## #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.)

From Question 10, we know that the density for M (minimum of a series of value of a series of independent variables) is exponential with mean µ/n. We know mean = 1000 and n = 100. The exepcted value of the min is equal to the mean of the distribution

1000/100
## [1] 10

#### (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 $fZ(z) = (1/2)λe^{−λ|z|}$

We can start with the following PDFs:

$f(x_1)=λe^{−λx1}$

$f(x_2)=λe^{−λx2}$

$f(x_1)*f(x_2) = λ2e^{−λ(x_1+x_2)}$

We know:

$Z=x_1−x_2$

$x_1=Z+x_2$

Therefore:

$λ2e−λ((z+x2)+x2)=λ2e−λ(z+2x2)$

When z is negative

$∫λ2e^{−λ(z+2x_2)}dx=\frac{λ}{2}e^{λz}$

When z is positive

$∫λ2e^{−λ(z+2x2)}dx=\frac{λ}{2}e^{−λ|z|}$

Therefore, together we have: $(1/2)λe^{−λ|z|}$

## 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:

$P(|X−μ|≥kσ)≤ \frac{1}{k^2}$ μ = 10
σ = (100/3)^(1/2)

#### (a) P(|X − 10| ≥ 2).

2 = k*σ
k = 2/σ

k = 2/((100/3)^(1/2))
1/k^2
## [1] 8.333333

#### (b) P(|X − 10| ≥ 5).

k = 5/σ

k = 5/((100/3)^(1/2))
1/k^2
## [1] 1.333333

#### (c) P(|X − 10| ≥ 9).

k = 9/σ

k = 9/((100/3)^(1/2))
1/k^2
## [1] 0.4115226

#### (d) P(|X − 10| ≥ 20).

k = 5/σ

k = 20/((100/3)^(1/2))
1/k^2
## [1] 0.08333333