Question

The exponential distribution with rate parameter μ > 0 is a continuous distribution on [0, ∞) with density

f(t) = μ exp(−μt), t > 0

1. Compute the cumulative distribution function defined by

F(t) := P(X ∈ [0,t]).

2. Compute P(s < X ≤ t).

3) Find P( X∈ [1,2] ∪[3,4] ).

4) Compute the conditional probability P (X ∈ [3, 4] | X ∈ [1, 4] )

5) Compute the conditional probability P (X > t + s | X > s ) for s, t ≥ 0.

6) Find the mean of the exponential distribution with rate parameter μ > 0.

Answer 1

Answer on Question #63548 – Math – Statistics and Probability

The exponential distribution with rate parameter $\mu > 0$ is a continuous distribution on $[0, \infty)$ with density

$$f(t) = \mu \exp(-\mu t), \quad t > 0$$

Question

1. Compute the cumulative distribution function defined by $F(t) := P(X \in [0, t])$.

Solution

The cdf is

$$F(t) = P(X \leq t) = \int_{-\infty}^{t} f(t) \, dt = \begin{cases} 1 - e^{(-\mu t)}, & t \geq 0 \\ 0, & t < 0 \end{cases}$$

Answer: $F(t) = \begin{cases} 1 - e^{(-\mu t)}, & t \geq 0 \\ 0, & t < 0 \end{cases}$

Question

2. Compute $P(s < X \leq t)$.

Solution

$$P(s < X \leq t) = P(X \leq t) - P(X \leq s) = 1 - e^{-\mu t} - (1 - e^{-\mu s}) = e^{-\mu s} - e^{-\mu t}.$$ $$P(s < X \leq t) = \int_{s}^{t} f(t) \, dt = \int_{s}^{t} \mu e^{(-\mu t)} \, dt = -e^{(-\mu t)} \Big| \begin{cases} t \\ s \end{cases} = e^{-\mu s} - e^{-\mu t}.$$

Answer: $P(s < X \leq t) = e^{-\mu s} - e^{-\mu t}$.

Question

3. Find $P(X \in [1, 2] \cup [3, 4])$.

Solution

$$P(X \in [1, 2] \cup [3, 4]) = P(X \leq 2) - P(X \leq 1) + P(X \leq 4) - P(X \leq 3) = \\ = 1 - e^{-2\mu} - (1 - e^{-\mu}) + 1 - e^{-4\mu} - (1 - e^{-3\mu}) = e^{-\mu} + e^{-3\mu} - e^{-2\mu} - e^{-4\mu} = \\ = \frac{e^{3\mu} - e^{2\mu} + e^{\mu} - 1}{e^{4\mu}} = \frac{(e^{\mu} - 1)(e^{2\mu} + 1)}{e^{4\mu}}$$

Answer: $\frac{(e^{\mu} - 1)(e^{2\mu} + 1)}{e^{4\mu}}$.

Question

4. Compute the conditional probability $\mathrm{P}(\mathrm{X} \in [3, 4] \mid \mathrm{X} \in [1, 4])$.

Solution

$$\begin{array}{l} P (X \in [ 3, 4 ] | X \in [ 1, 4 ]) = \frac {P (X \in [ 3 , 4 ] \text{ and } X \in [ 1 , 4 ])}{P (X \in [ 1 , 4 ])} = \frac {P (X \in [ 3 , 4 ])}{P (X \in [ 1 , 4 ])} = \\ = \frac {P (X \leq 4) - P (X \leq 3)}{P (X \leq 4) - P (X \leq 1)} = \frac {e ^ {- 3 \mu} - e ^ {- 4 \mu}}{e ^ {- \mu} - e ^ {- 4 \mu}} = \frac {e ^ {\mu} - 1}{e ^ {3 \mu} - 1} = \frac {1}{e ^ {2 \mu} + e ^ {\mu} + 1}. \end{array}$$

Answer: $\frac{1}{e^{2\mu} + e^{\mu} + 1}$.

Question

5. Compute the conditional probability $\mathrm{P}(\mathrm{X} > \mathrm{t} + \mathrm{s} \mid \mathrm{X} > \mathrm{s})$ for $\mathrm{s}, \mathrm{t} \geq 0$.

Solution

$$\begin{array}{l} P (X > t + s \mid X > s) = \frac {P (X > s + t \text{ and } X > s)}{P (X > s)} = \frac {P (X > s + t)}{P (X > s)} = \frac {e ^ {- \mu (t + s)}}{e ^ {- \mu s}} = \\ = e ^ {- \mu t}. \end{array}$$

Answer: $P(X > t + s \mid X > s) = e^{-\mu t}$.

Question

6. Find the mean of the exponential distribution with rate parameter $\mu > 0$.

Solution

The mean is

$$\begin{array}{l} E X = \int_ {0} ^ {\infty} t f (t) d t = \int_ {0} ^ {\infty} t \mu e ^ {(- \mu t)} d t = - t e ^ {(- \mu t)} \Big | _ {0} ^ {\infty} + \int_ {0} ^ {\infty} e ^ {(- \mu t)} d t = \\ = 0 - 0 - \frac {1}{\mu} e ^ {(- \mu t)} \Big | _ {0} ^ {\infty} = 0 + \frac {1}{\mu} = \frac {1}{\mu}. \end{array}$$

Answer: $E X = \frac{1}{\mu}$.

www.AsignmentExpert.com