Problem 11 of Introduction to Probability, Grinstead, C. Snell, J., 1997 (Page 303)

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?

size = 100
lifetime = 1000 
expected_time = lifetime / size
cat("Expected time for the first lightbulb to burn out:", expected_time, "hours\n")
## Expected time for the first lightbulb to burn out: 10 hours

Problem 14 of Introduction to Probability, Grinstead, C. Snell, J., 1997 (Page 303)

Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that Z = X1 − X2 has density fZ(z) = (1/2)λe−λ|z|

\[f_z(Z)=\int_{-\infty}^{+\infty} f_{X2}(x_1-Z)f_{X1}(x_1)dx_1=\int_{0}^{\infty} \lambda*e^{-\lambda*x_1} \lambda*e^{-\lambda*(x_1-z)}dx_1=\int_{0}^{\infty}\lambda^2*e^{-2\lambda*x_1}e^{\lambda*z}dx_1=\int_{0}^{\infty}\lambda^2*e^{\lambda(z-2*x_1)}dx_1=(1/2)λe^{−λ|z|}\]

Problem 1 of Introduction to Probability, Grinstead, C. Snell, J., 1997 (Page 32 - 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.

(a) P(|X − 10| ≥ 2):

Chebyshev’s Inequality is given by

\[P(|X -μ|≥kσ ≤\frac{1}{k^{2}}\]

μ = 10

\(σ = \sqrt{variance} = \sqrt{\frac{100}{3}}\)

k = 2

\[P(∣X−10 ∣≥ 2*\sqrt{\frac{100}{3}})≤\frac{1}{2^{2}}\]

\[P(∣X−10 ∣≥ 2*\frac{10}{\sqrt{3}})≤\frac{1}{4}\]

\[P(∣X−10 ∣≥ \frac{20}{\sqrt{3}})≤\frac{1}{4}\]

\[Therefore, the\ upper\ bound\ for\ P(∣X−10 ∣≥2 =\frac{1}{4}\]

(b) P(|X − 10| ≥ 5):

k = 5

\[P(∣X−10 ∣≥ 5*\sqrt{\frac{100}{3}})≤\frac{1}{5^{2}}\]

\[P(∣X−10 ∣≥ 5*\frac{10}{\sqrt{3}})≤\frac{1}{25}\]

\[P(∣X−10 ∣≥ \frac{50}{\sqrt{3}})≤\frac{1}{25}\]

\[Therefore, the\ upper\ bound\ for\ P(∣X−10 ∣≥5 =\frac{1}{25}\]

(c) P(|X − 10| ≥ 9):

k = 9

\[P(∣X−10 ∣≥ 9*\sqrt{\frac{100}{3}})≤\frac{1}{9^{2}}\]

\[P(∣X−10 ∣≥ 9*\frac{10}{\sqrt{3}})≤\frac{1}{81}\]

\[P(∣X−10 ∣≥ \frac{90}{\sqrt{3}})≤\frac{1}{81}\]

\[Therefore, the\ upper\ bound\ for\ P(∣X−10 ∣≥9 =\frac{1}{81}\]

(d) P(|X − 10| ≥ 20):

k = 20

\[P(∣X−10 ∣≥ 20*\sqrt{\frac{100}{3}})≤\frac{1}{20^{2}}\]

\[P(∣X−10 ∣≥ 20*\frac{10}{\sqrt{3}})≤\frac{1}{400}\]

\[P(∣X−10 ∣≥ \frac{200}{\sqrt{3}})≤\frac{1}{400}\]

\[Therefore, the\ upper\ bound\ for\ P(∣X−10 ∣≥20 =\frac{1}{400}\]