PS2 - Problem Set 2 Answer Key Econometrics 120A 1(a If we...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem Set 2 Answer Key - Econometrics 120A 1. (a) If we think of each roll of a die as a “trial” that is a success if a 6 is rolled and a failure if a 6 is not rolled, it is clear that X is a binomial random variable. The probability of success for each roll is π = 1 / 6 and the number of trials is n = 5. What we want then is the pmf p ( s ) for s = 0, 1, 2, 3, 4 and 5. Using either the formula for the pmf of a binomial or excel we get: p ( s ) = P ( X = s ) (pmf) X = 0 0.402 X = 1 0.402 X = 2 0.161 X = 3 0.0322 X = 4 0.0032 X = 5 0.0001 (b) To get the cdf we just use the relationship between the cdf and the pmf (e.g. P ( s 1) = p (0) + p (1)) to get: F ( s ) = P ( X s ) (pmf) X = 0 0.402 X = 1 0.804 X = 2 0.965 X = 3 0.996 X = 4 0.999 X = 5 1 (c) To find the mean we just use the formula μ = p ( x ) x and the answers to ( a ) to find μ = 0 . 833. (d) Similarly, for the variance we use the formula σ 2 = p ( x )( x - μ ) 2 to get σ 2 = 0 . 694. 2. Since P ( X 0) = 0, it must be that P ( X = 0) = 0. From here on we want to use the relationship between the cdf and the pdf. For example, P ( X 1) = P ( X = 0) + P ( X = 1) so plugging in: 0 . 1 = 0 + P ( X = 1) and we get P ( X = 1) = 0 . 1. Doing a similar exercise we find
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
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