Probability: Multivariate distribution change of variables

[Master1022]

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.

 

[pasmith]

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 $U_1$.

 

[Master1022]

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?