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+1. The commenter above also wanted cared about bijective mappings, and squaring a random variable in [-1, 1] is not bijective. Squaring a random variable defi
by omegalulw 4y ago
+1. The commenter above also wanted cared about bijective mappings, and squaring a random variable in [-1, 1] is not bijective. Squaring a random variable defined over positive real numbers would lead to a bijective mapping and the distribution would still remain uniform.
Actually, I find it hard to come up with a bijective mapping that leads to a non uniform distribution that's useful for anything practical.
- im3w1l 4y agoOk 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 4y 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 4y 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.