{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sol1 - STA3007 Applied probability(08-09 Chapter 1 Solution...

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

View Full Document Right Arrow Icon
STA3007 Applied probability (08-09) Solution to Assignment 1 Chapter 1 Problem 2.3 Let S r = ξ 1 + · · · + ξ r , where ξ k is the number of additional samples needed to observe the k th distinct element compared to observing the ( k - 1)th distinct element. Assume that the k - 1 distinct element have already been observed, then, let p k = Pr( ξ k = 1) = N - ( k - 1) N = 1 - k - 1 N , we have Pr( ξ k = n ) = p k (1 - p k ) n - 1 , n = 1 , 2 , · · · , which is a geometric distribution indexed by parameter p k . So, E( ξ k ) = 1 /p k . Finally, the given formula E( S r ) = E( ξ 1 ) + · · · + E( ξ r ) = 1 p 1 + · · · + 1 p r = N ( 1 N + 1 N - 1 + · · · + 1 N - r + 1 ) is established. Problem 2.7 We are given that Pr { U > u and W > w } = Pr { U > u } Pr { W > w } , which means that Pr { U > u and W > w } = [1 - F U ( u )][1 - F W ( w )] for all u, w . According to the definition for independence we should show that Pr { U u and W w } = F U ( u ) F W ( w ) for all u, w . Taking complements and using the fact that Pr( A B ) = Pr( A ) + Pr( B ) - Pr( A B ) , 1
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
we have Pr { U u and W w } = 1 - Pr { U > u or W > w } = 1 - [Pr { U > u } + Pr { W > w } - Pr { U > u and W > w } ] = 1 - [(1 - F U ( u )) + (1 - F W ( w )) - (1 - F U ( u ))(1 - F W ( w ))] = F U ( u ) F W ( w ) .
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