How to find $\lim_{n\to \infty} P(a≤(X_1X_2...X_n)^{-n/2}e^{n/2}≤b)$ where $X_1,X_2,...,X_n \sim U[0,1]$?

Question

I am trying to calculate $$\lim_{n\to \infty} P(a≤(X_1X_2...X_n)^{-n/2}e^{n/2}≤b)$$ in terms of $a,b$, where $$X_1,X_2,...,X_n \sim U[0,1]\,\,\,\,\,\,\,(i.i.d.)$$ and $$0≤a<b$$ My attempt is to take natural logarithm on both sides, and end up with a r.v $Y=\ln{X_1}+\ln{X_2}+...+\ln{X_n}$. However, I don't know how to estimate the p.d.f of $Y$ and calculate the integral. Can anyone gives me some hint?

Answer 1

Following snarski's hint, we have that $Y_i=-\log(X_i)$ has an exponential distribution with unit mean, hence: $$ \mathbb{P}\left[\log a \leq \frac{n}{2}\sum Y_i +\frac{n}{2}\leq \log b\right] $$ converges to zero since $\sum Y_i$ is concentrated around its mean value, $n$. Chebyshev's inequality is enough for proving it.

Something significant is the geometric mean of $X_1,\ldots,X_n$, given by $$ Z = \left(X_1\cdot X_2\cdot\ldots\cdot X_n\right)^{\frac{1}{n}}.$$ By the Central Limit Theorem, $$ \mathbb{P}[a\leq Z \leq b] = \mathbb{P}\left[\log a \leq \frac{1}{n}\sum(-Y_i) \leq \log b\right]$$ converges to the probability that a $\mathcal{N}(-1,1)$ random variable belongs to $[\log a,\log b]$.

Answer 2

I invite anyone who uses Mathematica to type in the following:

F[n_] := (Times @@ RandomReal[{0, 1}, n])^(-n/2) Exp[n/2]
Mean[Table[F[1000],{10^5}]]

This gives the sample mean of $10^5$ simulations for $n = 1000$. It is a huge number. Now try F[10000], F[10^6], etc. The number of digits of these is enormous. Tell me how this is at all consistent with the other answers posted here, because I have repeatedly suggested there is an error in how the question is posed.