This problem is from the book, "Introduction to Probability" by Hoel, Port and Stone. It is problem 22 on page 106.
Problem:
A box has $3$ red balls and $2$ black balls. A random sample of size $2$ is drawn without replacement. Let $U$ be the number of
red balls selected drawn and let $V$ by the number of black balls selected. Compute $\rho(U,V)$.
Answer:
Recall the formula:
$$ \rho(U,V) = \frac{ Cov(U,V) }{ \sigma_U \,\,\sigma_V } $$
So, the first step is to compute the mean of $U$ and the mean of $V$.
\begin{align*}
u_U &= 2\left( \frac{3}{5}\right) \left( \frac{2}{4} \right) +
1\left( \frac{2}{5} \left( \frac{2}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\
u_U &= \frac{2(3)(2)}{20} + \frac{4}{20}+ \frac{6}{20} = \frac{12 + 4 + 6}{20} = \frac{ 11 } {10} \\
u_V &=
2 \left( \frac{2}{4} \left( \frac{1}{3} \right) \right) + 1 \left( \frac{2}{5} \left( \frac{3}{4} \right)
+ \frac{3}{5} \left( \frac{2}{4} \right) \right) \\
u_V &= \frac{4}{12} + \frac{6}{20} + \frac{6}{20} \\
u_V &= \frac{1}{3} + \frac{3}{10} + \frac{3}{10} = \frac{1}{3} + \frac{3}{5} \\
u_V &= \frac{5 + 3(3)}{5(3)} \\
u_V &= \frac{14}{15} \\
E( U^2 ) &= 2^2 \left( \frac{3}{5}\right) \left( \frac{2}{4} \right)
+ 1^2 \left( \frac{2}{5} \left( \frac{2}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\
E( U^2 ) &= 4 \left( \frac{3}{5}\right) \left( \frac{2}{4} \right) + \left( \frac{2}{5} \left( \frac{2}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\
E( U^2 ) &= \frac{4(3)(2)}{20} + \frac{4}{20} + \frac{6}{26} = \frac{24}{20} + \frac{4}{20} + \frac{6}{20} \\
E( U^2 ) &= \frac{12}{10} + \frac{2}{10} + \frac{3}{10} \\
E(U^2) &= \frac{17}{10} \\
E( V^2 ) &= 2^2 \left( \frac{2}{4} \left( \frac{1}{3} \right) \right) + 1^2 \left( \frac{2}{5} \left( \frac{3}{4} \right)
+ \frac{3}{5} \left( \frac{2}{4} \right) \right) \\
E( V^2 ) &= 4 \left( \frac{2}{4} \left( \frac{1}{3} \right) \right) + \frac{2}{5} \left( \frac{3}{4} \right)
+ \frac{3}{5} \left( \frac{2}{4} \right) \\
E( V^2 ) &= \frac{8}{12} + \frac{6}{20} + \frac{6}{20} = \frac{2}{3} + \frac{3}{10} + \frac{3}{10} \\
E( V^2 ) &= \frac{2}{3} + \frac{3}{5} = \frac{19}{15}
\end{align*}
Now we can find $\sigma_U$ and $\sigma_V$.
