Data 605 HW8

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Exercise 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.)

Ans:

Let X₁, X₂, . . . , X_n be n independent random variables each of which has an exponential density with mean \(\mu\), the minimum value the X_j, M, is of exponential density function with mean \(\frac{\mu}{n}\).

In this case μ=1000 hours and n=100 so the expected time for the first of these bulbs to burn out is 1000/100=10 hours with the formula given above.

Exercise 7.14

Assume that X₁ and X₂ are independent random variables, each having an exponential density with parameter . Show that Z = X₁ - X₂ has density

\(f{_{Z}}(z) = (\frac{1}{2})\lambda\exp^{-\lambda\left | z \right |}\)

Ans:

Since both X₁ and X₂ are independent random variables, the distribution of X1 - X2 is equivalent to the distribution of X₂ - X₁ except with a sign change.

To start, let’s evaluate when X₂ \(\geq\) X₁,

\[ \begin{multline*} \begin{split} f{_{Z}}(z) &= \int_{-\infty }^{\infty } f{_{x_1}}(z+x_2)f_{x_2}(x_2)dx_2 \\ & = \int_{-\infty }^{0 } \lambda\exp^{-\lambda(z+x_2)}\lambda\exp^{-\lambda x_2}dx_2 \\ &= \int_{-\infty }^{0 } \lambda\exp^{-\lambda z} \lambda\exp^{-2\lambda x_2}dx_2 \\ &= \int_{-\infty }^{0 } \lambda\exp^{-\lambda z} \lambda\exp^{-2\lambda x_2}dx_2 \\ &= \lambda^{2}exp^{-\lambda z} (\int_{-\infty }^{0} \exp^{-2\lambda x_2}dx_2 ) \\ &= \lambda^{2}exp^{-\lambda z} \frac{-1}{2\lambda} \\ &= \frac{-\lambda}{2} exp^{-\lambda z} \end{split} \end{multline*} \]

\(f{_{Z}}(z) =\left\{\begin{matrix}\frac{-\lambda}{2} \exp^{-\lambda z} & z < 0 \\ \frac{\lambda}{2} \exp^{-\lambda z} & z >= 0\end{matrix}\right.\)

Then we proved that \(f(z) = \frac{\lambda}{2}\exp^{-\lambda \left | z \right |}\)

Exercise 1 (p. 320-321)

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.

\(P(|X - \mu| \geq \epsilon ) \leq \frac{\sigma^2}{\epsilon^2}\)

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

Ans: Let’s define \(\epsilon\) = 2, we get P(|X - 10| \(\geq\) 2) \(\leq \frac{100/3}{2^2} = \frac{25}{3}\)

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

Ans: Let’s define \(\epsilon\) = 5, we get P(|X - 10| \(\geq\) 5) \(\leq \frac{100/3}{5^2} = \frac{4}{3}\)

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

Ans: Let’s define \(\epsilon\) = 9, we get P(|X - 10| \(\geq\) 9) \(\leq \frac{100/3}{9^2} = \frac{100}{243}\)

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

Ans: Let’s define \(\epsilon\) = 20, we get P(|X - 10| \(\geq\) 20) \(\leq \frac{100/3}{20^2} = \frac{1}{12}\)