It has the value of a*x^2 if 1 <= x <= 2 and 0 <= y <= x, and 0 otherwise. What's the marginal PDF fY(y)?
To find the marginal PDF fY(y), we need to integrate the joint PDF over x for a given value of y:
fY(y) = ∫ f(x,y) dx
However, the limits of integration for x depend on the value of y:
- If 0 <= y <= 1, then the limits of integration for x are 0 and y.
- If 1 <= y <= 2, then the limits of integration for x are y and 2.
- Otherwise, f(x,y) = 0 for all values of x.
Therefore, we have:
fY(y) = ∫ f(x,y) dx
= ∫{0 to y} a*x^2 dy + ∫{y to 2} a*x^2 dy (if 1 <= y <= 2)
= (1/3)*a*y^3 + (1/3)*a*(2^3 - y^3)
= a*(8/3 - y^3/3) (if 1 <= y <= 2)
= 0 (otherwise)
Thus, the marginal PDF fY(y) is:
- fY(y) = a*(8/3 - y^3/3) if 1 <= y <= 2
- fY(y) = 0 otherwise.
To find the marginal PDF fY(y), we need to integrate the joint PDF over the range of x for each value of y.
Given that the joint PDF is defined as fXY(x, y) = ax^2 when 1 <= x <= 2 and 0 <= y <= x, and 0 otherwise, we can determine the marginal PDF fY(y) by integrating the joint PDF over the range of x.
For any given value of y, the range of x is from y to 2. Therefore, for y < x <= 2, the joint PDF fXY(x, y) is ax^2. For x <= y or x > 2, the joint PDF is 0.
To find the marginal PDF fY(y), we integrate the joint PDF over the range of x:
fY(y) = ∫[y,2] ax^2 dx
Now, let's evaluate this integral:
∫[y,2] ax^2 dx = [(ax^3)/3] from y to 2
= [(a*(2^3))/3] - [(a*y^3)/3]
= (8a/3) - (ay^3/3)
So, the marginal PDF fY(y) is given by:
fY(y) = (8a/3) - (ay^3/3) for y < x <= 2
fY(y) = 0 otherwise