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Let $0=X_0 \leq X_1 \leq X_2 \leq \cdots $ be an increasing sequence of random variables with $X_n - X_{n-1}$ i.i.d. and $\mathbb{E}(X_n - X_{n-1}) = 1$ for all integers $n > 0$.

I want to show that almost surely for each $m$, there is an $M>m$ such that $$ \frac{1}{M}X_M \leq \frac{1}{m}X_m. $$

I've computed $$ \mathbb{E}\left(\frac{T_{n}}{n} - \frac{T_{n-1}}{n-1} \right) = \frac{1}{n} - 1. $$ and $$ \mathbb{E}\left(\frac{T_{M}}{M} - \frac{T_{m}}{m} \right) = (M-m)\sum_{n=m+1}^{M} \left(\frac{1}{n} - 1\right). $$

I feel like I should be able to use this to show that $$ \mathbb{P}\left(\text{there exists $m$ such that for all $M > m$: } \frac{1}{M}X_M > \frac{1}{m}X_m \right) = 0 $$ by using independence and forming an infinite product.

I'm not sure how to make this work. I would appreciate any help.

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1 Answer 1

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I do not think that it is true.

According to the strong law of large numbers, we have: $$\frac{1}{n}\sum_{i=1}^n (X_i - X_{i-1}) = \dfrac{X_n - X_0}{n} \xrightarrow[n\to\infty]{\text{a.s.}} 1 $$

Suppose that $P(X_1<1) > 0$ (if not, we have $X_n = n$ for every $n$ a.s.). Then $$ P(\exists m,\;\forall M > m,\; \dfrac{X_M}{M} > \dfrac{X_m}{m}) \geq P(\min_{n \geq 1} \dfrac{X_n}{n} < 1) > 0 $$

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+1. There are several ways to express roughly the same idea, this one is strikingly concise. –  Did Mar 17 '13 at 8:50
    
Thanks. I was trying to use this an input into a theorem. In the end, it turned out I was trying to use the wrong theorem. Having the input result be false was a hint that something might be wrong. ;) –  RClark Mar 21 '13 at 21:04

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