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If $(X_n,\mathcal{F_n})_{n\in \mathbb{N}}$ is a martingale such that $\forall$ n $\in \mathbb{N}, \frac{X_{n+1}}{X_n}\in L^1$ How can be demonstrated that:

$\mathbb{E}[\frac{X_{n+1}}{X_n}]=1$ and that the random variables $\frac{X_{n+1}}{X_n}$ and $\frac{X_n}{X_{n-1}}$ are uncorrelated?

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    $\begingroup$ That it is finite, for instance. Also that you asked about it. How many more idiotic answers do you need before saying clearly what your actual problem is, what difficulty you have, and what estimate(s) you would like to get? $\endgroup$
    – fedja
    Commented Nov 18, 2017 at 1:48
  • $\begingroup$ you are right I've clarified my question $\endgroup$
    – cacy
    Commented Nov 18, 2017 at 10:08

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Since $\frac{X_{n+1}}{X_n}\in L^1$, then $\mathsf{E}[\bigl|\frac{X_{n+1}}{X_n} \bigr| 1_{\{X_n=0\}}]=0$ and $$ \mathsf{E}\Bigl[\frac{X_{n+1}}{X_n}1_{\{|X_n|>\varepsilon\}}\Bigr] =\mathsf{E}\Bigl[\frac{\mathsf{E}[X_{n+1}|X_n]}{X_n}1_{\{|X_n|>\varepsilon\}} \Bigr]=\mathsf{P}(|X_n|>\varepsilon).$$ Letting $\varepsilon\to0$ in above expression we get $$ \mathsf{E}\Bigl[\frac{X_{n+1}}{X_n}\Bigr]=\mathsf{P}(|X_n|>0).$$

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