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

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The discussion centers on the independence of the ratio of two IID exponential random variables and their sample average. It is established that the ratio, such as X_1 / X_2, is independent of the sample average 1/n * Σ X_i, with the ratio following a Pareto distribution. The reasoning behind this independence is linked to scale invariance, suggesting that the ratio does not depend on the overall scale of the sample average. However, when considering the independence of X/Y and X+Y, an example shows that they are not independent, as specific values of X and Y lead to dependent outcomes. This highlights the complexity of relationships between ratios and sums in probability theory.
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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?
 
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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.
 
Are X/Y and X+Y independent (given X and Y are)? I can't seem to show that in general...
 
e12514 said:
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.
 

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