Suppose that X and Y have the joint probability density function: f(x,y) = 30x^2y^2(1 - x^2 - y^2) for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. What is the marginal probability density function of X?Review LaterfX(x) = 15x(1 - x^2) for 0 ≤ x ≤ 1fX(x) = 15x^2(1 - x^2) for 0 ≤ x ≤ 1fX(x) = 10x(1 - x^2) for 0 ≤ x ≤ 1fX(x) = 10x^2(1 - x^2) for 0 ≤ x ≤ 1
Question
Suppose that X and Y have the joint probability density function: f(x,y) = 30x^2y^2(1 - x^2 - y^2) for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. What is the marginal probability density function of X?Review LaterfX(x) = 15x(1 - x^2) for 0 ≤ x ≤ 1fX(x) = 15x^2(1 - x^2) for 0 ≤ x ≤ 1fX(x) = 10x(1 - x^2) for 0 ≤ x ≤ 1fX(x) = 10x^2(1 - x^2) for 0 ≤ x ≤ 1
Solution
The marginal probability density function of X is obtained by integrating the joint probability density function over all possible values of Y.
The joint pdf is given by f(x,y) = 30x^2y^2(1 - x^2 - y^2) for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1.
To find the marginal pdf of X, we integrate the joint pdf over the range of Y, which is from 0 to 1.
So, fX(x) = ∫ from 0 to 1 of f(x,y) dy
= ∫ from 0 to 1 of 30x^2y^2(1 - x^2 - y^2) dy
= 30x^2 ∫ from 0 to 1 of y^2 - y^2(x^2 + y^2) dy
= 30x^2 [ (1/3)y^3 - (1/5)y^5 ] evaluated from 0 to 1
= 30x^2 [(1/3) - (1/5)]
= 30x^2 * (2/15)
= 4x^2
So, the marginal pdf of X is fX(x) = 4x^2 for 0 ≤ x ≤ 1.
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