Finding Variance and Covariance of Sampling Without Replacement

Question

Suppose that we have a box with 3 white marbles, 3 green marbles, and 3 red marbles in it. We choose three marbles without replacement from the box and let W = the number of white marbles chosen and G = the number of green marbles chosen

Calculate the Var(W) and Cov(W, G).

I know how to find variance and covariance when I'm given a probability mass function or joint density, but I'm having trouble figuring out how to find variance and covariance only given the what marbles were choosing without replacement.

Answer 1

1

I know how to find variance and covariance when I'm given a probability mass function or joint density, but I'm having trouble figuring out how to find variance and covariance only given the what marbles were choosing without replacement.

Then your first order of business is to find out what the joint and marginal probability mass functions are.

Your joint probability mass function is found by comparing: how many ways can you select $w$ from 3 white, $g$ from 3 green, and $3-w-g$ from 3 red marbles from the bag; verus how many ways to select any 3 from 9 marbles in the bag?

$$\mathsf P(W=w, G=g) = {\binom{3}{w} \binom{3}{g} \binom{3}{3-w-g}}\Big/{\binom{9}{3}} \qquad\big[w\in\{0,1,2,3\}, g\in\{0,..,3-w\}\Big]$$

(This is a bivariate hypergeometric distribution)

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Your marginal probability mass functions are found similarly:

$$\mathsf P(W=w) = {\binom{3}{w}\binom{6}{3-w}}/{\binom{9}{3}}\qquad\big[w\in\{0,1,2,3\}\Big]\\\mathsf P(G=g) = {\binom{3}{g}\binom{3}{3-g}}/{\binom{9}{3}}\qquad\big[g\in\{0,1,2,3\}\Big]$$

(These are just Hypergeometric distributions.)

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The conditionals may also be useful

$$\mathsf P(G=g\mid W=w) = \binom{3}{g}\binom{3}{3-w-g}\big/\binom{6}{3-w}\qquad\big[w\in\{0,1,2,3\}, g\in\{0,..,3-w\}\Big]\\\mathsf P(W=w\mid G=g) = \binom{3}{w}\binom{3}{3-w-g}\big/\binom{6}{3-g}\qquad\big[g\in\{0,1,2,3\}, w\in\{0,..,3-g\}\Big]$$

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2

An alternate and perhaps easier approach is to let $X_i$ be the indicator random variable that marble $i$ is chosen, with the first three white, the next three green, and the last red.

$$\begin{align}\mathsf E(W) ~=~& \sum_{i=1}^3\mathsf E(X_i) \\[1ex] \mathsf E(G)~=~&\sum_{i=4}^6\mathsf E(X_i) \\[1ex]\mathsf {Var}(W) ~=~& \sum_{i=1}^3\mathsf E(X_i^2)+\underset{i\neq j}{\sum_{i=1}^3\sum_{j=1}^3}\mathsf E(X_iX_j) - \mathsf E(W)^2 \\[1ex] \mathsf {Cov}(W,G) ~=~& \sum_{i=1}^3\sum_{j=4}^6 \mathsf E(X_iX_j)-\mathsf E(W)\mathsf E(G) \end{align}$$