DATA605_HW8

#11 and #14 on page 303 of probability text 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.)

Let Xi be the independent random variable for light bulb i. Then E[Xi]=1/λi=1000, because the expected lifetime of a bulb is 1000 hours. This means λi=1/1000.

Since Xi is given as exponential, we know that:

min{X1,X2,…,X100}∼exponential((100^∑_i=1)λi)

And (100^∑_i=1)λi=100×1/1000=1/10

Therefore, the expected value E[minXi]=1/1/10=10

So the expected time for the first of these bulbs to burn out is 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

fZ(z)=12λeλ|z|

I’ll start by finding the probability density function for X1 and X2 using the exponential distribution function:

f(x1)=λe−λx1 f(x2)=λe−λx2

Using convolutions, the joint density of X1 and X2 is (λe^{−λx1)×(λe}−λx2), or (λ^2)(e−λ(x1+x2)).

Because Z=X1−X2, we know that x1=z+x2. Then, we can use substition to get the joint density of Z and X2, which is (λ^2)(e−λ((z+x2)+x2)) or (λ^2)(e−λ(z+2x2)).

We know that x2=x1−z. Therefore, if z is negative, x2 is greater than −z, and if z is positive, then x2 is positive. So:

When z is negative, (∫^∞_−z)(λ^2)(e−λ(z+2x2))dx=λ/2(e^λz).

When z is positive, (∫^∞_0)(λ^2)(e−λ(z+2x2))dx=(λ/2)(e^−λz).

Or, fZ(z)=12λeλ|z|

#1 on page 320-321 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.

Chebyshev’s Inequality: P(|X−μ|≥kσ)≤(1/k^2)

σ=√(100/3)=10/√3

1. P(|X−10|≥2).

kσ=2, so k(10/√3)=2 or k=((2√3)/10). Therefore, 1/k^{2=1/(2√3/10)}2 or 1/0.12=8.333. But the probability can’t be more than 1, so the bound is 1.

2. P(|X−10|≥5).

kσ=5, so k(10/√3)=5 or k=√3/2. Therefore, 1/k^2=1/(√3/2)2 or (1/3/4)=1.333. But the probability can’t be more than 1, so the bound is 1.

3. P(|X−10|≥9).

kσ=9, so k(10/√3)=9 or k=((9√3)/10). Therefore, 1/k^{2=1/((9√3)/10)}2 or 1/2.43=0.412.

4. P(|X−10|≥20).

kσ=20, so k(10/√3)=20 or k=2√3. Therefore, 1/k^2=1/(2√3)2 or 1/12=0.083.