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 —, Xn be independently expoentially distributed random variables with rate parametesr λi…, λn. Then
min{Xi …, Xn} is also exponentially distributed with parameter
\[λ=λ_i+...+λ_n\]
\[Pr(k|X_k=minX_i...,X_n)=\frac{λ_k}{λ_i+...+λ_n}\]
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 = 1/0.1
Emin## [1] 1014 Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that Z = X1 − X2 has density \[fZ(z) = (\frac{1}{2})\lambda e^{−\lambda|z|}\]
For W = X + Y
\[f_w(w)=∫^∞_{−∞}f_x(x)f_y(w−x)dx\]
Therefore, for Z = X1 - X2, let us say Z = X + (-Y), so
\[f_z(z)=∫^∞_{−∞}f_x(x)f_{−y}(z−x)dx\]
We can say that
\[f_{−y}(z−x)=f_y(x−z)\]
So,
\[fz(z)=∫^∞_{−∞}f_x(x)f_y(x−z)dx\]
\[fz(z)=∫^∞_0λe^{−λx}λe^{−λ(x−z)}dx\]
\[=λeλz∫^∞_0λe^{−2λx}dx\]
\[=λe^{λz}(−\frac{1}{2}e^{−2λx}∣∞->0)\]
Which evaluates to the density below
\[fZ(z)=(\frac{1}{2})λe^{−λ|z|}\]
Exercises Let X be a continuous random variable with mean μ = 10 and variance = 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
\[P(|X−μ|≥kσ)≤\frac{σ^2}{k^2σ^2}=\frac{1}{k^2}\]
\[P(|X−10|≥2)≤\frac{1}{2^2}\]
mu = 10
var_nce = 100/3
sd = sqrt(100/3)
ksd = 2
k = 2/sqrt(100/3)
upper_cheb = 1/(k^2)
pmin(upper_cheb, 1)## [1] 1ksd = 5
k = 5/sqrt(100/3)
upper_cheb = 1/(k^2)
pmin(upper_cheb, 1)## [1] 1ksd = 9
k = 9/sqrt(100/3)
upper_cheb = 1/(k^2)
pmin(upper_cheb, 1)## [1] 0.4115226ksd = 20
k = 20/sqrt(100/3)
upper_cheb = 1/(k^2)
pmin(upper_cheb, 1)## [1] 0.08333333Confirmation
u = 10
o = sqrt(100/3)
x1 = c(2, 5, 9, 20)
k = (x1 )/o
p1 = round(1/(k^2), 4)
p1## [1] 8.3333 1.3333 0.4115 0.0833