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I know it's reasonable that $\frac{X}{E[X]}\approx 1$ , but how can I show it? I thought about Chebyshev's inequality but didn't get far.

Thanks.

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What do you mean by $\approx$ here? Certainly its expectation is 1 (provided $E[X] \ne 0$), which is a consequence of linearity. –  Clive Newstead Nov 15 '12 at 9:48
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It need not be "about 1"; in fact $X/E[X]$ can be arbitrarily far from $1$. What exactly do you mean? –  ShreevatsaR Nov 15 '12 at 9:52

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As ShreevatsaR already wrote: $\frac{X}{\mathbb{E}X}$ can be arbritarily large. What you are maybe thinking of is the Strong Law of Large Numbers: Let $X_n \in L^1$ ($n \in \mathbb{N}$) identically distributed and independent random variables, then

$$\frac{\sum_{i=1}^n X_i}{n} \to \mathbb{E}X_1 \quad \text{a.s.}$$

(So if you for example an experiment (like throwing a dice) and you repeat this (independently) very very often, the average of your results will be close to the expectation value.)

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(Or, rewriting the SLoLN: $\frac{\sum_{i=1}^n X_i}{E \left(\sum_{i=1}^n X_i\right)} \to 1 \quad \text{a.s.}$) –  TMM Nov 15 '12 at 10:20

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