How can I prove X = (X1, X2, X3) is independent?

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The discussion focuses on proving the independence of the variables X = (X1, X2, X3) derived from the random variables (Y1, Y2, Y3, Y4) distributed as D4(1,2,3,4;5). The variables X1, X2, and X3 are defined in terms of Z variables, where Z ~ N(0,1) and Z is IID G(1/2). The participants debate the validity of assuming that if X1 is independent of X2 and X2 is independent of X3, then X1 must also be independent of X3, highlighting that this assumption does not hold in general due to potential linear dependencies.

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For (Y1, Y2, Y3, Y4) ~ D4(1,2,3,4;5)
let Xk = [∑(from i=1 to k) Yi] / [∑(from i=1 to k+1) Yi] where k = 1,2,3

How can I prove X = (X1, X2, X3) is independent?
What I did was...

(Y1, Y2, Y3, Y4) ~ D4(1,2,3,4;5) = (Z1, Z2, Z3, Z4) / (Z1+Z2+Z3+Z4+Z5) where Z ~ N(0,1), Z IID G(1/2)

Now, we have
X1 = Z1 / (Z1+Z2)
X2 = (Z1+Z2) / (Z1+Z2+Z3)
X3 = (Z1+Z2+Z3) / (Z1+Z2+Z3+Z4)

I think if i can somehow show X1 and X2 are independent and X2 and X3 are independent then X1 and X3 are independent as well but how? this is a part I don't get T-T
 
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I can't understand your notation, hopefully someone will jump in, but in general, if you're talking about linear independence of vectors, it's not a transitive relation.
 


kkjs said:
I think if i can somehow show X1 and X2 are independent and X2 and X3 are independent then X1 and X3 are independent as well but how? this is a part I don't get T-T
That won't work in general. What if, for example, X3 turned out to be a multiple of X1? If X1 and X2 are independent, then X2 and X3 would also be independent, but X1 and X3 are definitely not independent.

Perhaps you can make it work because of the way the X's are constructed in your specific problem, but I'd guess it's not the right strategy.
 

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