# Probability: Multivariate distribution change of variables

Homework Statement:
Suppose that ## Y_1 ## and ## Y_2 ## are random variables with joint pdf:
$$f_{y_1, y_2} (y_1, y_2) = 8y_1 y_2$$ for ## 0 < y_1 < y_2 < 1 ## and 0 otherwise. Let ## U_1 = Y_1/Y_2 ##. Find the probability distribution ## p(u_1) ##.
Relevant Equations:
Jacobian
Hi,

I was attempting the problem above and got stuck along the way.

Problem:
Suppose that ## Y_1 ## and ## Y_2 ## are random variables with joint pdf:
$$f_{y_1, y_2} (y_1, y_2) = 8y_1 y_2$$ for ## 0 < y_1 < y_2 < 1 ## and 0 otherwise. Let ## U_1 = Y_1/Y_2 ##. Find the probability distribution ## p(u_1) ##.

Attempt:
We are not given a ## U_2 ##, but the problem provides a hint that we can define an arbitrary value for ## U_2 ##, for example, ## Y_2 ##. Then we can use that to find ## f(u_1, u_2) ## and then integrate with respect to ## u_2 ## to get ## f(u_1)##. It is the final step where I am confused as I am not completely sure about the limits for ## u_2 ##.

The working is as follows:

1. Define ## U_2 = Y_2 ##. Both transformations are one-to-one transformations so no extra steps are needed.

2. Find y1 and y2 in terms of u1 and u2. This yields ## y_1 = u_1 u_2 ## and ## y_2 = u_2 ##

3. Find the magnitude of the Jacobian, which turns out to be ## |J| = |u_2| ##

4. Find ## f_{u_1, u_2} (u_1, u_2) = f_{y_1, y_2} (y_1, y_2)|J| = 8u_1 u_2 ^3 ##

5. Then we can integrate to find the marginal distribution of ## u_1 ##. We need to use the inequality ## 0 < y_1 < y_2 < 1 ## to find the limits. Splitting it up we get ## u_1 u_2 > 0 ## and ## u_1 u_2 < u_2 < 1 ##. At this point, I am not quite sure how to use the inequalities. Should I just be using the limits ## 0 ## to ## 1 ##? I think I may be overthinking it as I seem to think there should be some use of ## u_1 ##...

Any help would be greatly appreciated.

## Answers and Replies

pasmith
Homework Helper
To find the marginal distribution of $U_1$, you integrate with respect to $u_2$. Since you have taken $u_2 = y_2$ its limits are indeed 0 and 1.

From $0 < y_1 < y_2$ you can determine the possible values of $U_1$.

To find the marginal distribution of $U_1$, you integrate with respect to $u_2$. Since you have taken $u_2 = y_2$ its limits are indeed 0 and 1.

You also have from $0 < y_1 < y_2$ that the possible values of $u_1 y_1/y_2$ lie in $(0,1)$.
Thank you @pasmith ! So that means that ## 0 < u_1 < 1 ##. Am I correct in thinking that this implies that the inequality leads to ## 0 < u_2 < 1 ## as well?