- #1
peripatein
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Hi,
I am having difficulties understand parts of the solution to the following problem in Statistics:
Let X and Y be two random, continuous variables, where X is distributed U(0,1) and Y|X=x is distributed U(x,x+1). I am asked to find the marginal density of Y.
For that I would first need to find the joint density. I know that the double integral of the joint density over the area defined by the lines y=x, y=x+1, x=1, x=0 (parallelogram) is 1. What I don't quite comprehend is why would the joint density then be 1/A where A is the area of the parallelogram in this case.
Next, suppose fX,Y(x,y) = 1, why do I need to separate fY(y) = ∫ (between -inf. and +inf.) dx into two integrals (one for 0 ≤ y ≤ 1 and one for 1 ≤ y ≤ 2)? Why couldn't I use one integral? And why are the integration boundaries for the second integral y-1 and 1? How could I have figured it out?
Homework Statement
I am having difficulties understand parts of the solution to the following problem in Statistics:
Let X and Y be two random, continuous variables, where X is distributed U(0,1) and Y|X=x is distributed U(x,x+1). I am asked to find the marginal density of Y.
Homework Equations
The Attempt at a Solution
For that I would first need to find the joint density. I know that the double integral of the joint density over the area defined by the lines y=x, y=x+1, x=1, x=0 (parallelogram) is 1. What I don't quite comprehend is why would the joint density then be 1/A where A is the area of the parallelogram in this case.
Next, suppose fX,Y(x,y) = 1, why do I need to separate fY(y) = ∫ (between -inf. and +inf.) dx into two integrals (one for 0 ≤ y ≤ 1 and one for 1 ≤ y ≤ 2)? Why couldn't I use one integral? And why are the integration boundaries for the second integral y-1 and 1? How could I have figured it out?