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Let $X_1, X_2, X_3, X_4$ be independent and $\operatorname{Exp}( \lambda )$. We have, $$Y_1 := X_1+X_2+ X_3+ X_4$$ $$Y_2:= \frac{X_1 }{X_1+X_2}$$ $$Y_3:= \frac{X_1 + X_2 }{X_1+X_2+ X_3}$$ $$Y_4:=\frac{X_1 + X_2 + X_3}{X_1+X_2+ X_3+ X_4}$$ We are now supposed to show that these, $Y_i$ are independent and determine their joint distribution (of $(Y_1, Y_2, Y_3, Y_4)$).

My initial approach is to use the fact that, $$Y_i = g_i(X_1,\dots,X_n)$$ $$X_i = h_i(Y_1,\dots,Y_n)$$ $$f_\mathbf{Y}(y_1,\dots,y_n) = f_\mathbf{X}(h_1(y_1,\dots,y_n),\dots,h_n(y_1,\dots,y_n)) \cdot |J|$$ $J$ being the Jacobian determinant. I start of with re-writing each $X_i$ as a function of $(Y_1,\dots, Y_1)$ and get, after some manipulations that, $$X_1 = Y_1 Y_2 Y_3 Y_4$$ $$X_2 = Y_1 Y_3 Y_4 (1-Y_2)$$ $$X_3 = Y_1 Y_4(1-Y_3)$$ $$X_4 = Y_1(1-Y_4)$$

Computing all partial derivatives,$\frac{\partial x_j}{\partial y_i}$ , yields me the following matrix,

\begin{bmatrix} y_2y_3y_4 & y_1y_3y_4 & y_1y_2y_4 & y_1y_2y_3 \\ y_3y_4(1-y_2) & -y_1y_3y_4 & y_1y_4(1-y_2) & y_1y_3(1-y_2) \\ y_4(1-y_3) & 0 & -y_1y_4 & y_1(1-y_3) \\ 1-y_4 & 0 & 0 & -y_1 \end{bmatrix}

My linear algebra is a little rusty, but I believe the doing the following is valid. The aim is to perform row/column operations to obtain a triangular determinant. First I swap the columns: 1. Swap column #1 and #4. 2. Swap column #2 and #1, 3. Swap column #2 and #3. This yields

\begin{bmatrix} y_1y_3y_4 & y_1y_2y_4 & y_1y_2y_3 & y_2y_3y_4 \\ -y_1y_3y_4 & y_1y_4(1-y_2) & y_1y_3(1-y_2)& y_3y_4(1-y_2) \\ 0 & -y_1y_4 & y_1(1-y_3) & y_4(1-y_3) \\ 0 & 0 & -y_1 & 1-y_4 \end{bmatrix}

I then: 1. Add row #1 to row #2. 2. Add row #2 to row #3. 3. Add row #3 to row #4. I then have,

\begin{bmatrix} y_1y_3y_4 & y_1y_2y_4 & y_1y_2y_3 & y_2y_3y_4\\ 0 & y_1y_4 & y_1y_3 & y_3y_4 \\ 0 & 0 & y_1 & y_4 \\ 0 & 0 & 0 & 1 \end{bmatrix} If I've done this correctly then, $$ |J| = y_1^3 y_4^2 y_3 $$ We have that the joint p.d.f of $(X_1,X_2,X_3,X_4)$ is, $$ f_\mathbf{X} = \lambda^4 e^{-\lambda(x_1+x_2+x_3+x_4)} $$ We then get, after some manipulation $(x_1 + x_2 + x_3 + x_4 = y_1)$, that, $$ f_\mathbf{Y} = \lambda^4 e^{-\lambda y_1} y_1^3 y_4^2 y_3 $$

The question is now, have I done this correctly (i.e. computed the joint distribution of $(Y_1, Y_2, Y_3, Y_4)$? If so, how do I show independence for the $Y_i$'s? Do I have to compute all the marginal distributions or is there a simpler approach?

Any hints on how to proceed would be greatly appreciated!

Thanks.

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    $\begingroup$ looks ok (to me), you might write supports of $y_1,\dots,y_4$. $\endgroup$ – MAN-MADE Sep 17 '17 at 16:29
  • $\begingroup$ I think it's OK, too. You might offer, as a check, that the integral of your joint density over the range in question works out to 1. $\endgroup$ – kimchi lover Sep 17 '17 at 16:31
  • $\begingroup$ So you have $y_1\ge0,\ 0\le y_2\le 1,\ 0 \le y_3\le 1,\ 0\le y_4\le 1.$ I wouldn't omit that from any write-up explaining the solution. $\qquad$ $\endgroup$ – Michael Hardy Sep 17 '17 at 16:33
  • $\begingroup$ Thank you!! I integrated all $y_i$ over $[ 0; \inf [$ and couldn't get the product to equal one. This solved it! $\endgroup$ – gabruomad Sep 17 '17 at 22:48
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If what you've done is correct (I haven't checked the details), then independence of $Y_1,Y_2,Y_3,Y_4$ follows from the factorization of their joint density: It's the product of a function of $y_1,$ a function of $y_2,$ a function of $y_3,$ and a function of $y_4.$ That's enough to infer independence. (If it depended on something like $y_2^2+y_2^2$ or $\cos(y_1+y_2),$ etc., then it would not be such a product. The function of $y_2$ in this case is constant, and from that you conclude that $Y_2$ is uniformly distributed.

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