Assignment 8

11 and #14 on page 303 of probability text & #1 on page 320-321

  1. A company buys 100 lightbulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the first of these bulbs to burn out? (See Exercise 10.) 10 - Let X1, X2, . . . , Xn be n independent random variables each of which has an exponential density with mean µ. Let M be the minimum value of the Xj β = µ(exponential density)/n(variables) In question 11, exponential lifetime of 1000 hours/ 100 lightbulbs = expected time for the first of these bulbs to burn out.
n <- 100
μ <- 1000
result <- μ / n
cat(result, "hours to burn out\n")
## 10 hours to burn out

14.Assume that \(X_1\) and \(X_2\) are independent random variables, each having an exponential density with parameter \(\lambda\). We want to show that \(Z = X_1 - X_2\) has the density:

\[ f_Z(z) = \frac{1}{2} e^{-\lambda |z|} \]

Rewrite this as: \[ f_Z(z) = \begin{cases} \frac{1}{2} e^{-\lambda z}, & \text{if } z \geq 0 \\ \frac{1}{2} e^{\lambda z}, & \text{if } z < 0 \end{cases}\]

PDF is below, given: \(X_1\) and \(X_2\) have exponential density \[ f_{X_1}(x) = f_{X_2}(x) = \begin{cases} \lambda e^{-\lambda x}, & \text{if } x \geq 0 \\ 0, & \text{otherwise} \end{cases}\] Next, find the PDF of \(Z\) by taking the derivative: \[ f_Z(z) = f_{X_1 - X_2}(z) = \int_{-\infty}^{\infty} f_{-X_2}(z - x_1) f_{X_1}(x_1) \, dx_1\] Distribute: \[ = \int_{-\infty}^{\infty} \lambda e^{-\lambda (z - x_1)} \cdot \lambda e^{-\lambda x_1} \, dx_1\] Simplify: \[ = \lambda^2 \int_{-\infty}^{\infty} e^{-\lambda z + \lambda x_1 - \lambda x_1} \, dx_1 \]

\[ = \lambda^2 \int_{-\infty}^{\infty} e^{\lambda z} \, dx_1 \]

We can rewrite the given: \(Z = X_1 - X_2\) and assume \(x_2 = x_1 - z\).

If Z, the difference is non negative, \(z \geq 0\), then \(x_2 = (x_1 - z) \geq 0\), and \(x_1 \geq z\):

\[ f_Z(z) = \frac{1}{2} e^{-\lambda z} \]

If the difference, Z, is negative, \(z < 0\), then \(x_2 = (x_1 - z) \geq 0\), and \(x_1 \geq 0\):

\[ f_Z(z) = \frac{1}{2} e^{\lambda z} \]

Combining both cases; the PDF of Z depends on the sign of z:

\[ f_Z(z) = \begin{cases} \frac{1}{2} e^{-\lambda z}, & \text{if } z \geq 0 \\ \frac{1}{2} e^{\lambda z}, & \text{if } z < 0 \end{cases} \]

  1. Let X be a continuous random variable with mean µ = 10 and variance \[ \sigma^2 = \frac{100}{3} \] Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
  1. P(|X − 10| ≥ 2).
  2. P(|X − 10| ≥ 5).
  3. P(|X − 10| ≥ 9).
  4. P(|X − 10| ≥ 20). The question is asking to use different standard deviation (a-d), to find the upper bound or maximum.

A. P(|X − 10| ≥ 2).

a <- 2
k_1 <- a/sqrt(100/3)
p_a <- 1/ (k_1)^2
p_a
## [1] 8.333333
  1. P(|X − 10| ≥ 5).
b <- 5
k_1 <- b/sqrt(100/3)
p_b <- 1/ (k_1)^2
p_b
## [1] 1.333333

C. P(|X − 10| ≥ 9).

c <- 9
k_1 <- c/sqrt(100/3)
p_c <- 1/ (k_1)^2
p_c
## [1] 0.4115226

D. P(|X − 10| ≥ 20).

d <- 20
k_1 <- d/sqrt(100/3)
p_d <- 1/ (k_1)^2
p_d
## [1] 0.08333333