{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

20065ee131A_1_EE131AF06HW5Solution

20065ee131A_1_EE131AF06HW5Solution - Solution of Assignment...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Solution of Assignment 5, EE 131A Due: Monday November 20, 2006 Problem 1. (a) First prove that the memoryless property of a random variable X P [ X > k + j vextendsingle vextendsingle X > j ] = P [ X > k ] , k, j > 0 , is equivalent to the following relation: P [ X > k + j ] = P [ X > k ] P [ X > j ] . Prove that the geometric random variable satisfies this property. (b) Suppose that X is a discrete random variable that satisfies the memoryless property. Show that this implies that X is a geometric random variable. Solution: (a) Since, for k, j > 0 , P [ X > k + j vextendsingle vextendsingle X > j ] = P [ { X > k + j } ∩ { X > j } ] P [ X > j ] = P [ X > k + j ] P [ X > j ] it follows that the memoryless property is equivalent to P [ X > k + j ] = P [ X > k ] P [ X > j ] . (b) Let p = P [ X = 1] and q = 1 - p . By induction, assume that P [ X k ] = 1 - q k , then the memoryless property implies that P [ X k + 1] = 1 - P [ X > k + 1] = 1 - P [ X > k ] P [ X > 1] = 1 - q k · q = 1 - q k +1 . Now, we have P [ X = k ] = P [ X k ] - P [ X k - 1] = 1 - q k - 1 + q k - 1 = q k - 1 (1 - q ) = q k - 1 p, which shows that X is a geometric random variable. Problem 2. The life of a certain type of automobile tire is normally distributed with mean m = 34 , 000 miles and standard deviation σ = 4000 miles. (a) What is the probability that such a tire lasts over 40,000 miles? (b) What is the probability that it lasts between 30,000 and 35,000 miles?
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(c) Given that it has survived 30,000 miles, what is the conditional probability that it survives another 10,000 miles?
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern