From Cumulative Return Bounds to Single-Period Return Bounds

Summary

The document asks how to derive bounds for a single-period return from bounds on cumulative returns. It starts with an exponential interval for cumulative growth based on a log-return mean and standard deviation, then invokes a central limit theorem approximation for the sample mean of log returns. The desired result is a symmetric interval around the single-period mean, expressed in terms of its standard deviation.

The key relation supplied is that the ratio of consecutive cumulative values equals one plus the single-period return. However, the document provides no derivation or answer, and the cumulative bound alone does not establish the proposed single-period bound: a bound on an aggregate does not automatically impose the same bound on each component. The notation and indexing also appear inconsistent, and a central limit theorem gives an asymptotic distributional approximation rather than a deterministic guarantee. These distinctions matter when interpreting the interval as a statistical confidence statement.

Key ideas

• The question seeks single-period return bounds from bounds on cumulative returns.
• It relates consecutive cumulative values to one plus the intervening return.
• A bound on cumulative growth does not by itself prove an equivalent bound for each period.
• The central limit theorem supports an asymptotic approximation rather than an unconditional deterministic limit.

Tags

# Boundaries on the single-period returns


# Boundaries on the single-period returns












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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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.