Week 8






P. 303 # 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.)

The question is asking about the expected or mean or average time of any of the first (minimum) of 100 bulbs.


Exercise 10 illustrated that the distribution of minimums is related to the original distributions.

\[\mu_M \ = \ \frac{\mu_X}{n}\]

and we know \(E(x) \ = \ \frac{1}{\lambda}\)

so the mean of the minimums or the expected minimum would therefore be \(\frac{1000}{100} = 10\)

ev<-1000
lambda<-1/ev
n<-100
ev/n
## [1] 10


The below simulation seems to support the above


# given lambda of .001 (one every thousand hours), rexp returns the number of hours to the next event
min(rexp(100,lambda))
## [1] 8.626458

P. 303 # 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

\[fZ(z) = (1/2)\lambda e^{-\lambda |z|}\]

We know that when Z = X1 + X2, then the integration …

\[\int_{-\infty}^{\infty} fx(z-y)fy(y)dy \ =\]

leads to..

\[fZ(z) = \lambda^2ze^{-\lambda z}\]

I presume to solve for X1-X2 we need to apply the integration algebra for negating the function of y.


I didnt see how this can be done, but Ill continue to look…


P. 320 # 1

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


The Chebyshev inequality is an alternative to the 68/95/99 convention of probability.


It is considered applicable to distributions that arent necessarily normal.


Its a looser metric that includes 75% of all values within 2 standard deviations.


It enables us to obtain bounds on probability when both the mean and variance of a random variable are known.


\[P(X \notin (lower,upper) ) \ = \ P(|X - E(X)| \ge c) \ = \ \frac{VAR(X)}{c^2}\]


where c > 0


If you factor the standard deviation out of c, then you have a simplified equation

\[P(|\sigma - \mu| > k\sigma ) \ \le \frac{1}{k^2}\]

\[(a) P(|X − 10| \ge 2)\]

mu<-10
sigma<-100/3


c<-2
sprintf("Bounds = %.4f", sigma^2/c^2)
## [1] "Bounds = 277.7778"

\[(b) P(|X − 10| \ge 5)\]

c<-5
sprintf("Bounds = %.4f", sigma^2/c^2)
## [1] "Bounds = 44.4444"

\[(c) P(|X − 10| \ge 9)\]

c<-9
sprintf("Bounds = %.4f", sigma^2/c^2)
## [1] "Bounds = 13.7174"

\[(d) P(|X − 10|\ge 20)\]

c<-20
sprintf("Bounds = %.4f", sigma^2/c^2)
## [1] "Bounds = 2.7778"


I had some trouble gaining clarity on this weeks questions especially what the definition of bounds is.


I did find this video to be a useful review of Markov and Chebyshev bounds and how they relate to some core concepts of distributions.


Also this python demonstration was also useful.