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$X_1,X_2,X_3,\ldots$ are IID random variable taking values in $(-1,\infty)$.

Also $t\in(0,1)$.

Define random variables $Y_1,Y_2,Y_3,\ldots$ recursively like

$$Y_1 = (1+tX_1)$$ $$Y_n = Y_{n-1}(1+tX_n)$$

Imagine this is money being partially invested someplace again and again.

How can I prove that

$$Y_n\to\infty\,\, \textrm{a.s.}\,\,\,\,\textrm{if and only if}\,\,\,\, E[\log(1+tX_1)] > 0 \,\,\,\,\,\textrm{ ?}$$

If you can help me solve this, I'd also appreacite help with a similar problem but generalized: Solving this random variable problem

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This is not a formal proof by any means, but more like an idea of how to do it. Maybe it'll help you to get started.


Note that you may write $$ Y_n=\prod_{i=1}^n (1+tX_i). $$ Furthermore, the log of this is $$ \log Y_n=\sum_{i=1}^n \log(1+tX_i). $$ By the law of large numbers, $$ \frac{\log Y_n}{n}=\frac{\sum_{i=1}^n\log(1+tX_i)}{n}\longrightarrow E\left[\log(1+tX_1)\right] \quad \text{as } n\to \infty, $$ where by the iid assumption we may use $X_1$ in the expectation.

For almost sure convergence, you want $$ Pr\left(\lim_{n\to \infty} Y_n=\infty\right)=1. $$ If $Y_n$ goes to infinity, then $\log Y_n$ will do so too. Hence, we can use $$Pr\left(\lim_{n \to \infty} \log Y_n = \infty\right)=1.$$

The key is to show that if $\frac{\log Y_n}{n}$ converges to a non-zero finite constant, then this implies that $\log Y_n$ tends to infinity.

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  • $\begingroup$ You start pretty well, rigorously and all, but when you begin decomposing limits, chaos ensues. Can you straighten out this latter part? $\endgroup$ – Did May 16 '14 at 21:17
  • $\begingroup$ My idea was to rewrite it in order to get $n$ in the denominator in order to use the LLN result, but perhaps that was the wrong way. Is it at the $\lim \lim$ part the chaos starts you mean, or is the step right before that where I go wrong? And by 'go wrong' I naturally mean 'start going wrong'. $\endgroup$ – hejseb May 16 '14 at 21:24
  • $\begingroup$ At the second $\Pr(\qquad)$ after "Rewrite the first term as...", when you start splitting the limits. $\endgroup$ – Did May 16 '14 at 21:33
  • $\begingroup$ @Did I see, thanks. I removed that part and just added what I think needs to be done, but I can't seem to be able to show it at the moment. Will try again tomorrow. $\endgroup$ – hejseb May 16 '14 at 22:01
  • $\begingroup$ Right. About "the key": assume that some sequence $(x_n)$ is such that $x_n/n\to\ell$ when $n\to\infty$, with $\ell\gt0$. Then $x_n/n\gt\ell/2$ for every $n$ large enough, say every $n\gt N$. Thus, $x_n\gt n\ell/2$ for every $n\gt N$. And you want to show that $x_n\to+\infty$. Well... $\endgroup$ – Did May 16 '14 at 22:09
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Note that

$$\log(Y_n) = \sum_{i=1}^{n}\log{(1+tX_i)}> \sum_{i=1}^{n}\frac{\log(1+tX_i)}{i} $$

By the SLLN, almost surely,

$$\frac{\log(Y_n)}{n} = \frac{1}{n}\sum_{i=1}^{n}{\log(1+tX_i)} \rightarrow E[\log(1+tX_1) > 0 \\ $$

An application of the Cesaro Mean Theorem says that if $\sum_{i=1}^{n}\frac{\log(1+tX_i)}{i}$ coverges to a finite limit, then $\frac{1}{n}\sum_{i=1}^{n}\log(1+tX_i) \rightarrow 0$, which is not the case.

Therefore $\log(Y_n) \rightarrow \infty$ and $Y_n \rightarrow \infty$.

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  • $\begingroup$ The trick of dividing by $i$ each $\log(1+tX_i)$ is both wrong (since $X_i$ can be negative) and unnecessary (since $\frac1n\log(Y_n)\to\ell$ with $\ell\gt0$ does imply $Y_n\to+\infty$). $\endgroup$ – Did May 16 '14 at 21:15
  • $\begingroup$ @Did: Thank you. I see. So $\log(Y_n) > n\ell/2$ for all $n$ sufficiently large. $\endgroup$ – RRL May 16 '14 at 23:32

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