Problem Statement: A box contains $N_1$ white balls, $N_2$ black balls, and $N_3$ red balls $(N_1+N_2+N_3=N).$ A random sample of $n$ balls is selected from the box (without replacement). Let $Y_1,Y_2,$ and $Y_3$ denote the number of white, black, and red balls, respectively, observed in the sample. Find the correlation coefficient for $Y_1$ and $Y_2.$ (Let $p_i=N_i/N$ for $i=1,2,3.$)

This is Exercise 5.110 in Mathematical Statistics with Applications, 5th Ed., by Wackerly, Mendenhall, and Scheaffer.

My Work So Far: Because the sampling is without replacement, the distribution is much like the hypergeometric distribution. In fact, we have the joint distribution function as the multivariate hypergeometric distribution: $$ p(y_1,y_2,y_3) =\frac{\displaystyle\binom{N_1}{y_1}\binom{N_2}{y_2}\binom{N_3}{y_3}} {\displaystyle\binom{N}{n}}. $$ Since we are not concerned about $Y_3,$ we can simplify a bit: $$ p(y_1,y_2) =\frac{\displaystyle\binom{N_1}{y_1}\binom{N_2}{y_2}\binom{N-(N_1+N_2)}{n-(y_1+y_2)}} {\displaystyle\binom{N}{n}}. $$ This is as far as I can get. The sums involved in calculating even something like $E(Y_1Y_2)$ are beyond my abilities (and even beyond the abilities of Mathematica). I know that the answer is $$\operatorname{Cov}(Y_1,Y_2)=-n\,\frac{N-n}{N-1}\,\frac{N_1}{N}\,\frac{N_2}{N}.$$

My Question: How do I move on from here?

Thanks for your time!

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    $\begingroup$ Are you familiar with indicator random variables? They are the method of choice in this kind of problem, because they reduce the algebra to calculations involving products of zeros and ones. $\endgroup$
    – whuber
    Commented Aug 11, 2020 at 15:54
  • 1
    $\begingroup$ I am familiar with indicator functions/random variables, but not with how to use them in this derivation. $\endgroup$ Commented Aug 11, 2020 at 17:06
  • $\begingroup$ Another approach to this sort of problem is from the polarisation identity: $4\mathrm{cov}[X_1, X_2]=\mathrm{var}[X_1+X_2]-\mathrm{var}[X_1-X_2]$, especially as you can probably think of $X_1+X_2$ as a single hypergeometric variable from a different distribution $\endgroup$ Commented Jun 12 at 2:08

1 Answer 1


Let the white balls form a set $\mathcal W$ of size $N_1$ and the black balls form a set $\mathcal B$ of size $N_2,$ both of which are disjoint subsets of the set of all $N$ balls, $\mathcal U.$

For any ball $i\in \mathcal U,$ let $X_i$ be the indicator that ball $i$ is included in the sample $\mathcal S:$ that is, it equals $1$ when $i\in S$ and otherwise equals $0.$ The process of sampling makes these $X_i$ into random variables, because associated with any sample $\mathcal S$ is the vector of values of all the $X_i$ corresponding to $\mathcal S.$

Indicators enjoy nice connections with the underlying probabilities. In particular, notice that since $X_i^2 = X_i,$

$$E[X_i^2] = E[X_i] = \Pr(X_i=1)1 + \Pr(X_i=0)0 = \Pr(X_i=1) = \Pr(i\in\mathcal{S}).$$

Let's pause to work out some important probabilities. To say the sample has size $n$ is equivalent to saying the sum of the indicators is $n:$

$$n = \sum_{i\in\mathcal U}X_i.$$

Since in simple random sampling all balls have equal chances $\pi_i = p$ (say) of being in the sample, by taking expectations we deduce

$$n = E\left[\sum_{i\in\mathcal U}X_i\right] = \sum_{i\in\mathcal{U}} E\left[X_i\right] = \sum_{i\in\mathcal{U}} \pi_i = \sum_{i\in\mathcal{U}} p = Np,$$


$$\Pr(i\in\mathcal S) = \pi_i = p = \frac{n}{N}.$$

Similarly, given two distinct balls $i$ and $j\ne i,$ the chance that both $i$ and $j$ are in the sample (which I will write as $\pi_{ij}=q,$ which also does not depend on $i$ or $j$) can be found by taking expectations of the product $n\times n$ expressed as sums of indicators:

$$\begin{aligned} n^2 &= E[n^2] = E\left[\sum_{i\in\mathcal{U}} X_i\, \sum_{j\in\mathcal{U}}X_j\right]\\ &= \sum_{i,\,j} E[X_iX_j]\\ &= \sum_i E[X_i^2] + \sum_{i\ne j}E[X_iX_j]\\ &= \sum_i E[X_i] + \sum_{i\ne j}\pi_{ij}\\ & = Np + N(N-1)q\\ &= n + N(N-1)q, \end{aligned} $$


$$\Pr(i\ne j\in\mathcal S) = \pi_{ij} = q = \frac{n^2-n}{N(N-1)} = \frac{\binom{n}{2}}{\binom{N}{2}},$$

the same result one would get from a combinatorial argument.

Returning to the question, we now have enough information to compute variances and covariances. For example, with $i\ne j,$

$$\operatorname{Cov}(X_i,X_j) = E[X_iX_j] - E[X_i]E[X_j] = q - p^2 = -\frac{n(N-n)}{N^2(N-1)}$$

and, similarly,

$$\operatorname{Var}(X_i) = E[X_i^2] - E[X_i]E[X_j] = p - p^2 = \frac{n(N-n)}{N^2}.$$

Finally, we may compute variances and correlations of the $Y$'s by expressing them as suitable sums of indicators and applying the bilinear property of covariance. For instance,

$$\begin{aligned} \operatorname{Cov}(Y_1,Y_2) &= \operatorname{Cov}\left(\sum_{i\in\mathcal W}X_i,\sum_{j\in\mathcal{B}} X_j\right)\\&= \sum_{i\in\mathcal{W},\,j\in\mathcal{B}}\operatorname{Cov}(X_i,X_j)\\&= N_1\,N_2\left(-\frac{n(N-n)}{N^2(N-1)}\right)\end{aligned}$$

because $\mathcal W$ and $\mathcal B$ are disjoint. This agrees with the formula quoted in the question.

You have now seen examples of all the techniques needed to complete the calculation of the correlation coefficient of $Y_1$ and $Y_2.$ (When computing $\operatorname{Var}(Y_k),$ you will need to split the sum into separate sums of variances $\operatorname{Cov}(X_i,X_i)=\operatorname{Var}(X_i)$ and covariances $\operatorname{Cov}(X_i,X_j),$ $j\ne i,$ much as I did in the calculation of $E[n^2],$ so make sure you're comfortable manipulating such double sums.)

In the spirit of self-study questions I will stop short of a complete answer.

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    $\begingroup$ Many thanks! I feel sure I can continue. $\endgroup$ Commented Aug 11, 2020 at 18:41

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