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I have known independence between events, between $\sigma$-algebras, between random variables, between any two of the above mentioned concepts. I wonder if independence between distributions (or independent distributions) make sense? If yes, how is it defined? Does it mean "independence between two probability measures" in some sense? Thanks and regards!

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For me, a distribution is basically the same as a probability. What is a "distribution" for you? –  André Caldas Oct 16 '11 at 1:27
    
@AndréCaldas: I think a "distribution" is a probability measure. –  Tim Oct 16 '11 at 1:32
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up vote 4 down vote accepted

Yes and no.

Yes: one can consider as a measurable space the set $\mathbb M$ of all the probability measures on a given measurable space $(S,\mathcal S)$, where $\mathbb M$ is endowed with a sigma-algebra $\mathfrak M$. Then a random probability measure is a random variable $M:(\Omega,\mathcal F,\mathrm P)\to(\mathbb M,\mathfrak M)$, that is, a function defined on a given probability space $(\Omega,\mathcal F,\mathrm P)$ such that, for every $B$ in $\mathfrak M$, $[M\in B]$ belongs to $\mathcal F$. In this setting, two random probability measures $M_1$ and $M_2$ are independent if and only if, for every $B_1$ and $B_2$ in $\mathfrak M$, $$ \mathrm P(M_1\in B_1,M_2\in B_2)=\mathrm P(M_1\in B_1)\cdot\mathrm P(M_2\in B_2). $$ For example, if $\mathfrak M$ is such that the function expectation, defined by $$ M\mapsto\langle M\rangle=\displaystyle \int_S s\mathrm dM(s), $$ is measurable as a function from $(\mathbb M,\mathfrak M)$ to $(\mathbb R,\mathcal B(\mathbb R))$, one can consider the events $$ B_1=\{M\in\mathbb M\mid\langle M\rangle\geqslant1\},\qquad B_2=\{M\in\mathbb M\mid\langle M\rangle\leqslant3\}. $$ The random distributions $M_1$ and $M_2$ being independent would imply, in particular, that $$ \mathrm P(\langle M_1\rangle\geqslant1,\langle M_2\rangle\leqslant3)=\mathrm P(\langle M_1\rangle\geqslant1)\cdot\mathrm P(\langle M_2\rangle\leqslant3). $$

No: most of the time, independent distributions refers, somewhat sloppily, to distributions of independent random variables.

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What is $P$ in the first centered formula? –  Rasmus Jan 27 '12 at 11:38
    
See edit. $ $ $ $ –  Did Jan 27 '12 at 12:00
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Yes, of course. Let x and y be two variables.

F(m) = 2x^4 + 3x + 3

F(n) = 5y^3 + 2y - 6

So functions F(m) and F(n) are independent with respect to x and y.

But let's have:

F(z) = 2x^2 + 3y^2

Now F(z) depends on both x and y and hence the two variables are not independent with respect to the function.

The functions I gave don't make sense as probability distributions but you should get the idea...


Perhaps some concrete examples would help.

I buy lottery tickets for two different games. The probability distribution for my winning one of the prizes in the first game has no effect on the probability distribution of my expected winning in the second game.

In shooting a gun at a target it is generally helpfully to consider horizontal and vertical errors separately. But since horizontal and vertical deflect are both caused where gun barrel is pointing this isn't really true. The big reason they are treated separately is because scope/sights adjust separately for horizontal and vertical changes.

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This answer seems to miss the point of the question. –  Rasmus Jan 27 '12 at 11:40
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