# Expected value problem (probability)

1. Mar 31, 2009

### compliant

1. The problem statement, all variables and given/known data
Given Y1 and Y2 are integer values, where 0$$\leq$$Y1$$\leq$$3, 0$$\leq$$Y2$$\leq$$3, 1$$\leq$$Y1+Y2$$\leq$$3

p(Y1, Y2) = $$\frac{{4 \choose y_1}{3 \choose y_2}{2 \choose {3-y_1-y_2}}}{{9 \choose 3}}$$

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

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

E1(Y1) = $$\int_{-{\infty}}^{\infty} y_1 f(y_1) dy_1$$

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 = $$\sum_{i=1}^n a_i*Y_i$$
E(U1) = $$\sum_{i=1}^n a_i*E_i (Y_i)$$

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

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

= $$\frac{{4 \choose 0}{3 \choose 1}{2 \choose {2}}}{{9 \choose 3}}$$ + $$\frac{{4 \choose 1}{3 \choose 0}{2 \choose {2}}}{{9 \choose 3}}$$

= $$\frac {3+4}{84}$$

= $$\frac {7}{84}$$

P(Y1+Y2=2)

= $$\frac{{4 \choose 0}{3 \choose 2}{2 \choose {1}}}{{9 \choose 3}}$$ + $$\frac{{4 \choose 1}{3 \choose 1}{2 \choose {1}}}{{9 \choose 3}}$$ + $$\frac{{4 \choose 2}{3 \choose 0}{2 \choose {1}}}{{9 \choose 3}}$$

= $$\frac {3(2)+4(3)(2)+6(2)}{84}$$

= $$\frac {42}{84}$$

P(Y1+Y2=3)

= $$\frac{{4 \choose 0}{3 \choose 3}{2 \choose {0}}}{{9 \choose 3}}$$ + $$\frac{{4 \choose 1}{3 \choose 2}{2 \choose {0}}}{{9 \choose 3}}$$ + $$\frac{{4 \choose 2}{3 \choose 1}{2 \choose {0}}}{{9 \choose 3}}$$ + $$\frac{{4 \choose 3}{3 \choose 0}{2 \choose {0}}}{{9 \choose 3}}$$

= $$\frac {1+4(3)+6(3)+4)}{84}$$

= $$\frac {35}{84}$$

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

= $${(1)}{\frac {7}{84}}+{(2)}{\frac {42}{84}}+{(3)}{\frac {35}{84}}$$

= $$\frac {196}{84}$$

= $$\frac {7}{3}$$

But I'm supposed to be using the formula I listed above. So I have no idea how to deal with that.

2. Mar 31, 2009

### ascapoccia

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