-100 \mu}} e^{-\mu t} \right]$$ Apply integration by parts:
$$\mathbb{E}[T] = \left[ -\frac{t}{1-e^{-100 \mu}} e^{-\mu t} \right]_0^{100} + \int_0^{100} \frac{1}{1-e^{-100 \mu}} e^{-\mu t} \, dt$$ Calculate:
$$\mathbb{E}[T] = \left( -\frac{100}{1-e^{-100 \mu}} e^{-100 \mu} + 0 \right) + \frac{1}{1-e^{-100 \mu}} \left[ -\frac{1}{\mu} e^{-\mu t} \right]_0^{100}$$ $$\mathbb{E}[T] = -\frac{100 e^{-100 \mu}}{1-e^{-100 \mu}} + \frac{1}{\mu(1-e^{-100 \mu})} (1 - e^{-100 \mu})$$ $$\mathbb{E}[T] = \frac{1}{\mu} - \frac{100 e^{-100 \mu}}{1-e^{-100 \mu}}$$ The median life expectancy is:
$$m = -\frac{1}{\mu} \ln(0.5 + 0.5e^{-100 \mu})$$ The mean life expectancy is:
$$\mathbb{E}[T] = \frac{1}{\mu} - \frac{100 e^{-100 \mu}}{1-e^{-100 \mu}}$$