0
$\begingroup$

$X_1$, $X_2$ 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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – StubbornAtom Mar 30 '18 at 20:07
  • $\begingroup$ 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 ? $\endgroup$ – Youssef Mar 30 '18 at 20:39
  • $\begingroup$ 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 $i$. $\endgroup$ – StubbornAtom Mar 30 '18 at 20:43
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ No problem, you had already done most of the work getting the joint distribution. Once you are certain you are happy with it as an answer do consider accepting it, or up voting if/when you are able $\endgroup$ – Nadiels Mar 31 '18 at 2:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.