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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.

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  • $\begingroup$ Hi and welcome. How do you want to approach this problem? How do you have to(?) approach this problem? I.e. could you run some simulation studies, or are approximations acceptable, e.g. assuming $1+\xi_t$ to be lognormally distributed, and $x_t=\log(1+\xi_t)$ to be normally distributed so that $P_t=P_0e^{\sum_{\tau=1}^t x_{\tau}}$ ? $\endgroup$ Aug 23 at 13:21
  • $\begingroup$ You can, of course, find the variance of $Z_t=\prod_{i=1}^{t}\xi_i$ via $Var(Z_t)=E(Z_t^2)-E(Z_t)^2$ and calculating $E(Z_t^2)$ as $\left((1+\mu)^2+\sigma^2\right)^t$. $\endgroup$ Aug 23 at 13:25
  • $\begingroup$ Thanks for your reply @Kermittfrog. Approximations are fine, though $\xi_t$ is meant to denote daily price returns (Mandelbrot proposed a Lévy alpha-stable distribution, but I have approximated with a Gaussian $\xi$). The variance of $Z_t$ is enough to determine the variance of $P_t$, so thanks again. $\endgroup$
    – UNOwen
    Aug 24 at 12:54
  • $\begingroup$ I have updated the question for time varying $\mu$ with my calculation of variance. $\endgroup$
    – UNOwen
    Aug 29 at 16:28
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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,

$$ \begin{align} \mathrm{Var}\left(Z_n\right)&=\mathrm{E}\left(Z_n^2\right)-\mathrm{E}\left(Z_n\right)^2\\ &=\left(\left(1+\tilde{\mu}\right)^2+\sigma^2\right)^{n}-\left(1 + \tilde{\mu}\right)^{2n} \end{align} $$

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  • $\begingroup$ Thank you for your answer. The mean $\mu_t$ (your $\tilde{\mu}$) varies with time. So, the random variables are independent but not identically distributed. I was wondering if a solution could be found for varying $\mu_t$ in terms of $\mathbb{E}[P_t]$. Nonetheless, your response is very helpful for constant $\mu$ (i.i.d random variables). $\endgroup$
    – UNOwen
    Aug 30 at 11:26
  • $\begingroup$ I've updated the answer accordingly. $\endgroup$ Aug 30 at 12:25
  • $\begingroup$ Sure, thanks again. $\endgroup$
    – UNOwen
    Aug 30 at 15:40

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