Variance of Cumulative Returns Under Independent Period Returns

Summary

The document derives the variance of a multi-period compounded return when single-period returns are independent and share the same mean and variance. It first expands the squared gross return, showing that its expectation equals the return variance plus the square of the mean gross return. It then applies independence to factor expectations of products across periods, both for the product of squared gross returns and for the product of gross returns themselves.

Subtracting the squared expected compounded gross return yields a closed-form expression for the variance over k periods. The derivation is useful for understanding how compounding changes return dispersion and why the result depends on both the single-period mean and variance. Its assumptions are restrictive: identical moments and independence across periods are required for the displayed simplification. Serial dependence, changing distributions, or non-identical periods would require a different calculation; the answer does not provide empirical evidence or address those cases.

Key ideas

• A compounded return over several periods is the product of single-period gross returns minus one.
• The variance of the compounded return is unchanged by subtracting one from the gross return product.
• For independent returns, expectations of products factor into products of expectations.
• The second moment of a gross return combines its variance with the square of its mean gross return.
• The closed-form result assumes independent periods with the same mean and variance.

Tags

# Derivation of arithmetic variation of a portfolio over multiple periods


# Derivation of arithmetic variation of a portfolio over multiple periods












I am very confused on how to derive the attached equation (15).

Would someone be kind enough to walk me through the proof?

## Answer by caverac (score 2)

https://quant.stackexchange.com/a/36321

$\newcommand{\E}{\mathbb{E}}$ $\newcommand{\V}{\mathbb{V}}$

First note that

\begin{eqnarray} \E[(r_{t + n} + 1)^2] &=& \E[r_{t+n}^2 + 2r_{t + n} + 1] = \E[r_{t + n}^2] + 2\mu + 1 \\&=& (\V[r_{t+n}] + \E^2[r_{t+n}]) + 2 \mu + 1 \\ &=& \sigma^2 + (\mu^2 + 2\mu + 1) = \sigma^2 + (\mu + 1)^2 \tag{1} \end{eqnarray}

Now, since $r_{t+n}$ are independent random variables we have

\begin{eqnarray} \V[r_{t,t+k}] &=& \V\left[\prod_{n=1}^k (r_{t+n} +1) - 1\right] = \V\left[\prod_{n=1}^k (r_{t+n} +1)\right]\\ &=& \E\left[\left(\prod_{n=1}^k (r_{t+n} +1)\right)^2\right] - \E^2\left[\prod_{n=1}^k (r_{t+n} +1)\right] \\ &=& \E\left[\prod_{n=1}^k (r_{t+n} +1)^2\right] - \left(\E\left[\prod_{n=1}^k (r_{t+n} +1)\right]\right)^2 \\ &=& \prod_{n=1}^k \E[(r_{t+n} +1)^2] - \left(\prod_{n=1}^k \E[r_{t+n} +1]\right)^2 \\ &\stackrel{(1)}{=}& \prod_{n=1}^k [\sigma^2 + (\mu + 1)^2] - \prod_{n=1}^k(\mu + 1)^2 \\ &=&[\sigma^2 + (\mu + 1)^2]^k - (\mu + 1)^{2k} \end{eqnarray}

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