Question

The random variables X and Y are distributed according to the joint PDF:

It has the value of a*x^2 if 1 <= x <= 2 and 0 <= y <= x, and 0 otherwise. What's the marginal fY(y) in two scenarios:

1. 0 <= y <= 1
2. 1 <= y <= 2

Just give me useful hints please

Answer 1

1. The result is 28/15
2. The result is 4/15*(8-y^3)

Answer 2

.

Answer 3

1. To find fY(y) for 0 <= y <= 1, you need to integrate the joint PDF over x for this range of y. In other words, you need to find:

fY(y) = ∫fXY(x,y)dx, where the limits of integration are 1 for the lower limit and y for the upper limit.

2. To find fY(y) for 1 <= y <= 2, you need to integrate the joint PDF over x for this range of y. In other words, you need to find:

fY(y) = ∫fXY(x,y)dx, where the limits of integration are y for the lower limit and 2 for the upper limit.

Answer 4

The result for the marginal fY(y) for 0 <= y <= 1 is correct, which is fY(y) = (28/15)y^3.

However, the result for the marginal fY(y) for 1 <= y <= 2 needs some clarification. Using the same formula, we have:

fY(y) = ∫ fXY(x,y) dx, with the limits being y <= x <= 2 because the joint PDF is defined for y <= x <= 2.

So we have:

fY(y) = ∫[0 to 2] a*x^2 dx if y <= x <= 2

fY(y) = a/3 (2^3 - y^3) if y <= 2,

where a is some constant that we need to determine by integrating the joint PDF over its entire domain, that is:

∫∫ fXY(x,y) dx dy = 1

Integrating fXY(x,y) over its support, we get:

∫[0 to 1] ∫[y to 2] a*x^2 dx dy = 1

This integral evaluates to:

a/3 (2^3 - 1) = 1

Therefore,

a = 3/7

Substituting this value of a in the expression for fY(y) above, we obtain:

fY(y) = (3/7) * (8 - y^3) if 1 <= y <= 2

So the correct expression for the marginal fY(y) for 1 <= y <= 2 is fY(y) = (3/7) * (8 - y^3).

Answer 5

You're welcome! I'm glad I could help. If you have any more questions, feel free to ask!