Marginal probability density function from joint distribution

$X_1$, $X_2$ and $X_3$ are three independent random variables with the same density function $f_{Xi}(x)= e^{-x},\, i\in{\{1,2,3\}}$.

We also have \begin{align} Y_1 &= \frac{X_1}{X_1+X_2} \\ Y_2 &= \frac{X_1+X_2}{X_1+X_2+X_3} \\ Y_3 &= {X_1+X_2+X_3}. \end{align}

Using the Jacobian transformation. We find $f_{Y_1,Y_2Y_3}(y_1,y_2,y_3)= y_2y_3^2e^{-y_3},\, y_1,y_2,y_3>0$

I want to find the marginal densities of $Y_1,Y_2,Y_3$. For this, I want to use the joint density function $f_{Y_1,Y_2Y_3}(y_1,y_2,y_3).$

Example: $$f_{Y1}(y_1)=\int_0^?\int_0^? f_{Y_1,Y_2Y_3}(y_1,y_2,y_3) \,dy_2dy_3$$ But I can't find the limits of integration.

Thank you

• Well, $x_1,x_2,x_3>0$ implies $y_1$ and $y_2$ should lie in $(0,1)$ as they are positive fractions; $y_3$ is simply positive. Mar 30 '18 at 20:07
• Yes, but how to be sure that there is no relation between $Y_1,Y_2$ and$Y_3$ that will affect the limits of integration ? Mar 30 '18 at 20:39
• If you have found the inverse solutions $x_i$'s in terms of the $y_i$'s, then you would get the ranges of $y_i$'s just from the fact that $x_i>0$ for all $i$. Mar 30 '18 at 20:43

You have your density but now need to keep track of the support of this density, clearly $Y_3$ has support $[0, \infty)$ and this support has no functional dependence on $Y_1$ or $Y_2$. So lets integrate out $Y_3$ to get $$f_{Y_1,Y_2}(y_1,y_2)=\int_0^{\infty}f_{Y_1,Y_2,Y_3}(y_1,y_2,y)\operatorname{d}y = 2y_2,$$ if we can just take the support of $Y_1$ and $Y_2$ to be $[0, 1]$ respectively - which is their maximum possible supports - then we are done, but your legitimate concern is whether there is a dependence between these two variables.
So let us explore that; the easiest option would be to ask if we have $Y_1 = c$ for some $c \in (0, 1)$ is there sufficient freedom for $Y_2$ to take any particular value or is it constrained? Well for the moment just treating the variables as symbols to be manipulated in algebraic expressions we have \begin{align} Y_1 = c \implies Y_2 = \frac{c^{-1}X_1}{c^{-1}X_1 + X_3} \end{align} so that for any particular value of $c, X_1$ there is sufficient freedom for $Y_2$ to take any value in $(0, 1]$ through the variable $X_3$ - indeed it takes the maximum at $X_3 = 0$ and then monotone decreases after that. Now since $Y_1$ is functionally, and as a random variable, independent of $X_3$ we can claim that the support of $Y_2$ is indeed independent of $Y_1$. So with this extra information we can write $$f_{Y_1,Y_2}(y_1, y_2) = 2y_2, \qquad y_1,y_2 \in [0, 1] \times [0, 1],$$ which should now be enough information for you to finish.