Solutions Set 7
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Probability and Statistics Hints/Solutions to Test Set 7 1. U,V U,V U,V U,V 11 3 1 19 p (1,1) ,p (1, 2) ,p (2,1) ,p (2,2) . 60 8 8 60 2. 2 . 2 2 1 2 1 2 Z (X X) (Y Y) 1 2 1 2 X X Y Y ~ N(0,1), ~ N(0,1). 2 2 So 2 2 2 Z ~ . 2 Hence Z 2 has a negative exponential distribution with mean 4. 3. v/ U,V 2 2 1 1 f (u,v) ve . ,u 0, v 0. The densities of U and V can be derived easily now. (1 u) 4. 1 1 2 y /2 Y ,Y 1 2 1 2 2 2 1 1 f (y ,y ) e . ,y 0, y . 2 (1 y) Clearly Y 1 and Y 2 are independent. 5. Similar. Y and Z are independent. Z and U are independent. 6. Similar. Y 1, Y 2 and Y 3 are independent. 7. Consider ln P and use linearity property of independent normal variables. Now use the symmetry of normal distribution to get values of L 1 and L 2 . 8. Use the properties of a bivariate normal population. 9. Use the linearity property of normal distribution.
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probability and statistics: test set 7 solutions
Transcript of Solutions Set 7
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Probability and Statistics Hints/Solutions to Test Set 7
1. U,V U,V U,V U,V11 3 1 19p (1,1) ,p (1,2) ,p (2,1) ,p (2,2) .60 8 8 60
2. 2. 2 21 2 1 2Z (X X ) (Y Y ) 1 2 1 2X X Y Y~ N(0,1), ~ N(0,1).
2 2 So
222
Z ~ .2
Hence Z2 has a negative exponential distribution with mean 4.
3. v/U,V 2 21 1f (u,v) ve . , u 0,v 0. The densities of U and V can be
derived easily now. (1 u)
4. 11 2
y /2Y ,Y 1 2 1 22
2
1 1f (y , y ) e . , y 0, y .2 (1 y )
Clearly Y1 and Y2 are
independent. 5. Similar. Y and Z are independent. Z and U are independent. 6. Similar. Y1, Y2 and Y3 are independent. 7. Consider ln P and use linearity property of independent normal variables. Now use
the symmetry of normal distribution to get values of L1 and L2. 8. Use the properties of a bivariate normal population. 9. Use the linearity property of normal distribution.
