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Expectation of discrete RV

  1. Jul 1, 2009 #1
    The distribution function of a random variable X is given by:

    F(x) =

    0 if x <-3
    3/8 if -3 <= x < 0
    1/2 if 0 <= x < 3
    3/4 if 3 <= x <4
    1 if x => 4

    Calculate E(X) and E(X2 - 2|X|)

    Well I'm at a loss of E(X) although once I know this the other should be fairly simple..

    Ive got E(X) = -3(3/8)-2(3/8)-1(3/8)+1(1/8)+2(1/8)+1(1/4)+2(1/4)+3(1/4)+4(1/4) = 5/8

    Is this the right method?

    Rather than making another post I also need a hand with the following.
    Let X be the number of different birthdays among four persons selected randomly. Find E(X).

    I know how to do this in principal E(X) = 0.p(0) + 1.p(1) + 2.p(2) + 3.p(3) + 4.p(4)

    but i don't know how to find the probability mass function p(x) I know it will takes values 0,1,2,3,4 and will probably involve the pick formula 365!/(365-x)! or something along those lines.

    Any help would be greatly appreciated!
     
  2. jcsd
  3. Jul 1, 2009 #2
    No. You were given the cdf F(x), so first you could find the pmf p(x). It looks like you tried to do that, but not correctly. The only x-values that give a positive value for p(x) are x=-3, 0, 3, and 4.


    You could treat finding p(x) as five different problems, for each of x=0, 1, 2, 3, and 4.

    Example: p(1)=prob they all have same birthday=P(1st person has any b-day)*P(2nd is same)*P(3rd is same)*P(4th is same)=1*(1/365)*(1/365)*(1/365).

    p(4)=prob they all have different b-days, which is similar.

    For p(2), there may be clever ways, but a boring safe way is to make a tree of possibilities. Similarly for p(3).

    If you use this approach (i.e. one person at a time), just remember the trick that for the first person, you always have P(1st person has any b-day)=1.
     
  4. Jul 1, 2009 #3
    p(x) = 3/8 for x = -3
    p(x) = 1/8 for x = 0
    p(x) = 2/8 for x = 3
    p(x) = 2/8 for x = 4

    is this right?

    E(X) = -9/8 + 6/8 + 8/8 = 5/8 (this gives the same answer as above I presume this is just coincidence if my first method is wrong)
     
  5. Jul 3, 2009 #4
    New method is right, and first was just coincidence.
     
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