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Homework Help: 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,

    = [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]


    = [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]


    = [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
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