Probability: Multivariate distribution change of variables

[Master1022]

Homework Statement

Suppose that ## Y_1 ## and ## Y_2 ## are random variables with joint pdf:
[tex] f_{y_1, y_2} (y_1, y_2) = 8y_1 y_2 [/tex] 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:
[tex] f_{y_1, y_2} (y_1, y_2) = 8y_1 y_2 [/tex] 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.

 

[pasmith]

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

From [itex]0 < y_1 < y_2[/itex] you can determine the possible values of [itex]U_1[/itex].

 

[Master1022]

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

You also have from [itex]0 < y_1 < y_2[/itex] that the possible values of [itex]u_1 y_1/y_2[/itex] lie in [itex](0,1)[/itex].

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?