DATA605 - Assignment 8

Ch 7 Ex 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.)

Answer

From problem 10, minimum value has the mean \(μ/n\)

Given, μ = 1000, n=100

Expected time for the first burn out is \(1000/100=10\)

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Ch 7 Ex 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 \(f_Z(z)=(1/2)λe^{−λ|z|}\)

Answer

Given, \(Z = X_1 − X_2\) with \(X_1\) and \(X_2\) being two positive numbers

Case 1: When \(X_1≤X_2\) and \(Z≤0\)

Probability here is from \(−∞\) to \(0\)

So, \(f_Z(z)=∫_{−∞}^0f_{X_1}(z+x_2)f(x_2)dx_2\\ =∫_{−∞}^0λe^{−λ(z+x_2)}λe^{−λx_2}dx_2\\ =λ^2e^{−λz}∫_{−∞}^0e^{−2λx_2}dx_2\\ =λ^2e^{−λz}\frac{−1}{2λ}\\ =\frac{-1}{2}λe^{−λz}\)

Case 2: When \(X_1≥X_2\) and \(Z≥0\)

Probability here is from \(0\) to \(∞\)

So, \(f_Z(z)=∫_0^∞f_{X_1}(z+x_2)f(x_2)dx_2\\ =∫_0^∞λe^{−λ(z+x_2)}λe^{−λx_2}dx_2\\ =λ^2e^{−λz}∫_0^∞e^{−2λx_2}dx_2\\ =λ^2e^{−λz}\frac{1}{2λ}\\ =\frac{1}{2}λe^{−λz}\)

Case 1 and 2 together, \(f_Z(z)=\begin{cases}\frac{-1}{2}λe^{−λz} Z≤0 \\ \frac{1}{2}λe^{−λz} Z≥0\end{cases}\)

Generalizing the above leads to \(f_Z(z)=\frac{1}{2}λe^{−λ|z|}\)

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Ch 8, Ex 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 (a) P(|X−10|≥2). (b) P(|X−10|≥5). (c) P(|X−10|≥9). (d) P(|X−10|≥20).

Answer

Given

Means \(μ = 10\)

Variance \(σ^2 = \frac{100}{3}\)

μ<-10
σ <-sqrt(100/3)

From Chebyshev’s Inequality we have

\(P(|X−μ|≥kσ)≤\frac{1}{k^2}\)

(a) P(|X−10|≥2)

kσ = 2
k=2/σ
paste(1/k^2)

## [1] "8.33333333333334"

(b) P(|X−10|≥5).

kσ = 5
k=5/σ
paste(1/k^2)

## [1] "1.33333333333333"

(c) P(|X−10|≥9)

kσ = 9
k=9/σ
paste(1/k^2)

## [1] "0.411522633744856"

(d) P(|X−10|≥20).

kσ = 20
k=20/σ
paste(1/k^2)

## [1] "0.0833333333333333"