91影视

Show that X and Y are uncorrelated.

Define X_i as the indicator random variable that marks if i th Ace has been drawn, i=$1$, ...., $4$ . Also, define Y_j as the indicator random variable that marks if j th Spade has been drawn, j=$1$, ..., $13$. Therefore, we can write

$X=\underset{i=1}{\overset{4}{鈭�}}{X}_i,Y=\underset{j=1}{\overset{13}{鈭�}}{Y}_j$

Using the linearity of the covariance in both arguments, we have that,

$\mathrm{Cov}鈦�(X,Y)=\mathrm{Cov}鈦�(\underset{i=1}{\overset{4}{鈭�}}{X}_i,\underset{j=1}{\overset{13}{鈭�}}{Y}_j)=\underset{i=1}{\overset{4}{鈭�}}\underset{j=1}{\overset{13}{鈭�}}\mathrm{Cov}鈦�(X_i,Y_j)$

This sum that contains $52$addends on the right side can be divided into two parts. The first part contains $51$addends that describe the covariance between indicator variables thattwo distinct Ace and Spade have been drawn. That covariance is equal to

$\mathrm{Cov}鈦�(X_i,Y_j)=E\left(X_i{Y}_j\right)鈭�E\left(X_i\right)E\left(Y_j\right)$

$\text{The common expectation}E\left(X_i{Y}_j\right)\text{is equal to}$

$E\left(X_i{Y}_j\right)=P(X_i{Y}_j=1)=P(X_i=1,Y_j=1)=\frac{\left(\begin{array}{c}50\\ 11\end{array}\right)}{\left(\begin{array}{l}52\\ 13\end{array}\right)}=\frac{1}{17}$

and means are equal to

$E\left(X_i\right)=E\left(Y_j\right)=\frac{1}{4}$

which implies

$\mathrm{Cov}鈦�(X_i,Y_j)=鈭�\frac{1}{272}$

The second part of the sum contains only and only one covariance. It is the variance of the indicator random variable that Spade Ace has been drawn. That is the consequence that there exists a card that is simultaneously Ace and Spade. The variance is equal to

$\mathrm{Var}鈦�\left(X_i\right)=\frac{1}{4}鈰�\frac{3}{4}=\frac{3}{16}$

Hence,

$\mathrm{Cov}鈦�(X,Y)=51鈰�(鈭�\frac{1}{272})+\frac{3}{16}=0$

So we have proved that X and Y are uncorrelated.

X and Y are uncorrelated.

Are they independent?

$\text{No, they are not independent. Observe that}P(X=4)>0\text{and}P(Y=13)>0\text{, but we have}$

$P(X=4,Y=13)=0$

Since it would imply that we have drawn $17$cards in total.

No, They are not independent.