Let $(X_{n})_{n}$ be independent random variables that are $\mathcal{U}{[1,2]}$

Prove $(\prod_{i=1}^{n}X_{n})^{\frac{1}{n}}$ exists for $n \to \infty$ and that $\exists c \in \mathbb R$ such that

$(\prod_{i=1}^{n}X_{n})^{\frac{1}{n}}\to c$, a.s. for $n \to \infty$

I honestly do not even know where to begin because I've never worked with the product of random variables.

Any ideas and tips?

  • 1
    $\begingroup$ Hint. Strong Law of Large Numbers. $\endgroup$ – Sangchul Lee Jan 7 at 13:49
  • $\begingroup$ But strong law of large numbers pertains to a sum of random variables $\sum X_{i}$ and not the Product $\prod X_{i}$ $\endgroup$ – SABOY Jan 7 at 13:51

It is unclear what $\mathcal{U}[1,2]$ means, so I'll presume $\mathcal{U}[1,2]$ refers to the uniform distribution on $[1,2]$.

The answer is straightforward once we note that $$\ln \left([\prod_{i=1}^n X_i]^{1/n}\right) = \frac{1}{n}\sum_{i=1}^n\ln X_i$$ is an averaged sum of i.i.d. random variables $(\ln X_n)_{n=1}^\infty$. By strong law of large numbers, it follows $$ \frac{1}{n}\sum_{i=1}^n\ln X_i \to E[\ln X_1]= \int_1^2 \ln x \;dx = (x\ln x-x)\big|^2_1=2\ln 2-1 $$almost surely as $n\to\infty$. Therefore, we have $$ \left[\prod_{i=1}^n X_i\right]^{1/n}=\exp\left(\frac{1}{n}\sum_{i=1}^n\ln X_i\right)\to \exp(2\ln 2-1) = \frac{4}{e}. $$

  • $\begingroup$ Would never have though of that $\operatorname{ln}$ trick thank you! $\endgroup$ – SABOY Jan 7 at 13:54
  • $\begingroup$ Just one question: How do we know that $\operatorname{ln}(X)$ ~ $\mathcal{U}{[1,2]}$ if $X$ ~ $\mathcal{U}{[1,2]}$? $\endgroup$ – SABOY Jan 7 at 17:38
  • $\begingroup$ Umm ... ? There's a misunderstanding, I guess. Let me explain. Of course $\ln X\sim \mathcal{U}[1,2]$ is false, and I didn't use it. What I used is just well-known expectation formula that $E[g(X)]=\int g(x)f(x)dx$ where $f(x)$ is a p.d.f. of $X$. Yes, over the same interval $[1,2]$ where $X$ is defined! I hope this makes it clear. $\endgroup$ – Song Jan 7 at 17:41
  • $\begingroup$ So over the same interval $[1,2]$ $\endgroup$ – SABOY Jan 7 at 17:42
  • $\begingroup$ Quick question: you say that $(\ln(X_{i}))_{i=1,...,n}$ are i.d.d. random variables if $(X_{i}))_{i=1,...,n}$ are i.i.d. But why would $(\exp(\lambda X_{i}))_{i=1,...,n}$ not be i.i.d. ? As stated here: math.stackexchange.com/a/3068631/512018 $\endgroup$ – SABOY Jan 10 at 19:47

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