Source: 1954 Miklos Schweitzer, Problem 5
Let $\xi _{1},\xi _{2},\dots ,\xi _{n},… $ be independent random variables of uniform distribution in $(0,1)$. Show that the distribution of the random variable
$$\eta _{n}= \sqrt[]{n}\prod_{k=1}^{n}(1-\frac{\xi _{k}}{k}) (n= 1,2,…)$$
tends to a limit distribution for $n \to \infty $.
Attempt: First I factor out a $\frac{1}{k}$ term from the product to get
$$\sqrt[]{n}\prod_{k=1}^{n}(1-\frac{\xi _{k}}{k}) = \frac{1}{\sqrt{n}(n-1)!}\prod\limits_{k=1}^n (k-\xi_k).$$
Here, we have $k-\xi_k\sim \text{Unif}(k-1,k)$ and so then with $Y_k=-\log(k-\xi_k)$ and $\eta_n=-\log(\sqrt{n}(n-1)!)-\sum_{k=1}^n Y_k$, we have
$$f_{Y_k}(y)=\frac{1}{k-(k-1)}\cdot |-e^{-y}|=e^{-y}.$$
I thought this may be promising as it looks exponential, but the support doesn't match up – here we have $y\in[-\log(k),-\log(k-1)]$. Additionally, I took the approach I did because I wanted to make this appear as exponentials whose sum is Gamma, but I realized this is dumb as we have an infinite number of them so even with aligned support, the resulting random variable is infinite WP1. The (n-1)! term on the denominator made me think of the Erlang distribution, but the exponential-looking random variables are not identically distributed and again this is limiting, not finite.
Alternatively, I figured by inspection, it looks like we could apply the CLT and obtain something normal-looking, but I got tired looking at the problem.
Edit: As per Brian's suggestion, we could take the log first
$$\log(\eta_n) = \log(\sqrt{n})+\sum\limits_{k=1}^n \log(1-\frac{\xi_k}{k}).$$
Then letting $X_k=\log(1-\frac{\xi_k}{k})$, we have for $x\in(\log(1-\frac{1}{k}),0)$,
$$f_{X_k}(x)=ke^x.$$
Taking the Laplace transform gives us
$$\mathcal{L}(f_{X_k})=\mathbb{E}\left[e^{-tX_k}\right]=\int\limits_\Omega e^{-tx}\cdot ke^x\;dx$$
$$=\frac{k}{t+1}\left[\left(1-\frac{1}{k}\right)^{-(t+1)}-1\right].$$
This looks a bit horrid (possible I messed up the algebra), then with this, the idea is we'd have
$$\mathbb{E}\left[e^{-t\sum_{k=1}^n X_k}\right]=\prod\limits_{k=1}^n \frac{k}{t+1}\left[\left(1-\frac{1}{k}\right)^{-(t+1)}-1\right]$$
by independence and we'd be able to identify this distribution. But two problems with this for me: 1) there's the lingering $\sqrt{n}$ term I don't know what to do with and 2) not sure what distribution this looks like.
Question: I am all for hints on how to approach this correctly – thanks.
Best Answer
Interesting question. First take log as you did. Note that $(\log{n}- \sum_{k=1}^{n}1/k)$ converges to the Euler constant. So it suffices to show the series $$\sum_{k=1}^{n}\left(\log\Big(1-\frac{\xi_{k}}{k}\Big)+\frac1{2k}\right) \tag{1}$$ converges. But the series $$\sum_{k=1}^{\infty}\frac{\xi_{k}-1/2}{k}$$ converges almost surely by the Kolmogrov one series theorem since $$\sum_{k=1}^{\infty}\frac{\operatorname{Var}(\xi_{k})}{k^2}<\infty.$$ The absolute convergence of the series $$\sum_{k=1}^{\infty}\left[\log\Big(1-\frac{\xi_{k}}{k}\Big)+\frac1{2k}+\frac{\xi_{k}-1/2}k\right]$$ follows from the fact that $|\log(1-z)+z|\le C|z|^{2}$ for $|z|$ small. So we have actually shown the series (1) converges almost surely and this implies convergence in distribution. Note that for indepedent sums, convergence in distribution and convergence a.s. is equivalent. See the question asked by myself if $e^{itx_n}$ converges for every t in a countable dense subset of $[-t0, t0]$, does $x_n$ converge?