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I'm trying to find the marginal statistic of 2 order stats from the full joint and I can't figure out the bounds on the following problem.

Consider $0\leq Y_1\leq Y_2\leq Y_3\leq Y_4 < 1$ from the distrbution $2x, 0 \leq x \leq <1$. What is the marginal distribution of $Y_3$ and $Y_4$.

So far I have: $f_{Y_1,Y_2,Y_3,Y_4} = 4!y_1y_2y_3y_4$

$Y_1$ and $Y_2$ need to be integrated out and I think the bounds should be: $\int_{y_1}^{y_3}\int_{0}^{y_2}f_{Y_1,Y_2,Y_3,Y_4}dy_1dy_2$ but this makes no sense. Can someone help me understand what the bounds should be?

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First off, the joint density as you specified it, it not normalized. It should be

$$ f_{Y_1, Y_2, Y_3, Y_4}\left(y_1, y_2, y_3, y_4\right) = 4! \cdot 2^4 \cdot y_1 y_2 y_3 y_4 \cdot [ 0 \leqslant y_1 \leqslant y_2 \leqslant y_3 \leqslant y_4 < 1 ] $$

Marginalizing: $$\begin{eqnarray} f_{Y_3, Y_4}\left(y_3, y_4\right) &=& \int_0^1 \int_0^1 f_{Y_1, Y_2, Y_3, Y_4}\left(y_1, y_2, y_3, y_4\right) \mathrm{d}y_1 \mathrm{d}y_2 \\ &=& 4! \cdot 2^4 \cdot y_3 y_4 \cdot [0 < y_3 \leqslant y_4 < 1] \cdot \int_0^{y_3} y_2 \left(\int_0^{y_2} y_1 \mathrm{d}y_1 \right) \mathrm{d}y_2 \end{eqnarray} $$

You should be able to take it from here.

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  • $\begingroup$ why does the y2 bound go from zero to y3? Isn't y2 between y1 and y3? $\endgroup$ – yoshi Apr 14 '16 at 3:57
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    $\begingroup$ Given $y_3$ and $y_4$ the range of values of $y_2$ is determined by $ 0 \leqslant y_2 \leqslant y_3$. For given value of $y2$, $y_3$ and $y_4$ the range of values of $y_1$ is determined by $ 0 \leqslant y_1 \leqslant y_2$. Does it make sense? $\endgroup$ – Sasha Apr 14 '16 at 13:18
  • $\begingroup$ ya that helps alot, thx! $\endgroup$ – yoshi Apr 14 '16 at 23:24

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