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.)

Solution:

According to the properties of exponential distribution, if\(X_i\sim Exp(\lambda_i)\) then \(minX_i \sim Exp(\Sigma\lambda_i)\)

According to Exercise 10, if \(X_i\sim Exp(\lambda)\) then \(min\left\{X1, ..., Xn\right\} \sim Exp(n\lambda)\)

\(\frac{1}{\lambda} = E[x] =1000\)

So

\(\lambda = \frac{1}{1000}\)

\(n\lambda = 100 * \frac{1}{1000} = \frac{1}{10}\)

\(P(min {X1, X2, . . . , Xn} <= x) = 1-e^{-\frac{1}{10}x}\)

\(E[minX_i]= \frac{1}{\frac{1}{10}}=10\)

So the expected time for the first of these bulbs to burn out is 10 h.

14. Assume that X1 and X2 are independent random variables, each having an exponential density with parameter \(\lambda\). Show that Z = X1 - X2 has density:

\(f_Z(z) = (1/2)\lambda e^{-\lambda|z|}\)

Solution:

First compute the cumulative distribution function (CDF):

\(P_r\left\{Z \leq z \right\}\)

Use the converlution formula, we can write:

\[\begin{equation} \int\limits_{-\infty}^\infty f_{X_1}(x_1)f_{X_2}(x_2)\; \mathrm{d}x = \int\limits_{-\infty}^\infty f_{X_1}(x_1)f_{-X_2}(z-x_1)\; \mathrm{d}x \end{equation}\]

for discrete random variable, P(-X=-1) = P(X=1), same is true for probability density function (PDF):

\(f_{-X_2}(z-x_1) = f_{X_2}(x_1-z)\)

So we can write the above equation as:

\[\begin{equation} \int\limits_{-\infty}^\infty f_{X_1}(x_1)f_{-X_2}(z-x_1)\; \mathrm{d}x_1 = \int\limits_{-\infty}^\infty f_{X_1}(x_1)f_{X_2}(x_1-z)\; \mathrm{d}x_1 \end{equation}\]

For z<0, the right part of above equation could be written according to the exponetial distribution function:

\[\begin{equation} \begin{split} f_Z(z) & = \int\limits_0^\infty \lambda e^{-\lambda x_1} \lambda e^{-\lambda (x_1-z)}\; \mathrm{d}x_1 & = \int\limits_0^\infty \lambda^2 e^{-2\lambda x_1}e^{\lambda z}\; \mathrm{d}x_1 \\ & = \lambda e^{\lambda z} \int\limits_0^\infty \lambda e^{-2\lambda x_1}\; \mathrm{d}x_1 \\ & = \lambda e^{\lambda z} (\left.-\frac{1}{2}e^{-2\lambda x_1}\right|_0^\infty) \\ & = \frac{1}{2}\lambda e^{\lambda z} \end{split} \end{equation}\]

\(Z = X_1 - X_2\), so we know \(-Z = X_2 - X_1\). \(X_1\) and \(X_2\) are independent and identically distributed, so we know \(X_1 - X_2\) has the same distribution as \(X_2 - X_1\). Z has the same distribution as -Z.That means the distribution of Z is symmetric around 0.

\(f_Z(z) = f_{-Z}(-z)\)

for z>=0,

\[\begin{equation} f_Z(z) = \frac{1}{2}\lambda e^{\lambda (-z)} \end{equation}\]

Taken together, we have

\[\begin{equation} f_Z(z) = (1/2)\lambda e^{-\lambda|z|} \end{equation}\]

1. Let X be a continuous random variable with mean \(\mu = 10\) and variance \(\sigma^2 = 100/3\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities.(on page 320-321)

  1. \(P(|X - 10| \geq 2)\)
  2. \(P(|X - 10| \geq 5)\)
  3. \(P(|X - 10| \geq 9)\)
  4. \(P(|X - 10| \geq 20)\)

Solution:

Chebyshev’s Inequality:

\(P(|X - \mu| \leq k\sigma) \leq \frac{1}{k^2}\)

we know:

\(\mu = 10\)

\(\sigma^2 = 100/3\)

(a) P(|X - 10| >= 2)

\(P(|X - 10| \geq 2)\leq \frac{1}{k^2}\)

\(k=2/\sqrt{100/3}\)

k <- 2/sqrt(100/3)
1/k^2
## [1] 8.333333

\(P(|X - 10| \geq 2) \leq 8.3333\)

(b) P(|X - 10| >= 5)

\(P(|X - 10| \geq 5)\leq \frac{1}{k^2}\)

\(k=5/\sqrt{100/3}\)

k <- 5/sqrt(100/3)
1/k^2
## [1] 1.333333

\(P(|X - 10| \geq 5) \leq 1.3333\)

(c) P(|X - 10| >= 9)

\(P(|X - 10| \geq 9)\leq \frac{1}{k^2}\)

\(k=9/\sqrt{100/3}\)

k <- 9/sqrt(100/3)
1/k^2
## [1] 0.4115226

\(P(|X - 10| \geq 9) \leq 0.4115\)

(d) P(|X - 10| >= 20)

\(P(|X - 10| \geq 20)\leq \frac{1}{k^2}\)

\(k=20/\sqrt{100/3}\)

k <- 20/sqrt(100/3)
1/k^2
## [1] 0.08333333

\(P(|X - 10| \geq 20) \leq 0.0833\)