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  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?
light_bulbs <- 100
lifetime <- 1000
  
lambda_rate <- light_bulbs/lifetime
1/lambda_rate
## [1] 10
  1. Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that Z = X1 − X2 has density \(f_Z(z) = (1/2)e^{-\lambda|z|}\).

\(f_Z(z) = (1/2)e^{-\lambda|z|}\) can be re-written as \(f_Z(z) = \begin{cases} (1/2)e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2)e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)

Since \(X_1\) and \(X_2\) have exponential density, their PDF is

\(f_{X_1}(x)=f_{X_2}(x)=\begin{cases} \lambda e^{-\lambda x}, & \mbox{if } x\ge 0, \\ 0, & \mbox{otherwise. }\end{cases}\)

\[ \begin{split} f_Z(z) &= f_{X_1+(-X_2)}(z) \\ &= \int_{-\infty}^{\infty} f_{-X_2}(z-x_1) f_{X_1}(x_1) dx_1 \\ &= \int_{-\infty}^{\infty} f_{X_2}(x_1-z) f_{X_1}(x_1) dx_1 \\ &= \int_{-\infty}^{\infty} \lambda e^{-\lambda(x_1-z)} \lambda e^{-\lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{-\lambda x_1 + \lambda z} e^{-\lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{\lambda z - \lambda x_1 - \lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 \end{split} \]

Page 320-321

  1. Let X be a continuous random variable with mean µ = 10 and variance \(σ^2\) = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
  1. \(P(|X − 10| ≥ 2)\).
mu <- 10
variance <- sqrt(100/3)
k_a <- 2/variance
upper_bound_a <- 1/k_a^2
upper_bound_a
## [1] 8.333333
  1. \(P(|X − 10| ≥ 5)\).
k_b <- 5/variance
upper_bound_b <- 1/k_b^2
upper_bound_b
## [1] 1.333333
  1. \(P(|X − 10| ≥ 9)\).
k_c <- 9/variance
upper_bound_c <- 1/k_c^2
upper_bound_c
## [1] 0.4115226
  1. \(P(|X − 10| ≥ 20)\).
k_d <- 20/variance
upper_bound_d <- 1/k_d^2
upper_bound_d
## [1] 0.08333333