Nothing beats a brute force confirmation.pic.twitter.com/VkBrX9HVmu
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Nothing beats a brute force confirmation.pic.twitter.com/VkBrX9HVmu
except maybe a succinct proof????
An alternative way to prove this would be to use the Irwin–Hall distribution - the sum of a number of independent random variables.pic.twitter.com/CRu6w7k8zf
For x<=1, you don’t even need the full Irwin-Hall generalization. Nice overview at http://nsgrantham.com/sum-over-one/
Write m(x) for m_x. Then the integral equation is m(x)=1+\int_0^x m(u)du. This gives: m'(x) = m(x) and m(0) =1, so m(x) = \exp(x).
Can you explain the step right before? Why does x<=1 imply the first integral is equivalent to the second?
This is incorrect. You conclude that m_x is e^x. However, asymptotically, m should be linear. Mistake is in casework with x vs x-mu < 1.
Apologies: we were wrong and @fermatslibrary was right. As @arunkris95 pointed out, the claim was not as strong as we thought it was.
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