I am looking for a concise example of three pairwise independent (non-degenerate) random variables that are not mutually independent.
Thanks.
I think I know what you mean. But it's not really possible imo because pairwise independent collections of random variables are sets that by definition are independent :)
Okay, but I'm now lost due to not being a mathematician lol :)
Dear George,
Thanks so much.
Then we let X1 = IA, X2 = IB, and X3 = IC, where IA is the indicator function of the set A.
@Juan Weisz,
The question is not contradictory at all if. Mutual independence implies pairwise independence, but the converse is not true.
Here is a good example:
Consider the unit square abcd with diagonals intersecting in e. Let A be the triangle acd, B the triangle bcd, and C the union of the two triangles abe and dec.
Now, consider the uniform distribution defined on the square. Let X = IA, Y = IB, and Z = IC. E(X) = P(A)=1/2, E(Y)=P(B)=1/2, E(Z)=P(C)=1/2.
E(XY)=P(AB)=1/4. E(XZ)= P(AC)=1/4, and E(YZ)=p(BC)=1/4, where AB is the intersection of A and B.
E(XYZ) = P(ABC) = P(triangle dec)=1/4 which is not equal to E(X)E(Y)E(Z) = 1/8.
Therefore, X, Y, and Z are pairwise independent, but the are not mutually independent.
Thanks, Mohamed and George
If you replaced the term random variable by operator and the term independent by commutating would you have an equivalent problem?
Im thinking in terms of contextuality in the Quantum theory.
in which case commutating operators are simultaneously measurable. For matrices this would be simultaneously diagonazable.
Would lie algebra identities apply?
Hi Juan,
I think the two problems conceptually are different. Although I am not familiar with quantum mechanics, I guess there is a kind of analogy between random variables and observables in many theoretical aspects.
Thanks for your answers and messages.
Mohamed
Well, Im not expecting the same answer in Clasical probability as in Quantum probability, because they are different, only some paralels of sort.
Independence in terms of random variables would be that the frequency envent A presents itself is not influenced by the frequency B presents itself.
I think this translates into P(A)=P(A given B)
In the Quantum theory this would be the operator translation of A commutes
with the operator translation of B, plain and simple. This garantes that they are simultaneously measurable and the measurability of one does not interfere with the measurement of the other.
Of course there are other requirements. The operators are Hermitian, and opérate within a common Hilbert space.
In the Quantum theory when they do not commute, as with x and linear momentum along the same direction p, the probabilities densities are functionally linked. In this case the amplitude of probability Psi (p) is a Fourier transform of the amplitude Psi(x). For example if the measurement of x gives a Sharp value of x0, then the distrubution of p is way spread out.
The probability denisty is obtained as the product of the probability amplitudes.
Psi complex conjugate times Psi, a real positive function.
On the other hand a measurement of x has nothing to do with a measurement of y, two orthogonal directions.
Of course when I talk about Independence Im thinking in terms of Physics and not mathematically. Hopefully within QM some
of those finer mathematical points have been adequately worked out. Maybe mathematicians do not like an infinite plane wave, because it gives a divergent probablility when integrated over.
but what we do is truncate it with boundary conditions, ie put it in a box., and normalize.One cannot simply do amplitudes without boundary conditions.
Dear all,
just to add to the answers: The classical and further examples of higher order dependence can be found in Section 7.1 of arXiv:1712.06532
https://arxiv.org/abs/1712.06532
In particular, it also provides examples where any subfamily of (n-1) random variables is independent but all n random variables are dependent.
Happy research.
BB
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