Is there a proof of this? I know that if we want a uniform PDF on $x^2+y^2 \leq 1$ then we want the PDF to be $1/\pi$ and when I check and integrate I do get 1 but is there a formal statement involving this?
2026-04-07 12:51:06.1775566266
Why is the PDF of uniform distribution in 1/2/3 dimensions 1/length of interval (or 1/area or 1/volume)?
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The integral (or double or triple integral) across the entire domain of the distribution must be one, since probability density is defined as probability/size of interval near a given point. Dividing both sides by length (or area or volume) gives that the average value of the PDF over the length (or area or volume) where it has a positive value is 1/length (or 1/area or 1/volume). Since the PDF is constant as long is it is nonzero for a uniform distribution, the property in the question title must be true.