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.)
The exponential equation:
R(t)=R0*e^(−t/c),
t = -c * log(R(t)/R0)
R0 = 100 (the initial number of light bulbs)
c = 1000 (time)
R(t) = 99 (as we need to find when the first bulb will burn out)
t <- -1000 * log(99/100)
t## [1] 10.05034The expected time for the first of these bulbs to burn out is 10h.
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|).
Z=X1−X2
X2=X1−Z
\[fZ(z)= \int_{-\infty}^{+\infty} fx1(X1)fx2(X1-Z)dX1 = \int_{Z}^{+\infty} \lambda \epsilon^{-\lambda X1} \lambda \epsilon^{-\lambda (X1-Z)}dX1 = \int_{Z}^{+\infty} \lambda^{2} \epsilon^{-2\lambda X1+\lambda Z}dX1 = \int_{-Z/2lambda}^{+\infty} (λ/2)e−λ|z| \]
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.
P(|X−μ|≥ϵ)≤σ2/ϵ2
upper bound = σ2/ϵ2
σ^2 = 100/3 ϵ^2 = 2^2
σ.sq = 100/3
up.bound.1<- σ.sq/2^2
up.bound.1## [1] 8.333333b.P(|X−10|≥5).
σ.sq = 100/3
up.bound.2<- σ.sq/5^2
up.bound.2## [1] 1.333333c.P(|X−10|≥9).
σ.sq = 100/3
up.bound.3<- σ.sq/9^2
up.bound.3## [1] 0.4115226d.P(|X − 10| ≥ 20).
σ.sq = 100/3
up.bound.4<- σ.sq/20^2
up.bound.4## [1] 0.08333333