Variance of a Multiplicative Random Walk with Time-Varying Drift
Summary
The document derives the moments of a multiplicative price process whose returns are independent Gaussian variables, then extends the calculation to a time-varying mean. Writing the price as a product of one-period gross returns makes the recursion clear: independence lets the expected product factor into the product of expected returns. The second moment similarly factors into the product of each period’s second gross-return moment.
Variance follows by subtracting the squared first moment from the second moment. With a constant mean and common volatility, this yields a difference of powers; with changing means, the corresponding moments are products across periods. The method provides a direct route to the variance without assuming that the mean grows at a constant rate. It relies on independent period returns and the stated Gaussian moments, and does not address dependence, parameter estimation, or whether normally distributed simple returns are realistic. The response’s time-varying moment expressions are stated compactly, so care is needed to distinguish first and second moments when applying them.
Key ideas
- A multiplicative price process can be expressed as a product of gross returns.
- Independence allows the first moment of the product to factor across periods.
- The second moment factors using each period’s second gross-return moment.
- Variance is obtained by subtracting the squared first moment from the second moment.
- Time-varying means replace powers of constant moments with products across periods.
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Full text
# Variance of Random Walk with Drift
# Variance of Random Walk with Drift
For Gaussian random variables $\xi_t$ with mean $\mu_t$ and standard deviation $\sigma$, consider the random walk with initial condition $P_0=100$, such that
\begin{equation} P_t=P_{t-1}(1+\xi_t). \end{equation}
I assume that variance of $P_t$ satisfies \begin{equation*} \begin{split} \text{Var}P_{t}&=\text{Var}\left(P_{0}\prod_{i=1}^{t}\left(1+\xi_{i}\right)\right),\\&=P_{0}^{2}\left(\mathbb{E}\left[\prod_{i=1}^{t}\left(1+\xi_{i}\right)^{2}\right]-\mathbb{E}\left[\prod_{i=1}^{t}\left(1+\xi_{i}\right)\right]^{2}\right),\\&=\mathbb{E}\left[P_{t}\right]^2\left(\left(1+\frac{\sigma^{2}}{\mathbb{E}\left[P_{t}\right]}\right)^{t}-1\right). \end{split} \end{equation*}
However, I would like to know whether this can be proven for a changing mean $u_t$, such that $\mathbb{E}[P_t]\neq P_0(1+\mu)^t$. Any help would be much appreciated.
## Answer by Kermittfrog (score 2, accepted)
https://quant.stackexchange.com/a/66682
### Setup
Let
$$Z_n\equiv \prod\limits_{i=1}^n(1+x_i)$$
where each $x_i$ is iid normally distributed as $x_i\sim \mathrm{N}\left(\tilde{\mu},\sigma\right)$. For simplicity, and with some abuse of notation, let $\mu = 1 + \tilde{\mu} $, i.e.
$$Z_n\equiv \prod\limits_{i=1}^n(1+x_i)\sim\prod\limits_{i=1}^n(\mu + \sigma\varepsilon_i)$$
where each $\varepsilon_i$ is iid standard normal, $\varepsilon_i\sim \mathrm{N}\left(0,1\right)$. Note that
$$Z_n\equiv \prod\limits_{i=1}^n(1+x_i)\sim\prod\limits_{i=1}^{n-1}(\mu + \sigma\varepsilon_i)\left(\mu+\sigma\varepsilon_n\right)=Z_{n-1}\left(\mu+\sigma\varepsilon_n\right)$$ which we will use below.
### First and second moments
$$ \begin{align} \mathrm{E}\left(Z_1\right)&=\mathrm{E}\left(\mu+\sigma\varepsilon_1\right)\\ &=\mu \end{align} $$ and then: $$ \begin{align} \mathrm{E}\left(Z_n\right)&=\mathrm{E}\left(Z_{n-1}\left(\mu+\sigma\varepsilon_n\right)\right)\\ &=\mu\mathrm{E}\left(Z_{n-1}\right) + \sigma\mathrm{E}\left(\varepsilon_{n}Z_{n-1}\right)\\ &=\mu\mathrm{E}\left(Z_{n-1}\right)\\ &=\mu\mathrm{E}\left(\left(\mu+\sigma\varepsilon_{n-1}\right)Z_{n-2}\right)\\ &=\ldots\\ &=\mu^{n-1}\mathrm{E}\left(Z_{1}\right)\\ &=\mu^n\\ &=\left(1 + \tilde{\mu}\right)^n \end{align} $$
If the mean $\tilde{\mu_t}$ is time-dependent, then $$ \mathrm{E}\left(Z_n^2\right)=\prod\limits_{t=1}^n\left(1+\tilde{\mu_t}\right) $$
Likewise,
$$ \begin{align} \mathrm{E}\left(Z_1^2\right)&=\mathrm{E}\left(\left(\mu+\sigma\varepsilon_1\right)^2\right)\\ &=\mathrm{E}\left(\mu^2+2\sigma\mu\varepsilon_1+\sigma^2\varepsilon_1^2\right)\\ &=\mu^2+\sigma^2 \end{align} $$ and then $$ \begin{align} \mathrm{E}\left(Z_n^2\right)&=\mathrm{E}\left(Z_{n-1}^2\left(\mu+\sigma\varepsilon_n\right)^2\right)\\ &=\mathrm{E}\left(\mu^2Z_{n-1}^2\right)+2\sigma\mathrm{E}\left(\varepsilon_nZ_{n-1}^2\right)+\sigma^2\mathrm{E}\left(\varepsilon_n^2Z_{n-1}^2\right)\\ &=\left(\mu^2+\sigma^2\right)\mathrm{E}\left(Z_{n-1}^2\right)\\ &=\left(\mu^2+\sigma^2\right)\mathrm{E}\left(\left(\mu+\sigma\varepsilon_{n-1}\right)^2Z_{n-2}^2\right)\\ &=\ldots\\ &=\left(\mu^2+\sigma^2\right)^{n-1}\mathrm{E}\left(Z_{1}^2\right)\\ &=\left(\mu^2+\sigma^2\right)^{n}\\ &=\left(\left(1+\tilde{\mu}\right)^2+\sigma^2\right)^{n} \end{align} $$
If the mean $\tilde{\mu_t}$ is time-dependent, then $$ \mathrm{E}\left(Z_n^2\right)=\prod\limits_{t=1}^n\left(\left(1+\tilde{\mu_t}\right)^2+\sigma^2\right) $$
### Variance
Finally,
$$ \begin{align} \mathrm{Var}\left(Z_n\right)&=\mathrm{E}\left(Z_n^2\right)-\mathrm{E}\left(Z_n\right)^2\\ &=\left(\left(1+\tilde{\mu}\right)^2+\sigma^2\right)^{n}-\left(1 + \tilde{\mu}\right)^{2n} \end{align} $$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.