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Let F(x, y) be the d.f. of X and Y
if : F(x, y) = 1, for x + 2y ≥ 1
F(x, y) = 0, for x + 2y < 1,
then
  • a)
    F(x, y) can be joint distribution function of variables X and Y
  • b)
    F(x, y) cannot be joint distribution function of variables X and Y
  • c)
    F(x, y) cannot be marginal distribution function of variables X and Y
  • d)
    None of the above
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Let F(x, y) be the d.f. of X and Yif : F(x, y) = 1, for x + 2y ≥...
Let us define the events :
 
In (1) let us take : a = 0, b = 1/2, ; c = 1/4, d = 3/4 s.t. a < b and c < d. Then using (2) we get :
F(b, d) = 1 ; F(b, c) = 1 ; F(a, d) = 1 ; F(a, c) = 0.
Substituting in (1) we get :
 P(a < X ≤ b ∩ c < Y < d) =1 – 1 – 1 + 0 = –1 
which is not possible since P( . ) ³ 0.
Hence F(x, y) defined in (2) cannot be the distribution function of variates X and Y.
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Most Upvoted Answer
Let F(x, y) be the d.f. of X and Yif : F(x, y) = 1, for x + 2y ≥...
Explanation:
To determine whether F(x, y) can be a joint distribution function of variables X and Y, we need to check if it satisfies the properties of a joint distribution function.

A joint distribution function must satisfy the following properties:

1. Non-negativity: The joint distribution function must be non-negative for all values of x and y.
- In this case, F(x, y) = 1 for x ≥ 2y + 1, which is non-negative. However, for x < 2y="" +="" 1,="" f(x,="" y)="0," which="" violates="" the="" non-negativity="" property.="" />

Therefore, F(x, y) does not satisfy the non-negativity property and cannot be a joint distribution function of variables X and Y.

2. Marginal properties: The marginal distribution functions of X and Y can be obtained by summing or integrating the joint distribution function over the other variable.
- In this case, we cannot obtain the marginal distribution functions of X and Y from F(x, y) because F(x, y) is not defined for all values of x and y.

Therefore, F(x, y) cannot be the marginal distribution function of variables X and Y.

Conclusion:
Based on the above analysis, we can conclude that F(x, y) cannot be a joint distribution function of variables X and Y (Option B).
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Let F(x, y) be the d.f. of X and Yif : F(x, y) = 1, for x + 2y ≥1F(x, y) = 0, for x + 2y < 1,thena)F(x, y) can be joint distribution function of variables X and Yb)F(x, y) cannot be joint distribution function of variables X and Yc)F(x, y) cannot be marginal distribution function of variables X and Yd)None of the aboveCorrect answer is option 'B'. Can you explain this answer?
Question Description
Let F(x, y) be the d.f. of X and Yif : F(x, y) = 1, for x + 2y ≥1F(x, y) = 0, for x + 2y < 1,thena)F(x, y) can be joint distribution function of variables X and Yb)F(x, y) cannot be joint distribution function of variables X and Yc)F(x, y) cannot be marginal distribution function of variables X and Yd)None of the aboveCorrect answer is option 'B'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Let F(x, y) be the d.f. of X and Yif : F(x, y) = 1, for x + 2y ≥1F(x, y) = 0, for x + 2y < 1,thena)F(x, y) can be joint distribution function of variables X and Yb)F(x, y) cannot be joint distribution function of variables X and Yc)F(x, y) cannot be marginal distribution function of variables X and Yd)None of the aboveCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let F(x, y) be the d.f. of X and Yif : F(x, y) = 1, for x + 2y ≥1F(x, y) = 0, for x + 2y < 1,thena)F(x, y) can be joint distribution function of variables X and Yb)F(x, y) cannot be joint distribution function of variables X and Yc)F(x, y) cannot be marginal distribution function of variables X and Yd)None of the aboveCorrect answer is option 'B'. Can you explain this answer?.
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