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Let $X_1$, $X_2$, $X_3$, $X_4$ have the joint pdf $f(X_1,X_2,X_3,X_4) = 24$ , $0 < X_l < X_2 < X_3 < X_4 < 1$ , $0$ elsewhere. Find the joint pdf of $Y_1 = X_1/X_2$, $Y_2 = X_2/X_3$ , $Y_3 = X_3/X_4$, $Y_4 = X_4$ and show that they are mutually independent.

I know they joint pdf is $24y_2(y_3^2)(y_4^2)$ but how do I show they are mutually independent?

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Their joint pdf has product form, so they are independent. Let $A, B, C \subseteq \mathbb [0,1)$ be Borel, then by Fubini \begin{align*} P(Y_2 \in A, Y_3 \in B, Y_4 \in C) &= \int_{A \times B \times C} 24y_2y_3^2y_4^2\, dy\\ &= \int_A 24y_2\, dy_2\int_By_3^2\,dy_3\int_C y_4^2\,dy_4\\ &= P(Y_2 \in A)P(Y_3 \in B)P(Y_4 \in C) \end{align*}

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I know I have to find the marginal pdf first, g1(y1), g2(y2), g3(y3), g4(y4). how do I find that? –  user48495 Nov 7 '12 at 10:32
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The OP wrote:

I know the joint pdf is $24y_2(y_3^2)(y_4^2)$ but ...

Actually, this is incorrect (and has been left uncorrected for over 10 months). The joint pdf is $24y_2(y_3^2)(y_4^3)$.

To see this, let $(X_1, ..., X_4)$ have joint pdf $f(x_1, ..., x_4)$:

Then the joint pdf of random variables $(Y_1, ..., Y_4)$, say $g(y_1, ..., y_4)$, can be derived automatically with:

where I am using the Transform function from the mathStatica add-on to Mathematica to do the grunt work for me (I am one of the developers of the former), and with domain of support:

As to independence: $(Y_1, ..., Y_4)$ are said to be mutually stochastically independent if and only if their joint pdf is equal to the product of the marginals. This is easily shown to be the case here:

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