Expected value problem (probability)

[compliant]

Homework Statement

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)

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

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.

 

[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