Suppose a fair standard die Ls rolled twice, and the random variables X_1 and X_

Question

Suppose a fair standard die Ls rolled twice, and the random variables X_1 and X_2 return the numbers on the faces that come up on the first and second rolls, respectively. Compare and contrast the possible values, probability distribution functions, expected values, and variances of the random variables Y = 2X_1 and Z = X_1 + X_2. Suppose that the continuous random variables X and Y have the probability den sity functions f(x) = {}1 + x - 1 lessthanorequalto x lessthanorequalto 0 1 - x 0 lessthanorequalto x lessthanorequalto 1 0 otherwise and g(y) = {y 0 lessthanorequalto y lessthanorequalto 1 2 - y 1 lessthanorequalto y lessthanorequalto 2, re- 0 otherwise spectively. Determine the probability density function h(z) of the random variable Z = X + Y.

Explanation / Answer

Q.1 Y = 2X1

possible values of Y = 2,4,6,8,10,12

PDF fy(y) = 1/6 ; Y = 2,4,6,8,10,12

= 0 ; otherwise

Expected value E(Y) = 1/6 * ( 2 + 4 + 6 + 8 + 10 + 12) = 7

variance Var(Y) = E[Y2 ] - E[Y]2

E[Y2 ] = 1/6 * ( 22 + 42 + 62 + 82 + 102 + 122 ] = 60.67

E[Y]2 = 7 * 7 = 49

so Var(Y) = E[Y2 ] - E[Y]2 = 60.67 - 49 = 11.67

Z = X1 + X2

possible values of Z = 2,3,4,5,6,7,8,9,10,11,12

PDF fZ(Z) = 1/36 ; Y = 2,

= 2/36; Y = 3

= 3/36; Y = 4

= 4/36; Y = 5

= 5/36 ; Y = 6

= 6/36 ; Y = 7

= 5/36 ; Y = 8

= 4/36; Y = 9

= 3/36 ; Y = 10

= 2/36 ; Y = 11

= 1/36 ' Y = 12

= 0 ; otherwise

Expected value E(Z) = 1/36 * ( 1*2 + 2*3 + 3*4 + 4 *5 + 5 * 6 + 6* 7 + 5 * 8 + 4 * 9 + 3 * 10 + 2* 11 + 1* 12 ) = 7

variance Var(Y) = E[Z2 ] - E[Z]2

E[Z2 ] = 1/36 * ( 1*22 + 2*32 + 3*42 + 4 *52 + 5 * 62 + 6* 72 + 5 * 82 + 4 * 92 + 3 * 102 + 2* 112 + 1* 122] = 60.67

E[Z]2 = 7 * 7 = 49

so Var(Y) = E[Z2 ] - E[Z]2 = 60.67 - 49 = 11.67

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