\begin{align*}
\sigma_U^2 &= E(U^2) - u_U^2 = \frac{19}{15} - \left( \frac{ 11 } {10} \right) ^ 2 \\
\sigma_U^2 &= \frac{19}{15} - \frac{121}{100} = \frac{ 19(100) - 121(15)} {15(100)} \\
\sigma_U^2 &= \frac{ 85} {15(100)} = \frac{17}{300} \\
\sigma_V^2 &= E(V^2) - u_V^2 = \frac{19}{15} - \left( \frac{14}{15} \right) ^ 2 \\
\sigma_V^2 &= \frac{19(15) - 14^2}{15^2} = \frac{ 89} {225} \\
Cov(U,V) &= E(UV) - E(U)E(V) \\
E(UV) &= 1(1) P(U = 1, V = 1) = P(U = 1, V = 1) \\
P(U = 1, V = 1) &= \frac{3}{5}\left( \frac{2}{4}\right) + \frac{2}{5}\left( \frac{3}{4 }\right) = \frac{6}{20} + \frac{6}{20} \\
P(U = 1, V = 1) &= \frac{3}{10} \\
Cov(U,V) &= \frac{3}{10} - \frac{ 11 } {10} \left( \frac{14}{15} \right) = \frac{3}{10} - \frac{154}{ 150 } \\
Cov(U,V) &= -\frac{109}{150} \\
\rho(U,V) &= \frac{ -\frac{109}{150} } { \sqrt{ \frac{17}{300} } \left( \sqrt{ \frac{ 89} {225} } \right) } \\
\rho(U,V) &= -\frac{109 \sqrt{300(225) }} { 150 \sqrt{ 17(89) } } = -\frac{1090 \sqrt{3(225) }} { 150 \sqrt{ 17(89) } } \\
\rho(U,V) &= -\frac{109(15) \sqrt{3 }} { 15 \sqrt{ 17(89) } } = -\frac{109(15) \sqrt{3 }} { 15 \sqrt{ 1513 } } \\
\rho(U,V) &= -\frac{109 \sqrt{3 }} { \sqrt{ 1513 } }
\end{align*}
Hence $\rho(U,V) < -1$ so my answer cannot be right. The book's answer is:
$$ -1 $$
Where did I go wrong?
Here is an updated answer based upon the comments made by Rob Pratt. Recall the formula: $$ \rho(U,V) = \frac{ Cov(U,V) }{ \sigma_U \,\,\sigma_V } $$ So, the first step is to compute the mean of $U$ and the mean of $V$. \begin{align*} u_U &= 2\left( \frac{3}{5}\right) \left( \frac{2}{4} \right) + 1\left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ u_U &= \frac{2(3)(2)}{20} + \frac{6}{20}+ \frac{6}{20} = \frac{12 + 6 + 6}{20} = \frac{ 6 } {5} \\ \end{align*} Now, observe that $u_U + u_V = 2$ \begin{align*} u_V &= 2 \left( \frac{2}{5} \left( \frac{1}{3} \right) \right) + 1 \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ u_V &= \frac{4}{20} + \frac{6}{20} + \frac{6}{20} = \frac{16}{20} \\ u_V &= \frac{4}{5} \\ E( U^2 ) &= 2^2 \left( \frac{3}{5}\right) \left( \frac{2}{4} \right) + 1^2 \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ E( U^2 ) &= 4 \left( \frac{3}{5}\right) \left( \frac{2}{4} \right) + \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ E( U^2 ) &= 4 \left( \frac{6}{20} \right) + \left( \frac{6}{20} + \frac{6}{20} \right) \\ E( U^2 ) &= \frac{24}{20} + \frac{12}{20} = \frac{9}{5} \\ % % E( V^2 ) &= 2^2 \left( \frac{2}{5} \left( \frac{1}{3} \right) \right) + 1^2 \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ E( V^2 ) &= 4 \left( \frac{2}{5} \left( \frac{1}{3} \right) \right) + \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ E( V^2 ) &= \frac{8}{15} + \frac{6}{20} + \frac{6}{20} = \frac{8}{15} + \frac{6}{10} \\ E( V^2 ) &= \frac{8(10) + 6(15)}{10(15)} \\ E( V^2 ) &= \frac{17}{15} \\ \end{align*} Now we can find $\sigma_U$ and $\sigma_V$. \begin{align*} \sigma_U^2 &= E(U^2) - u_U^2 = \frac{9}{5} - \left( \frac{ 6 } {5} \right) ^ 2 \\ \sigma_U^2 &= \frac{9(5)}{5} - \frac{36}{25} = \frac{9}{15} \\ \sigma_V^2 &= E(V^2) - u_V^2 = \frac{17}{15} - \left( \frac{4}{5} \right) ^ 2 \\ \sigma_V^2 &= \frac{17}{15} - \frac{16}{25} = \frac{17(25) - 16(15)}{15(25)} \\ \sigma_V^2 &= \frac{ 185} { 375 } = \frac{19}{ 75 }\\ Cov(U,V) &= E(UV) - E(U)E(V) \\ E(UV) &= 1(1) P(U = 1, V = 1) = P(U = 1, V = 1) \\ P(U = 1, V = 1) &= \frac{3}{5}\left( \frac{2}{4}\right) + \frac{2}{5}\left( \frac{3}{4 }\right) = \frac{6}{20} + \frac{6}{20} \\ P(U = 1, V = 1) &= \frac{3}{10} \\ Cov(U,V) &= \frac{3}{10} - \frac{ 6 } {5} \left( \frac{4}{5} \right) = \frac{3}{10} - \frac{24}{25} \\ Cov(U,V) &= \frac{3(25) - 24(10)}{250} = \frac{15 - 48}{50} \\ Cov(U,V) &= \frac{-33} {50} \\ \rho(U,V) &= \frac{\frac{-33} {50} }{ \sqrt{ \frac{9}{15} \left( \frac{19}{ 75 } \right) } } \\ \rho(U,V) &= -\frac{ 33 \sqrt{ 15(75) } } { \sqrt{ 9(19) } } = -\frac{ 33(5) \sqrt{ 3(15) } } { \sqrt{ 9(19) } } \\ \rho(U,V) &= -\frac{ 33(5) \sqrt{ 3(15) } } { 3 \sqrt{ 19 } } = -\frac{11(5) \sqrt{45}}{15} \\ \rho(U,V) &= -\frac{11(3)\sqrt{5}}{3} \\ \rho(U,V) &= -11\sqrt{5} \end{align*} Hence $\rho(U,V) < -1$ so my answer cannot be right. The book's answer is $$ -1 $$ Where did I go wrong?
Here is an updated answer based upon the comments made by Rob Pratt. Recall the formula: $$ \rho(U,V) = \frac{ Cov(U,V) }{ \sigma_U \,\,\sigma_V } $$ So, the first step is to compute the mean of $U$ and the mean of $V$. \begin{align*} u_U &= 2\left( \frac{3}{5}\right) \left( \frac{2}{4} \right) + 1\left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ u_U &= \frac{2(3)(2)}{20} + \frac{6}{20}+ \frac{6}{20} = \frac{12 + 6 + 6}{20} = \frac{ 6 } {5} \\ \end{align*} Now, observe that $u_U + u_V = 2$ \begin{align*} u_V &= 2 \left( \frac{2}{5} \left( \frac{1}{3} \right) \right) + 1 \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ u_V &= \frac{4}{20} + \frac{6}{20} + \frac{6}{20} = \frac{16}{20} \\ u_V &= \frac{4}{5} \\ E( U^2 ) &= 2^2 \left( \frac{3}{5}\right) \left( \frac{2}{4} \right) + 1^2 \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ E( U^2 ) &= 4 \left( \frac{3}{5}\right) \left( \frac{2}{4} \right) + \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ E( U^2 ) &= 4 \left( \frac{6}{20} \right) + \left( \frac{6}{20} + \frac{6}{20} \right) \\ E( U^2 ) &= \frac{24}{20} + \frac{12}{20} = \frac{9}{5} \\ % % E( V^2 ) &= 2^2 \left( \frac{2}{5} \left( \frac{1}{4} \right) \right) + 1^2 \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ E( V^2 ) &= 4 \left( \frac{2}{5} \left( \frac{1}{4} \right) \right) + \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ E( V^2 ) &= \frac{8}{20} + \frac{6}{20} + \frac{6}{20} \\ E( V^2 ) &= 1 \end{align*} Now we can find $\sigma_U$ and $\sigma_V$. \begin{align*} \sigma_U^2 &= E(U^2) - u_U^2 = \frac{9}{5} - \left( \frac{ 6 } {5} \right) ^ 2 \\ \sigma_U^2 &= \frac{9(5)}{5} - \frac{36}{25} = \frac{9}{15} \\ \sigma_U^2 &= \frac{3}{ \sqrt{15} } \\ \sigma_V^2 &= E(V^2) - u_V^2 = 1 - \left( \frac{4}{5} \right) ^ 2 \\ \sigma_V^2 &= 1 - \frac{16}{25} = \frac{9}{25} \\ \sigma_V &= \frac{3}{5} \\ Cov(U,V) &= E(UV) - E(U)E(V) \\ E(UV) &= 1(1) P(U = 1, V = 1) = P(U = 1, V = 1) \\ P(U = 1, V = 1) &= \frac{3}{5}\left( \frac{2}{4}\right) + \frac{2}{5}\left( \frac{3}{4 }\right) = \frac{6}{20} + \frac{6}{20} \\ P(U = 1, V = 1) &= \frac{3}{10} \\ % continue here Bob Cov(U,V) &= \frac{3}{10} - \frac{ 6 } {5} \left( \frac{4}{5} \right) = \frac{3}{10} - \frac{24}{25}\\ Cov(U,V) &= \frac{3(25) - 240}{250} \\ Cov(U,V) &= -\frac{16}{25} \\ \rho(U,V) &= \frac{ -\frac{16}{25} }{ \frac{3}{ \sqrt{15} } \,\, \left( \frac{3}{5} \right) } = -\frac{ 16(5)(\sqrt{15}) }{ 25(9) } \\ \rho(U,V) &= -\frac{16 \sqrt{15}} { 45 } \end{align*} Hence $\rho(U,V) < -1$ so my answer cannot be right. The book's answer is $$ -1 $$ Where did I go wrong?
Here is an updated answer based upon the comments made by Rob Pratt. Recall the formula: $$ \rho(U,V) = \frac{ Cov(U,V) }{ \sigma_U \,\,\sigma_V } $$ So, the first step is to compute the mean of $U$ and the mean of $V$. \begin{align*} u_U &= 2\left( \frac{3}{5}\right) \left( \frac{2}{4} \right) + 1\left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ u_U &= \frac{2(3)(2)}{20} + \frac{6}{20}+ \frac{6}{20} = \frac{12 + 6 + 6}{20} = \frac{ 6 } {5} \\ \end{align*} Now, observe that $u_U + u_V = 2$ \begin{align*} u_V &= 2 \left( \frac{2}{5} \left( \frac{1}{3} \right) \right) + 1 \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ u_V &= \frac{4}{20} + \frac{6}{20} + \frac{6}{20} = \frac{16}{20} \\ u_V &= \frac{4}{5} \\ E( U^2 ) &= 2^2 \left( \frac{3}{5}\right) \left( \frac{2}{4} \right) + 1^2 \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ E( U^2 ) &= 4 \left( \frac{3}{5}\right) \left( \frac{2}{4} \right) + \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ E( U^2 ) &= 4 \left( \frac{6}{20} \right) + \left( \frac{6}{20} + \frac{6}{20} \right) \\ E( U^2 ) &= \frac{24}{20} + \frac{12}{20} = \frac{9}{5} \\ % % E( V^2 ) &= 2^2 \left( \frac{2}{5} \left( \frac{1}{4} \right) \right) + 1^2 \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ E( V^2 ) &= 4 \left( \frac{2}{5} \left( \frac{1}{4} \right) \right) + \left( \frac{2}{5} \left( \frac{3}{4} \right) + \frac{3}{5} \left( \frac{2}{4} \right) \right) \\ E( V^2 ) &= \frac{8}{20} + \frac{6}{20} + \frac{6}{20} \\ E( V^2 ) &= 1 \end{align*} Now we can find $\sigma_U$ and $\sigma_V$. \begin{align*} \sigma_U^2 &= E(U^2) - u_U^2 = \frac{9}{5} - \left( \frac{ 6 } {5} \right) ^ 2 \\ \sigma_U^2 &= \frac{9(5)}{5} - \frac{36}{25} = \frac{9}{15} \\ \sigma_U^2 &= \frac{3}{ \sqrt{15} } \\ \sigma_V^2 &= E(V^2) - u_V^2 = 1 - \left( \frac{4}{5} \right) ^ 2 \\ \sigma_V^2 &= 1 - \frac{16}{25} = \frac{9}{25} \\ \sigma_V &= \frac{3}{5} \\ Cov(U,V) &= E(UV) - E(U)E(V) \\ E(UV) &= 1(1) P(U = 1, V = 1) = P(U = 1, V = 1) \\ P(U = 1, V = 1) &= \frac{ {3 \choose 1} {2 \choose 1} }{ {5 \choose 2} } = \frac{3(2)}{ \frac{5(4)}{2} } \\ P(U = 1, V = 1) &= \frac{6}{10} = \frac{3}{5} \\ % continue here Bob Cov(U,V) &= \frac{3}{5} - \frac{ 6 } {5} \left( \frac{4}{5} \right) = \frac{3}{5} - \frac{24}{25}\\ Cov(U,V) &= \frac{15}{25} - \frac{24}{25} = -\frac{9}{25} \\ \rho(U,V) &= \frac{ -\frac{9}{25}}{ \frac{3}{ \sqrt{15} } \,\, \left( \frac{3}{5} \right) } = -\frac{9(5)\sqrt{15}}{3(3)(25)} \\ \rho(U,V) &= -\frac{9\sqrt{15}}{3(3)(5)} \\ \rho(U,V) &= -\frac{ \sqrt{15} }{5} \\ \end{align*} The book's answer is $$ -1 $$ Where did I go wrong?
The means are wrong, and I didn't check the subsequent steps. We have: \begin{align} E[U]&= \frac{2\binom{3}{2}\binom{2}{0}+1\binom{3}{1}\binom{2}{1}}{\binom{5}{2}} = \frac{6}{5}\\ E[V]&= \frac{2\binom{3}{0}\binom{2}{2}+1\binom{3}{1}\binom{2}{1}}{\binom{5}{2}} = \frac{4}{5} \end{align} As a sanity check, note that $U+V=2$, so $2=E[2]=E[U+V]=E[U]+E[V]$, which your computation violates.
Similarly: \begin{align} E[U^2]&= \frac{2^2\binom{3}{2}\binom{2}{0}+1^2\binom{3}{1}\binom{2}{1}}{\binom{5}{2}} = \frac{9}{5}\\ E[V^2]&= \frac{2^2\binom{3}{0}\binom{2}{2}+1^2\binom{3}{1}\binom{2}{1}}{\binom{5}{2}} = 1 \end{align}
Also: $$E[UV]= \frac{1\cdot 1\binom{3}{1}\binom{2}{1}}{\binom{5}{2}} = \frac{3}{5}$$
So $$ \rho(U,V)=\frac{E[UV]-E[U]E[V]}{\sqrt{E[U^2]-E[U]^2}\sqrt{E[V^2]-E[V]^2}} =\frac{3/5-(6/5)(4/5)}{\sqrt{9/5-(6/5)^2}\sqrt{1-(4/5)^2}}=-1 $$
Update: Here's an alternative approach that uses properties of covariance and variance to shorten the computations, as suggested in the comment by @NCh. \begin{align} \text{Cov}(U,V) &= \text{Cov}(U,2-U) = \text{Cov}(U,-U) = -\text{Cov}(U,U) = -\text{Var}(U) \\ \text{Var}(V)&=\text{Var}(2-U)=\text{Var}(-U)=(-1)^2\text{Var}(U)=\text{Var}(U) \\ \rho(U,V) &= \frac{\text{Cov}(U,V)}{\sqrt{\text{Var}(U)}\sqrt{\text{Var}(V)}}=\frac{-\text{Var}(U)}{\sqrt{\text{Var}(U)}\sqrt{\text{Var}(U)}}=-1 \end{align}