DATA 605 - HW 8

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?

Let X_i —, X_n be independently expoentially distributed random variables with rate parametesr λ_i…, λ_n. Then

min{X_i …, X_n} is also exponentially distributed with parameter

\[λ = λ_i + ... + λ_n\]

\[Pr(k|X_k = min{X_i ..., X_n}) = \frac{λ_k}{λ_i + ... + λ_n}\]

λ is the rate in which bulbs burn out therefore with an exponential lifetime, we can say that \[λ_i = \frac{1}{1000} , ∑λ_i= \frac{1}{10}\]

Emin = 1/(1/10)
paste("E(X) = ", Emin, "hours")

## [1] "E(X) =  10 hours"

14. Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that Z = X1 − X2 has density

Two numbers at random from the interval [0,∞) with an exponential density has the following sum: Letting X, Y, and Z = X + Y, then

\[fx(x) = fy(x) = \begin{cases} λe^{−λx},& \text{if } x\geq 0\\ 0, & \text{otherwise} \end{cases} \]

For W = X + Y

\[fw(w) = \int_{-∞}^{∞} f_x(x)f_y(w-x) dx\]

Therefore, for Z = X1 - X2, let us say Z = X + (-Y), so

\[fz(z) = \int_{-∞}^{∞} f_x(x)f_{-y}(z-x) dx\] We can say that

\[f_{-y}(z-x) = f_{y}(x-z) \]

So,

\[fz(z) = \int_{-∞}^{∞} f_x(x)f_y(x-z) dx\]

\[fz(z) = \int_{0}^{∞} λe^{-λx}λe^{-λ(x-z)}dx\]

\[ = λe^{λz}\int_{0}^{∞} λe^{-2λx}dx\]

\[ = λe^{λz}(-\frac{1}{2}e^{-2λx}\Bigr|_{\substack{∞\\0}})\]

Which evaluates to the density below

\[fZ(z) = (1/2)λe^{−λ|z|}\]

This video was incredibly useful: https://www.youtube.com/watch?v=f8Nli1AfygM

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 states the following. Suppose that μ = E(X) and σ^2 = V(X), then for any positive number ϵ > 0 we have

\[P(|X-μ|≥kσ)≤ \frac{σ^2}{k^2σ^2} = \frac{1}{k^2}\]

1. P(|X − 10| ≥ 2)

\[P(|X-10|≥2) ≤ \frac{1}{2^2}\]

cheb <- ((100/3)^2)/(2^2)
paste("The upper bound is =", pmin(cheb, 1))

## [1] "The upper bound is = 1"

2. P(|X − 10| ≥ 5).

\[P(|X-10|≥5) ≤ \frac{1}{5^2}\]

cheb <- (1)/(5^2)
paste("The upper bound is =", pmin(cheb, 1))

## [1] "The upper bound is = 0.04"

3. P(|X − 10| ≥ 9).

\[P(|X-10|≥9) ≤ \frac{1}{9^2}\]

cheb <- (1)/(9)
paste("The upper bound is =", pmin(cheb, 1))

## [1] "The upper bound is = 0.111111111111111"

4. P(|X − 10| ≥ 20).

\[P(|X-10|≥9) ≤ \frac{1}{9^2}\]

cheb <- 1/(20^2)
paste("The upper bound is =", pmin(cheb, 1))

## [1] "The upper bound is = 0.0025"