Question

Let the joint pdf of \(X\) and \(Y\) be \(f(x, y)=\frac{12}{7} x(x+y), 0

Short answer

The joint pdf of \(U\) and \(V\) is \n\(f_{UV}(u, v) = \frac{12}{7}u^2(v+u)\) for \(0X\)) and \(f_{UV}(u, v) = \frac{12}{7}v^2(v+u)\) for \(0Y\)).

Step-by-step solution

Understanding the given pdf and variables

It is given that \(f(x,y)=\frac{12}{7}x(x+y)\) where \(0<x<1,\,0<y<1\), and zero elsewhere. This is the joint pdf of two random variables \(X\) and \(Y\). Moreover, two new variables \(U\) and \(V\) are defined as \(U=\min(X, Y)\) and \(V=\max(X, Y)\).

Regions Definition

Different regions can be defined for U and V: Region 1 is defined as \(U=X, V=Y\) when \(Y>X\), and Region 2 where \(U = Y, V = X\) when \(X>Y\). Integration limits would differ for these regions.

Joint pdf Derivation for Region 1

For Region 1 where \(X=U,Y=V\) we have: \(f_{UV}(u, v)=\left|\frac{\partial(x,y)}{\partial(u,v)}\right|f_{XY}(x,y)\). Now, here, \(x=u\) and \( y=v\). So, \(f_{UV}(u,v) = \left|\frac{\partial(x,y)}{\partial(u,v)}\right|\frac{12}{7}u(u+v) = \frac{12}{7}u^2(v+u)\) for \(0<u<v<1\).

Joint pdf Derivation for Region 2

For Region 2 where \(Y=U,X=V\) we have: \(f_{UV}(u, v)=\left|\frac{\partial(x,y)}{\partial(u,v)}\right|f_{XY}(x,y)\). Now, here, \(x=v\) and \( y=u\). So, \(f_{UV}(u,v) = \left|\frac{\partial(x,y)}{\partial(u,v)}\right|\frac{12}{7}v(v+u) = \frac{12}{7}v^2(v+u)\) for \(0<u<v<1\).

Joint pdf of \(U\) and \(V\)

The joint pdf of \(U\) and \(V\) is given as \(f_{UV}(u, v) = \frac{12}{7}u^2(v+u)\) for \(0<u<v<1\) (Region 1) and \(f_{UV}(u, v) = \frac{12}{7}v^2(v+u)\) for \(0<v<u<1\) (Region 2). The final joint pdf of U and V is thus the sum of these two, ensuring that the respective regions are taken into account.