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0 pointsbobbylarrybobby1y ago0 comments

Unless I'm missing something, this can just be directly verified, no “understanding” necessary. All you need to know is that probability distributions can be characterized by their probability density function (PDF).

If Y=-ln(X)/lambda, then P(Y<a) = P(-ln(X)/lambda<a) = P(X>exp(-lambda a)) = 1-exp(-lambda a).

And if Z is exponential with rate parameter lambda, then P(Z<a) = P(lambda exp(-lambda t)<a) = integral from 0 to a of lambda exp(-lambda t)dt, which can be directly computed to be 1-exp(-lambda a).

They have the same PDF, so they're the same distribution.

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timeinput1y ago

I mean if starting from scratch that seems like many years in most western education systems to get to probability, logarithms, exponentiation.

I would say If you knew 2+2=4, and not much else you're years away from 'understanding', if you know ln(exp(y)) = y, and P(x>0.5) = 0.5 for a uniform distribution on [0, 1) then you don't need any additional understanding.

I would bet the GP comment is somewhere inbetween the two extremes, but I think a random sampling of the population would likely result in people generally not knowing the log / exponentiation relation, or anything about the uniform distribution.

zwnow1y ago

Yea got many answers and I dont understand a single one. Good thing you barely need math in programming.

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