Problem 1 (Section 7.2 Exercise 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.)

For exercise 10, it says “Let \(X_{1}\), \(X_{2}\), · · · , \(X_{n}\) be \(n\) independent random variables each of which has an exponential density with mean \(\mu\). Let \(M\) be the minimum value of the \(X_{j}\). Show that the density for \(M\) is exponential with mean \(\frac{\mu }{n}\)”

\(E[X_{j}]=\frac{1}{\lambda_{j}}=1000\)

\(\lambda_{j}=\frac{1}{1000}\)

\(\sum_{1}^{n}\lambda_{j}=100\cdot\frac{1}{1000}=\frac{1}{10}\)

\(E[minX_{j}]=\frac{1}{\lambda}=\frac{1}{\frac{1}{10}}=10\)

Problem 2 (Section 7.2 Exercise 14)

Assume that \(X_{1}\) and \(X_{2}\) are independent random variables, each having an exponential density with parameter \(\lambda\). Show that \(Z=X_{1}-X_{2}\) has density \[f_{Z}(z)=\frac{1}{2}\lambda e ^{-\lambda\left | z \right |}\] \(Z=X_{1}-X_{2}\) —> Let \(X_{1}=x\) and \(X_{2}=Z-X_{1}=Z-x\)

\(f(X_{1})=\lambda e^{-\lambda X_{1}}=\lambda e^{-\lambda x}\)

\(f(X_{2})=\lambda e^{-\lambda X_{2}}=\lambda e^{-\lambda (Z-x)}\)

\(f_{Z}(z)=\int_{-\infty }^{\infty} f_{X_{1}}(X_{1})f_{X_{2}}(X_{2})dx\)

\(f_{Z}(z)=\int_{0}^{\infty} f_{X_{1}}(X_{1})f_{X_{2}}(Z-X_{1})dx\)

\(f_{Z}(z)=\int_{0 }^{\infty} \lambda e^{-\lambda x} \cdot \lambda e^{-\lambda (Z-x)}dx\)

\(f_{Z}(z)= \int_{0}^{\infty} \lambda^{2} e^{-\lambda (z+2x)} dx\)

\(f_{Z}(z)=\frac{1}{2}\lambda e ^{-\lambda\left | z \right |}\)

Problem 3 (Section 8.2 Exercise 1)

Let X be a continuous random variable with mean \(\mu=10\) and variance \(\sigma^{2}=\frac{100}{3}\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities.

Chebyshev’s Inequality

\(Pr(\left |X-\mu \right |\geq k\sigma)\leq \frac{\sigma ^{2}}{k^{2}\sigma ^{2}}=Pr(\left |X-\mu \right |\geq k\sigma)\leq\frac{1}{k^{2}}\)

\(\mu=10\)

\(\sigma^{2}=\frac{100}{3}\)

\(\sigma^=\sqrt{\frac{100}{3}}=\frac{10\sqrt{3}}{3}\)

mean <- 10
var <- 100/3
sd <- sqrt(var)

Part a

Using Chebyshev’s Inequality, find an upper bound for P(|X−10|≥2)

\(k\sigma=2\) \(k=\frac{2}{\sigma}=\frac{2}{\frac{10\sqrt{3}}{3}}=\frac{\sqrt{3}}{5}\)

\(Pr(\left |X-10 \right |\geq 2)\leq \frac{1}{(\frac{\sqrt{3}}{5})^{2}}=\frac{25}{3}\)

n <- 2 

k <- n/sd

upper_bound <- 1/(k^2)
upper_bound
## [1] 8.333333

Since the maximum probability is 1 and \(\frac{25}{3}>1\), the upper bound is 1.

Part b

Using Chebyshev’s Inequality, find an upper bound for P(|X−10|≥5)

\(k\sigma=5\) \(k=\frac{5}{\sigma}=\frac{5}{\frac{10\sqrt{3}}{3}}=\frac{\sqrt{3}}{2}\)

\(Pr(\left |X-10 \right |\geq 5)\leq \frac{1}{(\frac{\sqrt{3}}{2})^{2}}=\frac{4}{3}\)

n <- 5

k <- n/sd

upper_bound <- 1/(k^2)
upper_bound
## [1] 1.333333

Since the maximum probability is 1 and \(\frac{4}{3}>1\), the upper bound is 1.

Part c

Using Chebyshev’s Inequality, find an upper bound for P(|X−10|≥9)

\(k\sigma=9\) \(k=\frac{9}{\sigma}=\frac{9}{\frac{10\sqrt{3}}{3}}=\frac{9\sqrt{3}}{10}\)

\(Pr(\left |X-10 \right |\geq 9)\leq \frac{1}{(\frac{3\sqrt{3}}{10})^{2}}=\frac{100}{243}\)

n <- 9 

k <- n/sd

upper_bound <- 1/(k^2)
upper_bound
## [1] 0.4115226

The upper bound is \(\frac{100}{243}\).

Part d

Using Chebyshev’s Inequality, find an upper bound for P(|X − 10| ≥ 20)

\(k\sigma=20\) \(k=\frac{20}{\sigma}=\frac{20}{\frac{10\sqrt{3}}{3}}=2\sqrt{3}\)

\(Pr(\left |X-10 \right |\geq 20)\leq \frac{1}{(2\sqrt{3})^{2}}=\frac{1}{12}\)

n <- 20

k <- n/sd

upper_bound <- 1/(k^2)
upper_bound
## [1] 0.08333333

The upper bound is \(\frac{1}{12}\).