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ANSWER

Let Yi ~ U(0,1) be IID.
Yi ー1
y = erplln(y)) = erp(Xin(Yi)) = exp(X)
where,
Xi = ln(Yi)
X =\sum_{i=1}^{n} ln(Yi)
now we can assume, Xi has
mean \mu = E[Xi]
and
variance \sigma ^{^{2}} = V[Xi]
using central limit theorem,
n-+00
means,
X \xrightarrow[n\rightarrow \infty]{d } N (n\mu ,n\sigma ^{2})
Thus,
for large value of n, we get
where, X is approximately normally distributed with mean
E[X] = n\mu
and
standard deviation
V[X] = \sigma{\sqrt{}}n

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