Question:

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.)

Answer:

From exercise 10, we were told that the expected value for M is \(min \quad {X}_{j} = \mu / n = 1000 / 100 = 10\)

Question:

Assume that \({X}_{1}\) and \({X}_{2}\) are independent random variables, each having an exponential density with parameter \(\lambda\). Show that \(Z = {X}_{1} − {X}_{2}\) has density

\[{f}_{Z}\left( z \right) = \left( 1/2 \right)\lambda{e}^{-\lambda|z|}\]

Answer:

This was was quite a struggle, so here’s my best shot. Please note, I changed the variables from \({X}_{1}, {X}_{2}\) to \(x, y\) to try from confusing myself while comparing this scenario to the reading.

We can epress this function in an integral with respect to y after re-wrting \(x = z + y\).

\[{f}_{Z}\left( z \right) = \int _{ - \infty }^{ +\infty }{f\left(z+y\right) f\left(y \right)dy}\]

To find the integral range, we have to evaluate the new domain for our formula. The integral for \(\lambda{e}^{-\lambda y}\) is defined over the domain of the exponential density function \(x \epsilon \left( 0, \infty \right]\). Now, the integral for \(\lambda{e}^{-\lambda\left(z+y\right)}\) is \(0 \le z + y \le \infty\). For the left side of the domain, we find that \(y = -z\) and for the right side we find that \(y = \infty - z = \infty\)

Time to integrate:

\[{f}_{Z}\left( z \right) = \int _{ -z }^{ +\infty }{\lambda{e}^{-\lambda\left(z+y\right)}\lambda{e}^{-\lambda y}dy}\]

\[{f}_{Z}\left( z \right) = \int _{ -z }^{ \infty }{{\lambda}^{2}{e}^{-\lambda\left(z+2y\right)}dy}\]

\[{f}_{Z}\left( z \right) = \int _{ -z }^{ \infty }{{\lambda}^{2}{e}^{-\lambda z}{e}^{-\lambda 2y}dy}\]

\[{f}_{Z}\left( z \right) = {\lambda}^{2}{e}^{-\lambda z}\left[\frac{1}{-2\lambda} {e}^{-\lambda 2y}\right]_{-z}^{\infty}\]

\[{f}_{Z}\left( z \right) = \frac{1}{-2} {\lambda}{e}^{-\lambda z}\left( {e}^{-2 \lambda \infty} - {e}^{2 \lambda z} \right)\]

As \(z \rightarrow \infty, {e}^{-2 \lambda \infty} \rightarrow 0\)

\[{f}_{Z}\left( z \right) = \frac{1}{2} {\lambda}{e}^{-\lambda z}\left( {e}^{2 \lambda z} - 0\right)\]

\[{f}_{Z}\left( z \right) = \frac{1}{2} {\lambda}{e}^{\lambda z}\]

You can see something is missing here, I do not show this function using the absolute value of \(z\). This is where I became stuck so please let me know your thoughts. Thank you.

Question:

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)

Answer:

From the reading, we know the following:

\[P\left( \left| X-\mu \right| \ge \epsilon \right) \le \frac{{\sigma}^{2}}{{\epsilon}^{2}}\]

So, the upper bound of the probability would be when it is equal to \(\frac{{\sigma}^{2}}{{\epsilon}^{2}}\)

In this case, \(\epsilon = 2\) and \({\sigma}^{2} = \frac{100}{3}\)

(round(min( (100/3) / (2)^2,1) ,3))
## [1] 1

Question:

  1. P (|X − 10| ≥ 5)

Answer:

In this case, \(\epsilon = 5\) and \({\sigma}^{2} = \frac{100}{3}\)

(round(min( (100/3) / (5)^2,1) ,3))
## [1] 1

Question:

  1. P (|X − 10| ≥ 9)

Answer:

In this case, \(\epsilon = 9\) and \({\sigma}^{2} = \frac{100}{3}\)

(round(min( (100/3) / (9)^2, 1) ,3))
## [1] 0.412

Question:

  1. P (|X − 10| ≥ 20)

Answer:

In this case, \(\epsilon = 20\) and \({\sigma}^{2} = \frac{100}{3}\)

(round(min( (100/3) / (20)^2, 1) ,3))
## [1] 0.083