##From problem 10 from book
##the density of minimum value among n independent random variables with an exponential density has mean μ/n
n = 100
μ = 1000
##expected time for the first of the bulbs to burn out is
em <- μ/n
em## [1] 10# expected time for the first of the bulbs to burn out is 10 hrs##2.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|
Answer: For X1≥X2
f_Z(z) = {-}^{}f{X_1}(z+x_2)f_{X_2}(x_2)dx_2 \ f_Z(z) = _0{}e{-(z+x_2)} e^{-x_2}dx_2 \ f_Z(z) = _0{}e{-(z)} e^{-2x_2}dx_2 \ f_Z(z) = ^2 e^{-z}(_0^{} e^{-2x_2}dx_2) \ f_Z(z) = ^2 e^{-z}() \ f_Z(z) = (e^{-z})
For X1≥X2 f_Z(z) = {-}^{}f{X_1}(z+x_2)f_{X_2}(x_2)dx_2 \ f_Z(z) = {-}0e{-(z+x_2)} e^{-x_2}dx_2 \ f_Z(z) = {-}0e{-(z)} e^{-2x_2}dx_2 \ f_Z(z) = ^2 e^{-z}(_{-}^0 e^{-2x_2}dx_2) \ f_Z(z) = ^2 e^{-z}() \ f_Z(z) = (e^{-z})
combine these two separate cases we get the combined distribution of:
f(z)=λ2e−λ|z|
##Answer: For X1≥X2
## f_Z(z) = \int_{-\infty}^{\infty}f_{X_1}(z+x_2)f_{X_2}(x_2)dx_2 \\
## f_Z(z) = \int_0^{\infty}\lambda e^{-\lambda (z+x_2)} \lambda e^{-\lambda x_2}dx_2 \\
## f_Z(z) = \int_0^{\infty}\lambda e^{-\lambda (z)} \lambda e^{-2\lambda x_2}dx_2 \\
## f_Z(z) = \lambda^2 e^{-\lambda z}(\int_0^{\infty} e^{-2\lambda x_2}dx_2) \\
## f_Z(z) = \lambda^2 e^{-\lambda z}(\frac{ -1}{2\lambda}) \\
## f_Z(z) = \frac{\lambda}{2} (e^{-\lambda z})
## For X1≥X2
## f_Z(z) = \int_{-\infty}^{\infty}f_{X_1}(z+x_2)f_{X_2}(x_2)dx_2 \\
## f_Z(z) = \int_{-\infty}^0\lambda e^{-\lambda (z+x_2)} \lambda e^{-\lambda x_2}dx_2 \\
## f_Z(z) = \int_{-\infty}^0\lambda e^{-\lambda (z)} \lambda e^{-2\lambda x_2}dx_2 \\
## f_Z(z) = \lambda^2 e^{-\lambda z}(\int_{-\infty}^0 e^{-2\lambda x_2}dx_2) \\
## f_Z(z) = \lambda^2 e^{-\lambda z}(\frac{ -1}{2\lambda}) \\
## f_Z(z) = \frac{-\lambda}{2} (e^{-\lambda z})
## combine these two separate cases we get the combined distribution of:
## f(z)=λ2e−λ|z|##3.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. (a) P(|X − 10| ≥ 2) (b) P(|X − 10| ≥ 5) (c) P(|X − 10| ≥ 9) (d) P(|X − 10| ≥ 20
## P(|X−μ|≥kσ)≤σ^2/k^2σ^2=1/k^2
#μ=10
#σ^2=100/3
#k=kσ/σ
##(a) P(|X − 10| ≥ 2).
# P(|X−10|≥2)≤100/3/2^2
# P(|X−10|≥2)≤25/3
## (b) P(|X − 10| ≥ 5).
# P(|X−10|≥5)≤100/3/5^2
# P(|X−10|≥5)≤4/3
## (c) P(|X − 10| ≥ 9)
# P(|X−10|≥9)≤100/3/9^2
# P(|X−10|≥5)≤100/243
## (d) P(|X − 10| ≥ 20
# P(|X − 10| ≥ 20)≤100/3/20^2
# P(|X−10|≥20)≤100/1200