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For the Beta ( r , s ) distribution,

  1. Determine E [ X n ] , where n is a positive integer.
  2. Use the result of part (a) to determine E [ X ] and Var [ X ] .
E [ X n ] = Γ ( r + s ) Γ ( r ) Γ ( s ) 0 1 t r + n - 1 ( 1 - t ) s - 1 d t = Γ ( r + s ) Γ ( r ) Γ ( s ) · Γ ( r + n ) Γ ( s ) Γ ( r + s + n ) =
Γ ( r + n ) Γ ( r + s ) Γ ( r + s + n ) Γ ( r )

Using Γ ( x + 1 ) = x Γ ( x ) we have

E [ X ] = r r + s , E [ X 2 ] = r ( r + 1 ) ( r + s ) ( r + s + 1 )

Some algebraic manipulations show that

Var [ X ] = E [ X 2 ] - E 2 [ X ] = r s ( r + s ) 2 ( r + s + 1 )
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The pair { X , Y } has joint distribution. Suppose

E [ X ] = 3 , E [ X 2 ] = 11 , E [ Y ] = 10 , E [ Y 2 ] = 101 , E [ X Y ] = 30

Determine Var [ 15 X - 2 Y ] .

EX = 3; EX2 = 11;EY = 10; EY2 = 101;EXY = 30; VX = EX2 - EX^2VX = 2 VY = EY2 - EY^2VY = 1 CV = EXY - EX*EYCV = 0 VZ = 15^2*VX + 2^2*VYVZ = 454
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The pair { X , Y } has joint distribution. Suppose

E [ X ] = 2 , E [ X 2 ] = 5 , E [ Y ] = 1 , E [ Y 2 ] = 2 , E [ X Y ] = 1

Determine Var [ 3 X + 2 Y ] .

EX = 2; EX2 = 5;EY = 1; EY2 = 2;EXY = 1; VX = EX2 - EX^2VX = 1 VY = EY2 - EY^2VY = 1 CV = EXY - EX*EYCV = -1 VZ = 9*VX + 4*VY + 2*6*CVVZ = 1
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The pair { X , Y } is independent, with

E [ X ] = 2 , E [ Y ] = 1 , Var [ X ] = 6 , Var [ Y ] = 4

Let Z = 2 X 2 + X Y 2 - 3 Y + 4. .
Determine E [ Z ].

EX = 2; EY = 1;VX = 6; VY = 4;EX2 = VX + EX^2 EX2 = 10EY2 = VY + EY^2 EY2 = 5EZ = 2*EX2 + EX*EY2 - 3*EY + 4 EZ = 31
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(See Exercise 9 from "Problems on Mathematical Expectation"). Random variable X has density function

f X ( t ) = ( 6 / 5 ) t 2 for 0 t 1 ( 6 / 5 ) ( 2 - t ) for 1 < t 2 = I [ 0 , 1 ] ( t ) 6 5 t 2 + I ( 1 , 2 ] ( t ) 6 5 ( 2 - t )

E [ X ] = 11 / 10 . Determine Var [ X ] .

E [ X 2 ] = t 2 f X ( t ) d t = 6 5 0 1 t 4 d t + 6 5 1 2 ( 2 t 2 - t 3 ) d t = 67 50
Var [ X ] = E [ X 2 ] - E 2 [ X ] = 13 100
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For the distributions in Exercises 20-22

      Determine Var [ X ] , Cov [ X , Y ] , and the regression line of Y on X .

(See Exercise 7 from "Problems On Random Vectors and Joint Distributions", and Exercise 17 from "Problems on Mathematical Expectation"). The pair { X , Y } has the joint distribution (in file npr08_07.m ):

P ( X = t , Y = u )
t = -3.1 -0.5 1.2 2.4 3.7 4.9
u = 7.5 0.0090 0.0396 0.0594 0.0216 0.0440 0.0203
4.1 0.0495 0 0.1089 0.0528 0.0363 0.0231
-2.0 0.0405 0.1320 0.0891 0.0324 0.0297 0.0189
-3.8 0.0510 0.0484 0.0726 0.0132 0 0.0077
npr08_07 Data are in X, Y, P jcalc- - - - - - - - - - - EX = dot(X,PX);EY = dot(Y,PY); VX = dot(X.^2,PX) - EX^2VX = 5.1116 CV = total(t.*u.*P) - EX*EYCV = 2.6963 a = CV/VXa = 0.5275 b = EY - a*EXb = 0.6924 % Regression line: u = at + b
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(See Exercise 8 from "Problems On Random Vectors and Joint Distributions", and Exercise 18 from "Problems on Mathematical Expectation"). The pair { X , Y } has the joint distribution (in file npr08_08.m ):

P ( X = t , Y = u )
t = 1 3 5 7 9 11 13 15 17 19
u = 12 0.0156 0.0191 0.0081 0.0035 0.0091 0.0070 0.0098 0.0056 0.0091 0.0049
10 0.0064 0.0204 0.0108 0.0040 0.0054 0.0080 0.0112 0.0064 0.0104 0.0056
9 0.0196 0.0256 0.0126 0.0060 0.0156 0.0120 0.0168 0.0096 0.0056 0.0084
5 0.0112 0.0182 0.0108 0.0070 0.0182 0.0140 0.0196 0.0012 0.0182 0.0038
3 0.0060 0.0260 0.0162 0.0050 0.0160 0.0200 0.0280 0.0060 0.0160 0.0040
-1 0.0096 0.0056 0.0072 0.0060 0.0256 0.0120 0.0268 0.0096 0.0256 0.0084
-3 0.0044 0.0134 0.0180 0.0140 0.0234 0.0180 0.0252 0.0244 0.0234 0.0126
-5 0.0072 0.0017 0.0063 0.0045 0.0167 0.0090 0.0026 0.0172 0.0217 0.0223
npr08_08 Data are in X, Y, P jcalc- - - - - - - - - - - - EX = dot(X,PX);EY = dot(Y,PY); VX = dot(X.^2,PX) - EX^2VX = 31.0700 CV = total(t.*u.*P) - EX*EYCV = -8.0272 a = CV/VXa = -0.2584 b = EY - a*EXb = 5.6110 % Regression line: u = at + b
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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