Question 7.8: Recall the urn model we introduced in Chap. 5. Consider an u......

(a) The random variables X and Y are independent if the balls are drawn with replacement. This becomes clear by understanding that drawing with replacement implies that for both the draws, the same balls are in the urn and the conditions in each draw remain the same. The first draw has no implications for the second draw.

If we were drawing the balls without replacement, then the first draw could possibly have implications for the second draw: for instance, if the first ball drawn was red, then the second one could not be red because there is only one red ball in the urn. This means that drawing without replacement implies dependency of X and Y . This can also be seen by evaluating the independence assumption (7.27):

P(X = x_{i} , Y = y_{j} ) = P(X = x_{i} )P(Y = y_{j} ) (7.27)

P(X = 2, Y = 2) = 0 ≠ P(X = 2) · P(Y = 2) = \frac{1}{8} \cdot \frac{1}{8}.

(b) The marginal probabilities P(X = x_{i} ) can be obtained from the given information. For example, 3 out of 8 balls are black and thus P(X = 1) = 3/8. The conditional distributions P(Y |X = x_{i} ) can be calculated easily by realizing that under the assumed dependency of X and Y , the second draw is always based on 7 balls (8 balls minus the one drawn under the condition X = x_{i})—e.g. if the first ball drawn is black, then 7 balls, 2 of which are black, remain in the urn and P(Y = 1|X = 1) = 2/7. We thus calculate

P(Y = 1, X = 1) = P(Y = 1|X = 1)P(X = 1) = \frac{2}{7} \cdot \frac{3}{8}=\frac{6}{56}

P(Y = 1, X = 2) = P(Y = 1|X = 2)P(X = 2) =\frac{3}{7} \cdot \frac{1}{8}=\frac{3}{56}

. . .

P(Y = 3, X = 3) = P(Y = 3|X = 3)P(X = 3) =\frac{3}{7} \cdot \frac{4}{8}=\frac{12}{56}

and obtain

(c) The expectations are

E(X) = 1 \cdot \frac{3}{8} + 2 \cdot \frac{1}{8}+ 3 \cdot \frac{4}{8}=\frac{17}{8}

E(Y ) = E(X) =\frac{17}{8}.

To estimate ρ(X, Y ), we need Cov(X, Y ) as well as Var(X) and Var(Y ):

E(XY) = 1 \frac{6}{56}+ 2 \frac{3}{56}+ 3 \frac{12}{56}+2 \frac{3}{56} + 4\cdot0 + 6\frac{4}{56}+3 \frac{12}{56}+6 \frac{4}{56}+9 \frac{12}{56}=\frac{246}{56}

E(X²) = E(Y²) = 1^{2} \frac{3}{8}+ 2^{2} \frac{1}{8}+3^{2} \frac{4}{8}=\frac{43}{8}

Var(X) = E(X²) − [E(X)]² = \frac{43}{8}- \left(\frac{17}{8}\right)^{2}=\frac{55}{64}

Var(Y ) = Var(X) = \frac{55}{64}.

Using (7.38) and Cov(X, Y ) = E(XY) − E(X)E(Y ), we obtain

\rho \left(X, Y\right)=\frac{Cov\left(X,Y\right) }{\sqrt{Var\left(X\right)  Var\left(Y\right) } }.  (7.38)

\rho =\frac{Cov\left(X,Y\right) }{\sqrt{Var\left(X\right)\cdot Var\left(Y\right) } } =\frac{\frac{246}{56}-\frac{289}{64} }{\sqrt{\frac{55}{64}\cdot \frac{55}{64} } } =-0.143.