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Suppose $Ui$ is independently uniformed distributed between [0,b], $Y = -\Sigma_1^n log(U_i)$. what is the pdf of Y? I tried used characteristic function but it doesn't match each of usual distribution.

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Note: the following argument assumes $b=1$. To generalise, add $\ln b$ to each $\ln U_i$ term, i.e. $-n\ln b$ to $y$ so its pdf shifts.

You probably already worked out $-\ln U_i\sim\operatorname{Exp}(1)$, because $$P(-\ln U\le x)=P(U\ge\exp -x)=1-\exp -x.$$Of course, this implies $-\ln U_i$ has characteristic function $1/(1-it)$, so $Y$ has cf $1/(1-it)^n$. Now, what distribution is that? Spoiler: it's

a Gamma distribution with $k=n,\,\theta=1$, so the pdf is $\frac{y^{n-1}}{(n-1)!}\exp -y$ for $y\ge 0$.

If you can guess that by thinking about integrals, the inversion theorem guarantees it's right because it gives the desired cf.

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  • $\begingroup$ Thanks. I have already worked out the case when $b=1$, however I couldn't figure out the generalised part. Why can we just add $-nlnb$ to y? How can you prove it is right regularization? $\endgroup$ – T.y Jan 19 at 9:11
  • $\begingroup$ @T.y Because $U_i\sim U(0,\,b)$ iff $U_i/b\sim U(0,\,1)$. $\endgroup$ – J.G. Jan 19 at 9:13
  • $\begingroup$ Still confused. I mean I already worked out the cf of general b version and it is %(b^(nit))/((1-it)^n)%. How can I match the pdf? $\endgroup$ – T.y Jan 19 at 9:30
  • $\begingroup$ @T.y Scaling $U_i$ to $bU_i$ changes the characteristic function $\varphi(t)$ to $\varphi(bt)$, not $b^{it}\varphi(t)$. $\endgroup$ – J.G. Jan 19 at 9:32

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