1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expected value problem (probability)

  1. Mar 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Given Y1 and Y2 are integer values, where 0[tex]\leq[/tex]Y1[tex]\leq[/tex]3, 0[tex]\leq[/tex]Y2[tex]\leq[/tex]3, 1[tex]\leq[/tex]Y1+Y2[tex]\leq[/tex]3

    p(Y1, Y2) = [tex]\frac{{4 \choose y_1}{3 \choose y_2}{2 \choose {3-y_1-y_2}}}{{9 \choose 3}}[/tex]

    Find E(Y1+Y2) and V(Y1+Y2)

    2. Relevant equations
    E(Y1+Y2) = E1(Y1)+E2(Y2)

    E1(Y1) = [tex]\int_{-{\infty}}^{\infty} y_1 f(y_1) dy_1[/tex]

    But that's for continuous variables, so I have no idea how to deal with discrete. Some guy tried explaining it to me, but it was just so unclear I didn't understand any of it.

    Required use of these theorems:

    Given U1 = [tex]\sum_{i=1}^n a_i*Y_i[/tex]
    E(U1) = [tex]\sum_{i=1}^n a_i*E_i (Y_i)[/tex]

    V(U1) = [tex]{\sum_{i=1}^n {a_i}^2*E_i (Y_i)}+2{\sum{\sum_{1{\leq{i}}<j{\leq{n}}} {a_i}^2*E_i (Y_i)}}[/tex]


    3. The attempt at a solution
    Well, first I did it the slow counting way involving counting,
    P(Y1+Y2=1)

    = [tex]\frac{{4 \choose 0}{3 \choose 1}{2 \choose {2}}}{{9 \choose 3}}[/tex] + [tex]\frac{{4 \choose 1}{3 \choose 0}{2 \choose {2}}}{{9 \choose 3}}[/tex]

    = [tex]\frac {3+4}{84}[/tex]

    = [tex]\frac {7}{84}[/tex]

    P(Y1+Y2=2)

    = [tex]\frac{{4 \choose 0}{3 \choose 2}{2 \choose {1}}}{{9 \choose 3}}[/tex] + [tex]\frac{{4 \choose 1}{3 \choose 1}{2 \choose {1}}}{{9 \choose 3}}[/tex] + [tex]\frac{{4 \choose 2}{3 \choose 0}{2 \choose {1}}}{{9 \choose 3}}[/tex]

    = [tex]\frac {3(2)+4(3)(2)+6(2)}{84}[/tex]

    = [tex]\frac {42}{84}[/tex]

    P(Y1+Y2=3)

    = [tex]\frac{{4 \choose 0}{3 \choose 3}{2 \choose {0}}}{{9 \choose 3}}[/tex] + [tex]\frac{{4 \choose 1}{3 \choose 2}{2 \choose {0}}}{{9 \choose 3}}[/tex] + [tex]\frac{{4 \choose 2}{3 \choose 1}{2 \choose {0}}}{{9 \choose 3}}[/tex] + [tex]\frac{{4 \choose 3}{3 \choose 0}{2 \choose {0}}}{{9 \choose 3}}[/tex]

    = [tex]\frac {1+4(3)+6(3)+4)}{84}[/tex]

    = [tex]\frac {35}{84}[/tex]


    E(Y1+Y2) = P(Y1+Y2=1)+2P(Y1+Y2=2)+3P(Y1+Y2=3)

    = [tex]{(1)}{\frac {7}{84}}+{(2)}{\frac {42}{84}}+{(3)}{\frac {35}{84}}[/tex]

    = [tex]\frac {196}{84}[/tex]

    = [tex]\frac {7}{3}[/tex]


    But I'm supposed to be using the formula I listed above. So I have no idea how to deal with that.
     
  2. jcsd
  3. Mar 31, 2009 #2
    well, expectation is the same for discrete as for random variables. If you think about it, integration is a really awesome way to take sums (ie riemann sums), so the expectation of discrete random variables is just the sum over all i of x(i)times the probability of x(i)

    also, you need to define U(i) as a new variable Y(1)+Y(2), and do the calculation on U. then you can use your formulas
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Expected value problem (probability)
Loading...