Question

Two ideal dice are thrown . Let X₁ be the score on the first die and X₂ be the score on the second die. Let Y = max(X₁, X₂) then

(i) Evaluate Corr(Y, X₁).

(ii) Evaluate Corr(Y, X₂).

Answer 1

Let us list all the possible outcomes when the two ideal dice are rolled.

The possible outcomes are listed below.

$\begin{bmatrix} (1,1) & (2,1) & (3,1) & (4,1)& (5,1)&(6,1) \\ (1,2) & (2,2)&(3,2)&(4,2)&(5,2)&(6,2)\\ (1,3)&(2,3)&(3,3)&(4,3)&(5,3)&(6,3)\\ (1,4)&(2,4)&(3,4)&(4,4)&(5,4)&(6,4)\\ (1,5)&(2,5)&(3,5)&(4,5)&(5,5)&(6,5)\\ (1,6)&(2,6)&(3,6)&(4,6)&(5,6)&(6,6) \end{bmatrix}$

We are given that,

$X_1$ is the score on the first die,

$X_2$ is the score on the second die and

$Y = max(X_1, X_2)$

The probability distribution of $X_1$ is given below.

X₁ 1 2 3 4 5 6

P (X₁) ${1\over 6}$ ${1\over 6}$ ${1\over 6}$ ${1\over 6}$ ${1\over 6}$ ${1\over 6}$

The probability distribution of $X_2$ is given below.

X₂ 1 2 3 4 5 6

P (X₂) ${1\over 6}$ ${1\over 6}$ ${1\over 6}$ ${1\over 6}$ ${1\over 6}$ ${1\over 6}$

The probability distribution of $Y$ is given below.

$Y$ 1 2 3 4 5 6

$P(Y)$ ${1\over 36}$ ${3\over36}$ ${5\over36}$ ${7\over36}$ ${9\over 36}$ ${11\over 36}$

The joint distribution of $Y$and $X_1$ is,

Y/X₁ 1 2 3 4 5 6

1 ${1\over36}$

2 ${1\over36}$ ${2\over36}$

3 ${1\over 36}$ ${1\over36}$ ${3\over36}$

4 ${1\over36}$ ${1\over 36}$ ${1\over36}$ ${4\over36}$

5 ${1\over36}$ ${1\over36}$ ${1\over36}$ ${1\over36}$ ${5\over36}$

6 ${1\over36}$ ${1\over36}$ ${1\over36}$ ${1\over36}$ ${1\over36}$ ${6\over36}$

The joint distribution of $Y$and $X_2$ is,

Y/X₂ 1 2 3 4 5 6

1 ${1\over36}$

2 ${1\over36}$ ${2\over36}$

3 ${1\over 36}$ ${1\over36}$ ${3\over36}$

4 ${1\over36}$ ${1\over 36}$ ${1\over36}$ ${4\over36}$

5 ${1\over36}$ ${1\over36}$ ${1\over36}$ ${1\over36}$ ${5\over36}$

6 ${1\over36}$ ${1\over36}$ ${1\over36}$ ${1\over36}$ ${1\over36}$ ${6\over36}$

We determine the following,

$E(X_1)=\sum X_1*P(X_1)=(1*{1\over6})+(2*{1\over6})+(3*{1\over6})+(4*{1\over6})+(5*{1\over6})+(6*{1\over6})=3.5$

$E(X_1^2)=\sum X_1^2*P(X_1)=(1*{1\over6})+(4*{1\over6})+(9*{1\over6})+(16*{1\over6})+(25*{1\over6})+(36*{1\over6})={91\over6}=15.17(2dp)$

Variance of the random variable $X_1$ is,

$Var(X_1)=E(X_1^2)-(E(X_1))^2={91\over6}-(3.5^2)={70\over24}$

$E(X_2)=\sum X_2*P(X_2)=(1*{1\over6})+(2*{1\over6})+(3*{1\over6})+(4*{1\over6})+(5*{1\over6})+(6*{1\over6})=3.5$

$E(X_2^2)=\sum X_2^2*P(X_2)=(1*{1\over6})+(4*{1\over6})+(9*{1\over6})+(16*{1\over6})+(25*{1\over6})+(36*{1\over6})={91\over6}=15.17(2dp)$

Variance of the random variable $X_1$ is,

$Var(X_2)=E(X_2^2)-(E(X_2))^2={91\over6}-(3.5^2)={70\over24}$

$E(Y)=\sum Y(P=Y)=(1*{1\over36})+(2*{3\over36})+(3*{5\over36})+(4*{7\over36})+(5*{9\over36})+(6*{11\over36})={161\over36}$

$E(Y^2)=\sum Y^2(P=Y)=(1*{1\over36})+(4*{3\over36})+(9*{5\over36})+(16*{7\over36})+(25*{9\over36})+(36*{11\over36})={791\over36}$

Variance of the random variable $Y$ IS,

$Var(Y)=E(Y^2)-(E(Y))^2={791\over36}-({161\over36})^2={2555\over1296}$

Also, we need to determine the following,

$E(X_1Y)=\sum (X _1Y)*P(X_1\cap Y)=(1*{1\over36})+(2*{1\over36})+(3*{1\over36})+(4*{1\over36})+(5*{1\over36})+(6*{1\over 36})+(4*{2\over36})+(6*{1\over36})+(8*{1\over36})+(10*{1\over36})+(12*{1\over36})+(9*{3\over36})+(12*{1\over36})+(15*{1\over36})+(18*{1\over36})+(16*{4\over36})+(20*{1\over36})+(24*{1\over36})+(25*{5\over36})+(30*{1\over36})+(36*{6\over36})={616\over36}$

$E(X_2Y)=\sum (X _2Y)*P(X_2\cap Y)=(1*{1\over36})+(2*{1\over36})+(3*{1\over36})+(4*{1\over36})+(5*{1\over36})+(6*{1\over 36})+(4*{2\over36})+(6*{1\over36})+(8*{1\over36})+(10*{1\over36})+(12*{1\over36})+(9*{3\over36})+(12*{1\over36})+(15*{1\over36})+(18*{1\over36})+(16*{4\over36})+(20*{1\over36})+(24*{1\over36})+(25*{5\over36})+(30*{1\over36})+(36*{6\over36})={616\over36}$

We use the following formula to find $corr(Y,X_1)$,

$corr(Y,X_1)={cov(X_1,Y)\over \sqrt{var(X_1)*var(Y)}}$ where $cov(X_1,Y)=E(X_1Y)-E(X_1)*E(Y)={616\over36}-({161\over36}*{21\over6})=1.45833331$

Therefore,

$corr(X_1,Y)={1.45833331\over2.39792917}=0.6082(4dp)$

To find $corr(Y,X_2)$ we use the formula below,

$corr(Y,X_2)={cov(X_2,Y)\over \sqrt{var(X_2)*var(Y)}}$ where $cov(X_2,Y)=E(X_2Y)-E(X_2)*E(Y)={616\over36}-({161\over36}*{21\over6})=1.45833331$

Therefore,

$corr(X_2,Y)={1.45833331\over2.39792917}=0.6082(4dp)$

Therefore, $corr(Y,X_1)=corr(Y,X_2)=0.6082$