56 Expected Values 35 However you rationalize it though not only are lottery

56 expected values 35 however you rationalize it

  • University of Toronto
  • ECO 227
  • Notes
  • fewmoments
  • 57
  • 100% (1) 1 out of 1 people found this document helpful

This preview shows page 35 - 37 out of 57 pages.

5.6 Expected Values 35 However you rationalize it though, not only are lottery players guaranteed to lose in the long run, the odds are that they will lose in the short run too. The Kansas State instant lottery of Review Exercises 4 Problem 19, returns about 47 cents in the dollar to the players. Wasserstein and Boyer [1990] worked out the chances of winning more than you lose after playing this game n times. The chances are best after 10 games at which point roughly one person in 6 will be ahead of the game. But after 100 games, only one person in 50 will have made a profit. Exercises on Section 5.6.1 1.Suppose a random variableXhas probability functionx2357pr(x)0.20.10.30.4FindμX, the expected value ofX.2.ComputeμX=E(X) whereXBinomial(n= 2, p= 0.3).5.6.2Population standard deviation 2]Just as with “population mean” and “sample mean”, we use the wordpopula-tionin the term “population standard deviation” to distinguish it from “samplestandard deviation”, which is the standard deviation of a batch of numbers.Thepopulationstandard deviation is the standard deviation of a distribution.It tells you about the spread of the distribution, or equivalently, how variablethe random quantity is.To compute sd(X), we need to be able to compute E[(X-μ)2]. How are weto do this? For discrete distributions, just as we compute E(
Image of page 35
36 Discrete Random Variables [This is part of a general pattern, E( X 2 ) = Σ x 2 i pr( x i ), E( e X ) = Σ e x i pr( x i ), or generally for any function h ( X ) we compute E[ h ( X )] = Σ h ( x i )pr( x i ).] We now illustrate the use of the formula for E[( X - μ ) 2 ]. Example 5.6.3 In Example 5.6.1, X has probability function x 0 1 2 3 pr( x ) 1 8 5 8 1 8 1 8 and we calculated that μ = E( X ) = 1 . 25. Thus, E[( X - μ ) 2 ] = X ( x i - μ ) 2 pr( x i ) = (0 - 1 . 25) 2 1 8 + (1 - 1 . 25) 2 5 8 + (2 - 1 . 25) 2 1 8 + (3 - 1 . 25) 2 1 8 = 0 . 6875 and sd( X ) = p E[( X - μ ) 2 ] = 0 . 6875 = 0 . 8292. Example 5.6.4 If X Binomial( n = 3 , p = 0 . 1) then X can take values 0 , 1 , 2 , 3 and we calculated in Example 5.6.2 that μ = E( X ) = 0 . 3. Then E[( X - μ ) 2 ] = X ( x i - μ ) 2 pr( x i ) = (0 - 0 . 3) 2 pr(0) + (1 - 0 . 3) 2 pr(1) + (2 - 0 . 3) 2 pr(2) + (3 - 0 . 3) 2 pr(3) = 0 . 27 . Thus, sd( X ) = p E[( X - μ ) 2 ] = 0 . 27 = 0 . 5196. [Note that the formulae for pr(0), pr(1), etc. are given in Example 5.6.2.] Exercises 5.6.2
Image of page 36
Image of page 37

You've reached the end of your free preview.

Want to read all 57 pages?

  • Spring '13
  • yu
  • Probability theory, Binomial distribution, Discrete probability distribution, Hypergeometric

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture