# Marginal probability distribution

 P: 44 I have a question I need to answer. $$f\left(x,y\right)= 6x^2y$$ such that $$0 Sci Advisor P: 3,553  Quote by _joey Things start falling apart with marginal density for [tex]Y$$ variable $$f_y\left(x,y\right)=\int\limits_{0}^{1}6x^2y\,dx =$\left. 2{{x}^{3}}{{y}} \right|_{0}^{1}$=2y,\;x Shouldn't you integrate x rather from 0 to y instead of from 0 to 1? Also note that for arbitrary value of y, x can range from 0 to 2 so your first integral for the marginal density of X is also not correct. P: 44  Quote by DrDu Shouldn't you integrate x rather from 0 to y instead of from 0 to 1? Also note that for arbitrary value of y, x can range from 0 to 2 so your first integral for the marginal density of X is also not correct. We have a set of inequalities: 0<x<y and x+y<2. If you solve the inequalities, you will get 0<x<1. 0<x+x<y+(2-y) If set inequality constraints on a plane, you will see a triangle with points (0,0) (1,0) and (0, 2). x is bounded above by 1 If we integrate from 0 to y, as you suggest, and then from x to 2-x with respect to y (y is bounded)? The volume will not be equal to 1. This is a property of marginal density  Sci Advisor P: 3,553 Marginal probability distribution Ok, I was to quick. Note that the second corner of the triangle must be (1,1), not (1,0) as x must be smaller than y. The marginal density for y is then [tex] \int_0^y dx 6x^2y$$ for y<1 and $$\int_0^{2-y} dx 6x^2y$$ for $$y \ge 1$$. I'm in a hurry, hope I didn't make an error this time.
P: 44
 Quote by DrDu Ok, I was to quick. Note that the second corner of the triangle must be (1,1), not (1,0) as x must be smaller than y. The marginal density for y is then $$\int_0^y dx 6x^2y$$ for y<1 and $$\int_0^{2-y} dx 6x^2y$$ for $$y \ge 1$$. I'm in a hurry, hope I didn't make an error this time.
Yes, point (1,1), it was a typo.

If we integrate the marginal density you obtained ($$12y^3-24y^2+16y$$) for 2nd time to see if it is valid (and the joint distribution from which density was obtained is valid), we don't get volume equal to 1. Unless the bounds are mixed in the 2nd integration.
 P: 44 This problem is giving me a headache.:(
 Sci Advisor P: 3,553 I did the integrals and got 1. The integral over y from 0 to 1 gives 2/5 and the second one over y from 1 to 2 gives 1-2/5.
 P: 44 Could you please elaborate on how and why you are choosing these bounds or refer me to relevant information. Thanks for your replies.
 P: 44 DrDu, What do you think of my solution for marginal density of Y that is, for f(x,y) over (0, 1) and then over (x, 2-x)
 Sci Advisor P: 3,553 The ranges of integration are easy to find. You already identified the relevant triangle. To get the marginal distribution of y, you have to integrate over x with y hold fixed. If y is smaller 1, x can range from 0 t0 y. If y is greater than 1, you are on the other side of the triangel and x can range from 0 to 1-y. Hence f(y)=2y^4 for y<1 and 2(2-y)^3 y for y>1. To confirm that the marginal density is normalized, you integrate over y. The first integral with y ranging from 0 to 1 is easy and yields 2/5, the second integral with y ranging from 1 to 2 becomes easy after introduction of z=2-y as a new variable of integration. It yields 1-2/5. So the total integral is 1.
P: 44
 Quote by DrDu The ranges of integration are easy to find. You already identified the relevant triangle. To get the marginal distribution of y, you have to integrate over x with y hold fixed. If y is smaller 1, x can range from 0 t0 y. If y is greater than 1, you are on the other side of the triangel and x can range from 0 to 1-y. Hence f(y)=2y^4 for y<1 and 2(2-y)^3 y for y>1. To confirm that the marginal density is normalized, you integrate over y. The first integral with y ranging from 0 to 1 is easy and yields 2/5, the second integral with y ranging from 1 to 2 becomes easy after introduction of z=2-y as a new variable of integration. It yields 1-2/5. So the total integral is 1.
 P: 44 Another question on bounds. It is a simple question but I lost confidence in setting up the bounds. If I need to find conditional probability f(y|x). Then, $$f(y|x) = \frac{f(x,y)}{f_X(x)}$$. In our case, [tex]f_X(x)}=\int\limits_{x}^{2-x}6x^2y dy = 12x^2-12x^3, 0