An urn contains $30$balls, of which$10$are red and 8 are blue. From this urn, 12 balls are randomly withdrawn. Let X denote the number of red and Y the number of blue balls that are withdrawn. Find Cov(X, Y)

(a) by defining appropriate indicator (that is, Bernoulli) random variables

$X_i,Y_j$such that $X=\sum _{i=1}^{10} X_i,Y=\sum _{j=1}^8 Y_j$

(b) by conditioning (on either X or Y) to determine$E\left[XY\right]$

Given the pointer irregular variable $X_i$that demonstrates regardless of whether $i$the red ball has been picked and $Y_j$that shows regardless of whether $j$the blue ball has been picked. Then, at that point, the number of red balls drawn $X$and blue balls drawn $Y$can be composed as

$X=\sum _{i=1}^{10} X_i,Y=\sum _{j=1}^8 Y_j$

Then we have that

$\mathrm{Cov}(X,Y)=\mathrm{Cov}\left(\sum _{i=1}^{10} X_i,\sum _{j=1}^8 Y_j\right)=\sum _i \sum _j \mathrm{Cov}\left(X_i,Y_j\right)=80\mathrm{Cov}\left(X_1,Y_1\right)$

we have that

$E\left(X_1{Y}_1\right)=P\left(X_1=1,Y_1=1\right)=\frac{12\cdot 11}{30\cdot 29}=\frac{22}{145}$

and

$E\left(X_1\right)=E\left(Y_1\right)=\frac{12}{30}$

so we have

$\mathrm{Cov}\left(X_1,Y_1\right)=E\left(X_1{Y}_1\right)-E\left(X_1\right)E\left(Y_1\right)=-\frac{12}{1450}$

which implies,

$\mathrm{Cov}(X,Y)=-\frac{96}{145}$

$\mathrm{Cov}(X,Y)=-\frac{96}{145}$

From the idea of the issue, we have that$X$is a Hypergeometric distribution with boundaries $N=30$, $K_x=10$,$n=12$. Likewise, we realize that $Y$is additionally a Hypergeometric distribution with boundaries $N=30$, $K_Y=8$, $n=12$. It infers that

$E\left(X\right)=n\frac{K_X}{N}=4,E\left(Y\right)=n\frac{K_Y}{N}=3.2$

Utilizing the law of the total expectation, we have that$E\left(XY\right)=E\left(E\right(XY\mid X\left)\right)=E\left(XE\right(Y\mid X\left)\right)$

Assuming we are given that there have been drawn $X$ red balls, we stay with$20$ non-red balls and we draw$n-X$ of them. In this way, all things considered, $Y$ has Hypergeometric distribution with boundaries $\hat{N}=20,K_Y=8,\hat{n}=12-X$. Thus, we have that

$E(Y\mid X)=\widehat{n}\frac{K_Y}{\widehat{N}}=4.8-0.4X$

Hence

$E\left(XE\right(Y\mid X\left)\right)=E\left(X\right(4.8-0.4X\left)\right)=4.8E\left(X\right)-0.4E\left(X^2\right)$

we know E(X)=4 and we have that

$\mathrm{Var}\left(X\right)=n\frac{K_X}{N}\frac{\left(N-K_X\right)}{N}\frac{N-n}{N-1}=\frac{48}{29}$

hence

$E\left(X^2\right)=\mathrm{Var}\left(X\right)+E(X{)}^2=\frac{48}{29}+16=\frac{512}{29}$

so, we have that

$E\left(XY\right)=4.8E\left(X\right)-0.4E\left(X^2\right)=\frac{352}{29}$

The covariance is then equivalent to

$\mathrm{Cov}(X,Y)=E\left(XY\right)-E\left(X\right)E\left(Y\right)=\frac{352}{29}-4\cdot 3.2=-\frac{96}{145}$

$E\left(XY\right)=4.8E\left(X\right)-0.4E\left(X^2\right)=\frac{352}{29}$