3 ms·
Ok so first to have a uniform distribution we have to have a bounded set. Maybe you can do something clever with limits but lets not overcomplicate things. Lets
by im3w1l 5y ago
Ok so first to have a uniform distribution we have to have a bounded set. Maybe you can do something clever with limits but lets not overcomplicate things. Lets say we have 0 <= v < 10. Define E = v^2. Then 0 <= e < 100
Uniformity of v would mean that p(0 <= v < 1) = 1 / 10
Uniformity of E would mean that p(0 <= E < 1) = 1 / 100
But by construction p(0 <= v < 1) = p(0 <= E < 1). So it's not possible for both to be uniform.
- omegalulw 5y agoIt's not necessary to have p(0 <= v < 1) = p(0 <= E < 1). Only that P(f(X)) is uniform. But this does bring up a good point. H(X||Y) = H(f(X)||f(Y)) for any bijective f if the distributions are discrete. When they are continuous this is not true, even with a bijective f. For example f = x^2 doesn't work even though it yields a binary distribution. Interestingly however, affine transformations work.
- im3w1l 5y ago(For non-negative v) v^2 is less than one if and only if v is less than one. That's why the probabilities have to agree in our specific case.