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Let $X_1,X_2,X_3,\ldots$ be IID r.v. with

\begin{equation} P(X_i<-1)=0 \end{equation} \begin{equation} P(X_i<0)>0 \end{equation} \begin{equation} P(X_i>0)>0. \end{equation}

Define \begin{equation} F_t = \prod_{i=1}^t(1+\frac{1}{2}X_i) \end{equation} \begin{equation} G_t = \prod_{i=1}^t(1+\frac{1}{4}X_i). \end{equation}

How can we show, for some integer $S>0$, that \begin{equation} P\left(\sup_{s\in[1,S]} \left[\sup_{t\in[1,s]} \frac{F_t-F_s}{F_t}\right] > \frac{1}{3}\right) > P\left(\sup_{s\in[1,S]} \left[\sup_{t\in[1,s]} \frac{G_t-G_s}{G_t}\right] > \frac{1}{3}\right) \end{equation}

Thanks in advance for any hints to get me started, or possibly a draft of a solution. I simply have no clue about how to proceed.


I have been thinking. Instinctvly, this problem seems to hold true because for any negative $X_i$, this holds:

\begin{equation} (1+\frac{1}{2}X_i) < (1+\frac{1}{4}X_i). \end{equation}

This should mean that $F_t$ will usually be dropping faster than $G_t$. I still wonder how to formalize these instincts.

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Hint: For every $x_k\gt-\frac12$, $1+2x_k\leqslant(1+x_k)^2$ hence $\prod\limits_k(1+2x_k)\leqslant \left(\prod\limits_k(1+x_k)\right)^2$.

Application: The event in the LHS is $\bigcup\limits_{t<s}B^a_{t,s}$ and the event in the RHS is $\bigcup\limits_{t<s}C^a_{t,s}$ with $$ B^a_{t,s}=\left[\prod_{k=t+1}^s(1+2x_k)< a\right], \quad C^a_{t,s}=\left[\prod_{k=t+1}^s(1+x_k)< a\right],\quad a=\tfrac23,\quad x_k=\tfrac14X_k. $$ For every $t<s$, $C^a_{t,s}\subseteq B^{a^2}_{t,s}$ and $B^{a^2}_{t,s}\subseteq B^{a}_{t,s}$ since $a<1$. The result follows (and the number $\frac13$ in the LHS may be replaced by $\frac59$).

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Thanks, I will have to think about your hint. I'm also wondering if the problem can be boiled down to a simpler number theory problem, as here:… – godel68 Aug 1 '12 at 17:01
Indeed, the proof above shows that the result is entirely deterministic (what you call number theory, I guess). – Did Aug 1 '12 at 18:49
Do you know how the result could be made more general, using not just 1/2 vs. 1/4, but rather any number $q$ vs. any other number $p<q$? (with $0<p,q<1$) – godel68 Aug 1 '12 at 23:09
Yes, start from $0\leqslant1+qx_k\leqslant(1+px_k)^{q/p}$ for every $x_k\geqslant-1/q$ and proceed as above. – Did Aug 2 '12 at 7:43

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