install.packages(“combinat”) require(combinat)

Chapter 7

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.)

From Exercise 10, we know that the CDF for the minimum function for the sum of n random varialbes is
P(min(\(X_1\),…,\(X_n\) > x) = (P(X1 > x))\(^n\)
since the probability density for X1 to Xn are all equal.

Converting this equation to exponential form by substituting X1 > x with X1 > e\(^(-x/u)\), we get
P(min (e\(^(-x/µ)^n\))) = P(e\(^-(n/u)*x\)).

Solving for PDF of the CDF, we get the deriative f(x) = n/u*e\(^(-(n/u)*x)\).

The mean of the minimum function for the exponential distribution is then
1/\(\lambda\) = 1/(n/u) or u/n.
The expected time then for the first (minimum) of the 100 (n) lighbulbs to burn out where u = 1000 hours (mean) is
u/n = 1000/100 = 10 hours.


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 fZ(z) = (1/2)\(\lambda\)e^(-\(\lambda\)|z|).

First we note that Z = X1 - X2 is the same as Z = X1 + (-X2) or setting X2 = Y, Z = X + (-Y).
This means that for the convulution formula fz(z) = \(\int_{-\infty}^{\infty} fx(x) f-y(z-x) dx\) = \(\int_{-\infty}^{\infty} fx(x) fy(x-z) dx\) since f-y(z-x) = fy(x-y)

Computing for the CDF then of an expontial density function given the equation above -
for z < 0, fx(x) = [0 for x< 0 and \(\lambda\)e^(-\(\lambda\)x) for x >= 0]

= \(\int_{0}^{\infty} \lambda e^(-\lambda*x) \lambda e^(-x(x-z)) dx\) = \(\lambda\)e^(\(\lambda\)z) * (-1/2 * e^(-2x)\(\lambda\))
= \(\lambda\)/2 * e^(\(\lambda\)*z) for z < 0

for z >= 0

we note that Z = X - Y and thus -Z = Y - X, or the density function fz(z) = -f-z(z)

Which means for z>=0, the CDF density function is the symmetrical opposite of Z < 0

thus for z >= 0

fz(z) = \(\lambda\)/2 * e^(-\(\lambda\)*z)

joining the 2 equations then all z values =

(1/2)\(\lambda\)e^(-\(\lambda\)|z|)


Chapter 8

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.

  1. P(|x - 10|>=2) <=
min(1,(100/3)/2^2)
## [1] 1
  1. P(|x - 10|>=5) <=
min(1,(100/3)/5^2)
## [1] 1
  1. P(|X - 10|>=9) <=
min(1,(100/3)/9^2)
## [1] 0.4115226
  1. P(|X - 10|>=20) <=
min(1,(100/3)/20^2)
## [1] 0.08333333