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 burn out?

We know that if

\[ X_{i} \sim Exp(\lambda_{i}) \]

then

\[ \min X_{i} \sim Exp({\sum\lambda_{i}}) \]

and

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

so

\[ {\lambda_{i}} = \frac{1}{1000} \]

this gets us to

\[ \min X_{i} \sim Exp({\sum\lambda_{i}}), with \lambda = \lambda_{1} + ... + \lambda_{100} = \frac{100}{1000} = \frac{1}{10} \]

and finally

\[ E[minX_{i}] = \frac{1}{\lambda} = \frac{1}{\frac{1}{10}} = 10 \]

  1. Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that Z = X1 − X2 has density \(fz(z)=\frac{1}{2}\lambda e ^{-\lambda \mid z \mid}\)

The PDF of:

\(W = X + Y\) is \(\int_{-\infty}^{\infty}f_X(x)f_Y(W - x)dx\).

Therefore we can say

\(Z = X + (-Y)\).

Using the convolution formula:

\(f_Z(z) = \int_{-\infty}^{\infty}f_X(x)f_{-Y}(z - x) dx\).

We can rewrite:

\(f_{-Y}(z - x)\) as \(f_Y(x - z)\).

Our pdf is:

\[ f_X(x) = f_Y(x) = \begin{cases} \lambda e^{\lambda x} & x \geq 0,\\ 0 & x < 0. \end{cases} \]

\(f_Z(z) = \int_{0}^{\infty}\lambda e^{-\lambda x}\lambda e^{-\lambda(x - z)} dx\).

\(f_Z(z) = \lambda e^{\lambda z}\int_{0}^{\infty}\lambda e^{-2\lambda x} dx\).

\(f_Z(z) = \lambda e^{\lambda z}\Big(-\frac{1}{2}e^{-2\lambda x}\Big|_{0}^{\infty}\Big) = \frac{1}{2}\lambda e^{\lambda z}\).

This can be restated as:

\(Z = X - Y\), as \(-Z = Y - X\)

Since they are iid this will have the same distribution as:

\(X - Y\), i.e. \(f_Z(z) = f_Z(-z)\)

Therefore the PDF is:

\[ f_Z(z) = \begin{cases} \frac{1}{2}\lambda e^{\lambda z} & z < 0,\\ \frac{1}{2}\lambda e^{-\lambda z} & z \geq 0. \end{cases} \]

or

\[ \frac{1}{2}\lambda e^{-\lambda|z|} \]

  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.

\(A. P(|X - 10| \geq 2)\).

## [1] 8.333333

\(B. P(|X - 10| \geq 5)\).

## [1] 1.333333

\(C. P(|X - 10| \geq 9)\).

## [1] 0.4115226

\(D. P(|X - 10| \geq 20)\).

## [1] 0.08333333