We have an urn with 6 red balls and 4 green balls. We draw balls from the urn one by one without replacement, noting the order of the colors, until the urn is empty. Let X be the number of red balls in the first five draws, and Y the number of red balls in the last five draws. Compute the covariance Cov(X,Y).

My work:



where Ii=1 if red and Jj=1 if green.

Cov(X,Y)= E(XY)-E(X)E(Y) E(XY)=E(I1J6)+E(I1J7)+...+E(I5J10)=25*E(I1J6)




So, Cov(X,Y)=25(6/10)(4/9)-25(6/10)(4/10)=2/3

I don't know I am doing it right. Are there other ways to solve this?

Thank you.


The simplest approach is to observe that $X + Y = 6$, as there are only $10$ balls in the urn, hence one is guaranteed to have drawn all $6$ red balls. Thus $$\begin{align*} \operatorname{Cov}[X,Y] &= \operatorname{Cov}[X, 6-X] \\ &= \operatorname{E}[X(6-X)] - \operatorname{E}[X]\operatorname{E}[6-X] \\ &= \operatorname{E}[6X] - \operatorname{E}[X^2] - \operatorname{E}[X](6-\operatorname{E}[X]) \\ &= 6 \operatorname{E}[X] - \operatorname{E}[X^2] - 6 \operatorname{E}[X] + \operatorname{E}[X]^2 \\ &= - \left(\operatorname{E}[X^2] - \operatorname{E}[X]^2\right) \\ &= - \operatorname{Var}[X]. \\ \end{align*}$$ Then note that the distribution of $X$ is hypergeometric; namely, $$\Pr[X = x] = \frac{\binom{6}{x}\binom{4}{5-x}}{\binom{10}{5}}, \quad x \in \{1, 2, 3, 4, 5\}.$$ (It is impossible to not get a red ball in the first five draws, as there are only four green balls; and it is obviously not possible to get six red balls as there are only five.) Do you recall the variance of a hypergeometric distribution? If not, it is not hard to derive the formula. And even then, if you cannot, with the above distribution having only $5$ possible outcomes, it is not difficult to explicitly compute $\operatorname{E}[X]$ and $\operatorname{E}[X^2]$.


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