Csdrhrt

The problem below is from the book Introduction to Probability Theory
by Hoel, Port and Stone. The answer I
computed is different from the back of the book. I am wondering where I went wrong or if the book is wrong.

Problem:

A box has $3$ red balls and $2$ black balls. A random sample of size
$2$ is drawn
without replacement. Let $U$ by the number of red balls selected and let $V$ be
the number of black balls selected. Compute $rho(U,V)$.

Answer:
begin{eqnarray*}
rho(U,V) &=& frac{ Cov(U,V) } {sigma_u sigma_v } \
u_u &=& 2 P(U = 2) + 1P(U = 1) \
P(U = 2) &=& frac{3}{5} left( frac{2}{4} right) = frac{3}{10} \
%
P(U = 1) &=& frac{2}{5} left( frac{3}{4} right) + frac{3}{5} left( frac{2}{4} right) \
P(U = 1) &=& frac{6}{20} + frac{6}{20} = frac{6}{5} \
%
u_u &=& 2 left( frac{3}{10} right) + frac{1}{2} = frac{6}{10} + frac{5}{10} \
u_u &=& frac{11}{10} \
u_v &=& 2 P(V = 2) + 1P(V = 1) \
P(V = 2) &=& frac{2}{5} left( frac{1}{4} right) = frac{1}{10} \
P(V = 1) &=& frac{2}{5} left( frac{3}{4} right) + frac{3}{5} left( frac{2}{4} right) =
frac{6}{20} + frac{6}{20} \
P(V = 1) &=& frac{3}{5} \
u_v &=& 2 left( frac{1}{10} right) + frac{3}{5} = frac{1}{5} + frac{3}{5} \
u_v &=& frac{4}{5} \
E( U^2 ) &=& 2^2 P(U = 2) + 1^2 P(U = 1) \
E( U^2 ) &=& 4 left(frac{3}{10}right) + frac{1}{2} = frac{6}{5} + frac{1}{2} \
E( U^2 ) &=& frac{17}{10} \
E( V^2 ) &=& 2^2 P(V = 2) + 1^2 P(V = 1) = 4 P(V = 2) + P(V = 1) \
E( V^2 ) &=& 4 left( frac{1}{10} right) + frac{3}{5} = frac{4}{10} + frac{6}{10} \
E( V^2 ) &=& 1 \
sigma_u^2 &=& E( U^2 ) - u_u^2 = frac{17}{10} - ( frac{6}{5} ) ^ 2\
sigma_u^2 &=& frac{17}{10} - frac{36}{25} = frac{17(5) - 36(2)}{50} \
sigma_u^2 &=& frac{105 - 72}{50} = frac{32}{50} = frac{16}{25} \
sigma_u &=& frac{4}{5} \
end{eqnarray*}
begin{eqnarray*}
sigma_v^2 &=& E( V^2 ) - u_v^2 = 1 - left( frac{4}{5} right)^2 = 1 - frac{16}{25} \
sigma_v^2 &=& frac{9}{25} \
sigma_v &=& frac{3}{5} \
Cov(U,V) &=& E(UV) - u_u u_v \
E(UV) &=& P(U=1)P(V=1)(1)(1) = left( frac{6}{5} right) left( frac{3}{5} right) \
E(UV) &=& frac{18}{25} \
Cov(U,V) &=& frac{18}{25} - left( frac{11}{10} right) left( frac{4}{5} right)
= frac{18}{25} - frac{44}{50} = frac{36-44}{50}\
Cov(U,V) &=& -frac{4}{25} \
rho(U,V) &=& frac{ -frac{4}{25} } { left( frac{4}{5} right) left( frac{3}{5} right) } =
frac{ -frac{4}{25} } { left( frac{12}{25} right) } \
rho(U,V) &=& -frac{1}{3} \
end{eqnarray*}
The book's answer is $-1$.

TIP: Never begin calculations until you have simplified equations.

(It reduces both workload and error accumulation .)

Here, make use of the fact that the total number of balls selected is two, so clearly $V=2-U$ .

$$begin{align}rho &=dfrac{mathsf{Cov}(U,V)}{sqrt{mathsf{Var}(U)~mathsf{Var}(V)~}}\&=dfrac{mathsf{Cov}(U,2-U)}{sqrt{mathsf{Var}(U)~mathsf{Var}(2-U)~}}\&=dfrac{-mathsf {Cov}(U,U)}{sqrt{mathsf{Var}(U)mathsf{Var}(U)}}\ &=-1end{align}$$

Which should be anticipated, because $U,V$ are perfectly linearly correlated, and negatively so.   The coefficient is a measure of linearity of correlation.

Remark : The error you have done is in the calculation of the mean $E(U)$ of $U$. Check your computations, you will find that it is not $11/10$ but $2 times 3/10 + 1 times 6/10 = 12/10$, spoiling the rest of your computations.

The main thing I would like to show you is that computations are not needed here, because $U$ and $V$ are related by $U+V=2$ whence $V=2-U$ (technical name : antithetic variables). See remark at the bottom.

How is it possible without computations ? Here is an explanation that could be considered a little as handwaving but I think that it is essential to develop some intuition in these issues.

Let us take a statistical point of view : all the samples you will have in a $(u,v)$ plane are $(U,V)=(0,2)$ or $(1,1)$ or $(2,0)$ (with resp probabilities $0.3, 0.6, 0.1$, but this is unimportant for the following). The corresponding points are aligned on the straight line with equation $v=2-u$ (note already the negative slope).

If you are accustomed to see samples of bivariate distributions as clouds of points (see for example http://www.analytictech.com/mb313/correlat.htm), you should know that the correlation coefficient $rho$ gives a measure of narrowness of these clouds : the narrower the cloud, the closer $|rho|$ to $1$, the sign of $rho$ being positive or negative according to the sign of the slope of the principal axis of the cloud.

Here the narrowness is extreme with a negative slope ; thus $rho=-1$.

Remark : $U+V=2 implies E(U)+E(V)=2$. You have here a fire alarm that rings when you try to check $11/10+8/10 neq 2$...