Pg. 303 #11
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.)
  • Exercise 10 tells us that \(M\), which represents the minimum value for each \(X_j\) is exponential, with a mean of \(\frac{μ}{n}\).
  • The lifetime expectancy of any bulb i is \(E[X_i] = \frac{1}{λ_i} = 1000\)
  • We’re given 100 bulbs, so n = 100, and the density for the first bulb to burn out (the minimum) =
  • \(E[X_i (min)]=\frac{μ}{n}=\frac{1000}{100}\) = 10 hours
Pg. 303 #14
Assume that \(X_1\) and \(X_2\) are independent random variables, each having an exponential density with parameter \(λ\). Show that \(Z = X_1 - X_2\) has density \(f_Z(z) = \frac{1}{2}λe^{-λ|z|}\).

  • We’re given that \(X_1\) and \(X_2\) are independent random variables with exponential densities, then we know the following (Prob. book Pg. 292);

  • We know the probability density for an exponentially distributed random variable is \(f(X_1) = λ e^{-λx_1}\) and \(f(X_2) = λ e^{-λx_2}\) if \(x \geq 0\) and 0 otherwise.

  • Using what we know about independent exponential random variables and the density function of their sum (Prob. book Pg. 293), we know;

  • \(Z = X_1 - X_2\) and thus \(x_1 = z + x_2\)

  • \(f_Z(z) = f_{X_1} + f_{X_2}\)

  • \(= f_{x_1 + -x_2}(z)\)

  • \(= \int_{-∞}^{∞} f_{-x_2}(z-x_1) f_{x_1}(x_1) dx_1\)

  • \(= \int_{-∞}^{∞} f_{x_2}(x_1-z) f_{x_1}(x_1) dx_1\)

  • \(= \int_{-∞}^{∞} λe^{-λ(x_1 -z)} λe^{-λx_1} dx_1\)

  • \(= \int_{-∞}^{∞} λ^2 e^{-λx_1 +λz} e^{-λx_1} dx_1\)

  • \(= \int_{-∞}^{∞} λ^2 e^{λz-λx_1-λx_1} dx_1\)

  • \(= \int_{-∞}^{∞} λ^2 e^{λ(z-2x_1)} dx_1\)

  • Returning to what we know; \(z = x_1 - x_2\) and \(x_1 = z + x_2\)

  • If \(z \geq 0\), then \(x_1 = z + x_2 \geq 0\), so \(x_1 \geq z\), and

  • If \(z \geq 0\), then \(f_Z(z) = \int_{z}^{∞} λ^2 e^{λ(z-2x_1)} dx_1 = \frac{1}{2}λe^{-λz}\)

  • If \(z < 0\), then \(x_1 = z + x_2 < 0\) and \(x_1 \geq 0\), and

  • If \(z < 0\), then \(f_Z(z) = \int_{0}^{∞} λ^2 e^{λ(z-2x_1)} dx_1 = \frac{1}{2}λe^{λz}\)

  • So we can see that the PDF \(f_Z(z)\) =

\[ \left\{\begin{matrix} \frac{1}{2}e^{- \lambda z} \texttt{,if z >= 0 } \\ \frac{1}{2}e^{\lambda z} \texttt{, if z <0} \end{matrix}\right. \]

Pg. 320 #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.
  • Chebyshev’s Inequality says that \(P(|X-μ| \geq kσ) \leq \frac{σ^2}{k^2 σ^2} = \frac{1}{k^2}\)
    1. \(P(|X-10| \geq 2)\)
    1. \(P(|X-10| \geq 5)\)
    1. \(P(|X-10| \geq 9)\)
    1. \(P(|X-10| \geq 20)\)
mu = 10
var = 100/3
sd = sqrt(var)

# (a)P(|X−10|≥2
# P(|X−10|≥2)≤ (100/3)(1/(2*2)) = 25/3
ka <- 2/sd
UBa <- 1/ka**2


# (b) P(|X−10|≥5)
# P(|X−10|≥5)≤ (100/3)(1/(5*5)) = 4/3
kb <- 5/sd
UBb <- 1/kb**2

# (c) P(|X−10|≥9)
# P(|X−10|≥9)≤ (100/3)(1/(9*9)) = 100/243
kc <- 9/sd
UBc <- 1/kc**2

# (d) P(|X − 10| ≥ 20)
# P(|X−10|≥20)≤ (100/3)(1/(20*20)) = 1/12
kd <- 20/sd
UBd <- 1/kd**2

cat("The upper bound for a is: ", UBa)
## The upper bound for a is:  8.333333
cat("The upper bound for b is: ", UBb)
## The upper bound for b is:  1.333333
cat("The upper bound for c is: ", UBc)
## The upper bound for c is:  0.4115226
cat("The upper bound for d is: ", UBd)
## The upper bound for d is:  0.08333333