Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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*}

share|cite|improve this answer
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

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:

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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