Let $X_{1}$ and $X_{2}$ be two random variables with jpdf:
$f(X_{1}, X_{2}) = 4X_{1}X_{2};$ for $0<X_{1}<1, 0<X_{2}<1$
Find the probability distribution of $Y_{1} = X_{1}^{2}$ and $Y_{2} = X_{1}X_{2}$. Hence show that $E(Y_{1}) = 1 - y_{2}^{2}$
Come across this in an exam today. Any idea on how to begin working on this? Can anyone recommend a resource that can help me solve similar problems? I use the texbook by Montgomery, but it lightly skims through this topic.
Let $Y_{1} = X_{1}^{2}$ and $Y_{2} = X_{1}X_{2}$. The inverse of this transformation is $x_1=\sqrt{y_1}$ and $x_2=\frac{y_2}{\sqrt{y_1}}$. The Jacobian is $$ J=\frac{\partial(x_1,x_2)}{\partial(y_1,y_2)}= \left|\matrix{\frac{\partial x_1}{\partial y_1} & \frac{\partial x_1}{\partial y_2}\\ \frac{\partial x_2}{\partial y_1} & \frac{\partial x_2}{\partial y_2}}\right|=\frac{1}{2y_1} $$ so that the distribution of $Y_1$ and $Y_2$ is $$ g(y_1,y_2)=f(x_1,x_2)\big|_{x_1=\sqrt{y_1},x_2=\frac{y_2}{\sqrt{y_1}}}\cdot|J|=\cases{\frac{2y_2}{y_1} & $y_2^2<y_1<1,\,0<y_2<1$\\ o& elsewhere.} $$ Finally the expected value of $Y_1$ is $$ \mathbb{E}(Y_1)=\int_{y_2^2}^1\int_0^1 y_1 g(y_1,y_2)\operatorname{d}y_2\operatorname{d}y_1=\int_{y_2^2}^1 \underbrace{\int_0^1y_1\frac{2y_2}{y_1}\operatorname{d}y_2}_{=1}\operatorname{d}y_1=\int_{y_2^2}^1\operatorname{d}y_1=1-y_2^2. $$
