Are Ratios of IID Exponential Variables Independent of Their Sample Average?

[e12514]

Suppose I have a sample X_1, ..., X_n of independently, identically distributed exponential random variables.

One result I deducted was that the ratio of any two of them (eg. X_1 / X_2) is independent of the sample average 1/n * \sum_{i=1}^{n} X_i.
(Aside: that ratio, as a random variable, has a Pareto distribution)

What's the reasoning/ intuitive appeal behind that? I know that any datapoint from an independently, identically distributed sample is in general not independent of the sample average unless there is zero variance, so.. How do we interpret this result here? Why is the ratio independent of the sample average?

 

[Hurkyl]

My gut feeling is that this is a manifestation of some sort of scale invariance. Incidentally, we can simplify the situation by doing away with all of the other variables -- we're just looking at the independence of X/Y and X+Y.

 

[e12514]

Are X/Y and X+Y independent (given X and Y are)? I can't seem to show that in general...

 

[gel]

Are X/Y and X+Y independent (given X and Y are)? I can't seem to show that in general...

No. Just to pick a simple example, suppose that X and Y are IID taking the values 1,2 each with a 50% probability.

X/Y=1/2 or X/Y = 2/1 <=> X+Y = 3
X/Y = 1 <=> X+Y = 2 or 4

so they aren't independent.