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I know that $e^{t\mu_{\operatorname{log}}-\Gamma\sqrt{t}\sigma_{\operatorname{log}}}\leq \widetilde{R}_t^S \leq e^{t\mu_{\operatorname{log}}+\Gamma\sqrt{t}\sigma_{\operatorname{log}}}$, with $\mu_{\operatorname{log}},\sigma_{\operatorname{log}},\Gamma \in \mathbb{R}^+$ and $\widetilde{R}_t^S:=\prod_{i=1}^{t-1}(1+\widetilde{r}_t^S)$. Note that this interval results from $\left | \frac{\frac{1}{t}\operatorname{log}\widetilde{R}_t^S-\mu_{\operatorname{log}}}{\frac{\sigma_{\operatorname{log}}}{\sqrt{t}}} \right | \leq \Gamma$, in its turn obtained from the Central Limit Theorem $\left | \frac{\frac{1}{n}\sum_{i=1}^{n}X_i-\mu}{\frac{\sigma}{\sqrt{n}}} \right |\overset{d}{\rightarrow} \operatorname{N}(0,1)$. How can I prove that the interval for $\widetilde{r}_t^S$ is

$\left | \widetilde{r}_t^S-\mu \right |\leq \Gamma\sigma \space \Rightarrow \space \mu-\Gamma\sigma \leq \widetilde{r}_t^S \leq \mu+\Gamma\sigma$

knowing that $\frac{\widetilde{R}_{t+1}^S}{\widetilde{R}_t^S}=1+\widetilde{r}_t^S$?

Thanks for any help.

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  • $\begingroup$ Probably using a Taylor expansion of the exponent or log... $\endgroup$ – steveo'america Nov 24 '20 at 17:31
  • $\begingroup$ @steveo'america That is to say? $\endgroup$ – Marco Pittella Nov 25 '20 at 8:48

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