We determined that f(y1, y2) = 6(1 − y2), 0 ≤ y1 ≤ y2 ≤ 1, 0, elsewhere is a valid joint probability density function. (a) Find
E(Y1|Y2 = y2). E(Y1|Y2 = y2) = (b) Use the answer derived in part (a) to find E(Y1). (Compare this with E(Y1) found using the marginal density function f1(y1) = 3(1 − y1)2.) E(Y1) =
1 answer:
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Answer: fourth option
Explanation:
1) the pair x = 3 f(x) = 0, leads you to probe this:
f(3) = 0 = A [4 ^ (3 - 1) ] + C = 0
=> A [4^2] = - C
A[16] = - C
if A = 1/4
16 / 4 = 4 => C = - 4
That leads you to the function f(x) = [1/4] 4 ^( x - 1) - 4
2) Now you verify the images for that function for all the x-values of the table:
x = 2 => f(2) + [1/4] 4 ^ (2 - 1) - 4 = [1/4] 4 - 4 = 4 / 4 - 4 = 1 - 4 = - 3 => check
x = 3 => f(3) = [1/4] 4^ (3 - 1) - 4 = [1/4] 4^2 - 4 = 16 / 4 - 4 = 4 - 4 = 0 => check
x = 4 -> f(4) = [1/4] 4^ (4-1) - 4 = [1/4] 4^(3) - 4 = (4^3) / 4 - 4 = 4^2 - 4 = 16 - 4 = 12 => check.
Therefore, you have proved that the answer is the fourth option.
Explanation:
1) the pair x = 3 f(x) = 0, leads you to probe this:
f(3) = 0 = A [4 ^ (3 - 1) ] + C = 0
=> A [4^2] = - C
A[16] = - C
if A = 1/4
16 / 4 = 4 => C = - 4
That leads you to the function f(x) = [1/4] 4 ^( x - 1) - 4
2) Now you verify the images for that function for all the x-values of the table:
x = 2 => f(2) + [1/4] 4 ^ (2 - 1) - 4 = [1/4] 4 - 4 = 4 / 4 - 4 = 1 - 4 = - 3 => check
x = 3 => f(3) = [1/4] 4^ (3 - 1) - 4 = [1/4] 4^2 - 4 = 16 / 4 - 4 = 4 - 4 = 0 => check
x = 4 -> f(4) = [1/4] 4^ (4-1) - 4 = [1/4] 4^(3) - 4 = (4^3) / 4 - 4 = 4^2 - 4 = 16 - 4 = 12 => check.
Therefore, you have proved that the answer is the fourth option.