11 and 14 on page 303 of probability text

  1. 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 burnout? (See Exercise 10.)

Since x1,x2…xn are independent exponential random variable … n= 100 \(λi=1/1000\)

so,\(E[Xi]=\frac{1}{λi}=1000\)

min\({X1,X2,...,X100}∼exponential(100∑i=1 λi)\)

and \(100 ∑ λi=100×\frac{1}{1000}=\frac{1}{10}\)

\(E(min) = \frac{1}{1/10} = 10\)

Ans = 10 hours

  1. Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that \(Z = X_1 − X_2\) has density \(fZ(z) = \frac{1}{2}λe−^{λ|z|}\)

Show the probability density function for \(X1\) and \(X2\) using exponential distribution function.

\(f(X1) = \lambda e ^{λ - x1}\) \(f(X2) = \lambda e ^{λ - x2}\)

The joint density function of X1 and X2 is \(\lambda e ^(λ - x1) * \lambda ^2 e ^ (-λ(x1 + x2))\)

We can rearrage \(Z = X_1 - X_2\) to \(X_1 = Z + X_2\) and \(X_2 = X_1 - Z\)

If Z is negative, X2 is greater than -Z. If Z is positive, X2 is positive.

Negative Z is \[\int_{-z}^{\infty} \lambda^2 e^{-\lambda(z+2x_2)}dx = \frac{\lambda}{2}e^{\lambda z}\] Positive Z is \[\int_{-z}^{\infty} \lambda^2 e^{-\lambda(z+2x_2)}dx = \frac{\lambda}{2}e^{-\lambda z}\]

  1. 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.
  1. P(|X − 10| ≥ 2)
  2. P(|X − 10| ≥ 5)
  3. P(|X − 10| ≥ 9)
  4. P(|X − 10| ≥ 20)

Chebyshev’s Inequality is \(P(|X - μ| ≥ kσ) ≤ 1/k^2\)

\(σ=√(100/3) = 10/C3\)

  1. \(kσ = 2\) so \(k = 2/√100/3\) \(P(|X − 10| ≥ 2) = 1/k^2\) = 8.3333 Probability can’t be more than 1, so the upper bound is 1.

  2. \(kσ = 5\) so \(k = 5/√100/3\) \(P(|X − 10| ≥ 5) = 1/k^2\) = 1.33333 Probability can’t be more than 1, so the upper bound is 1.

  3. \(kσ = 9\) \(k = 9/√100/3\) \(P(|X − 10| ≥ 9) = 1/k^2\) = 0.4115 0.4115 is the upper bound.

  4. \(kσ = 20\) \(k = 20/√100/3\) \(P(|X − 10| ≥ 20) = 1/k^2 = 0.083\) 0.083 is the upper bound.