Answer in Statistics and Probability for Buzz #270915

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).

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(ii) Evaluate Corr(Y, X2).

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

The possible outcomes are listed below.

“\begin{bmatrix}n (1,1) & (2,1) & (3,1) & (4,1)& (5,1)&(6,1) \\n (1,2) & (2,2)&(3,2)&(4,2)&(5,2)&(6,2)\\n (1,3)&(2,3)&(3,3)&(4,3)&(5,3)&(6,3)\\n (1,4)&(2,4)&(3,4)&(4,4)&(5,4)&(6,4)\\n (1,5)&(2,5)&(3,5)&(4,5)&(5,5)&(6,5)\\n (1,6)&(2,6)&(3,6)&(4,6)&(5,6)&(6,6)n\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\over36}” “{2\over36}”

“{1\over 36}” “{1\over36}” “{3\over36}”

“{1\over36}” “{1\over 36}” “{1\over36}” “{4\over36}”

“{1\over36}” “{1\over36}” “{1\over36}” “{1\over36}” “{5\over36}”

“{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\over36}” “{2\over36}”

“{1\over 36}” “{1\over36}” “{3\over36}”

“{1\over36}” “{1\over 36}” “{1\over36}” “{4\over36}”

“{1\over36}” “{1\over36}” “{1\over36}” “{1\over36}” “{5\over36}”

“{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,


“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,


“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,


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”



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(Y,X_1)=corr(Y,X_2)=0.6082”

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