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.)

From problem 10 from book

the density of minimum value among n independent random variables with an exponential density has mean μ/n,

where μ is mean of exponential density of individual variable.

Here n = 100

μ=1000,

so the expected time for the first of the bulbs to burn out is

E(M)=μ/n

1000/100 = 10 hours

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

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

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

Consider \(z=x_1-x_2\), then \(x_2=x_1-z\).

If \(z \ge 0\), then \(x_2=(x_1-z) \ge 0\), and \(x_1 \ge z\), and, using WolframAlpha, \(f_Z(z) = \int_{z}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 = \frac{1}{2} \lambda e^{-\lambda z}\).

If \(z < 0\), then \(x_2=(x_1-z) \ge 0\), and \(x_1 \ge 0\), and \(f_Z(z) = \int_{0}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 =\frac{1}{2} \lambda e^{\lambda z}\).

Combining two sides we get \(f_Z(z) = \begin{cases} (1/2)e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2)e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)

P-320 1. Let X be a continuous random variable with mean μ=10 and variance σ2=100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.

P(|X−10|≥2) P(|X−10|≥5) P(|X−10|≥9) P(|X−10|≥20)

here mean μ=10

variance \(σ^2=100/3\)

Standarddeviation σ = sqrt(100/3)

\(P(|X−μ|≥kσ)≤1/k^2\)

  1. P( | X - 10 | >= 2)

\(kσ = 2\)

\(k = \frac2{\sqrt{100/3}}\)

\(P(|X−10|⩾2)=1/k^2 = 8.3333≈1\)

  1. P( | X - 10 | >= 5)

\(kσ = 5\)

\(k = \frac5{\sqrt{100/3}}\)

\(P(|X−10|⩾5)=1k2=1.3333≈1\)

  1. P( | X - 10 | >= 9)

\(kσ=9\)

\(k = \frac9{\sqrt{100/3}}\)

\(P(|X−10|⩾9)=1k2=0.4115\)

  1. P( | X - 10 | >= 20)

\(kσ=20\)

\(k = \frac{20}{\sqrt{100/3}}\)

\(P(|X−10|⩾20)=1k2=0.0833\)