# What is $V(X^t)$ for any $t$ when only $E(X)$ and $Var(X)$ are known and $X$ is assumed normal?

Summary I'm trying to calculate $Var(X^t)$ where $t$ is the number of periods using only the following known parameters:

$E(X)$ and $Var(X)$. $X$ is a random variable and is the return factor $(1 + r)$ where $r$ is normally distributed with finite, non-zero mean and variance.

Background I'm attempting to determine the parameters of a probability distribution of the ending dollar value of an investment portfolio assuming that each time period's returns are normally distributed with a known mean and known variance.

To determine the ending distribution, I need the variance of the compounded returns at the ending period. The ending expected value is easy to calculate, but the ending variance is determined indirectly by higher moments which are not directly known without using more exotic methods.

An approximation of the variance of compounded return factors at any time $t$ can be observed through a Monte Carlo simulation, however these estimates are noisy without running large numbers of trials and are not feasible for my particular application.

• Does the title of the question actually match what you really want to know? Because if $X \sim \mathcal N(0, \sigma^2)$ then for any $p \in \mathbb N$ $E(X^p) = 0$ if $p$ is odd, and $\sigma^p (p-1)!!$ if $p$ is even.
– jld
Apr 23 '16 at 22:21
• See here for more details about that: en.wikipedia.org/wiki/Normal_distribution#Moments
– jld
Apr 23 '16 at 22:24
• The problem is that E(X) is not 0, so the more complicated versions of the formulas apply. I'm not exactly familiar with the implementation of the non-zero methods. The question title indirectly poses the ultimate question since V(X^t) can be determined from E(X^p). Apr 23 '16 at 22:40
• Is $r$ constant throughout the investment or does it vary from period to period? If the latter, are you assuming the values in the different periods are independent?
– whuber
Apr 27 '16 at 2:27

Let $X \sim \mathcal N(\mu, \sigma^2)$ and $p \in \mathbb N$. Noting that $E(X^p) = E\left( [X - \mu + \mu]^p \right)$, we can use the binomial theorem to obtain $$E\left( [X - \mu + \mu]^p \right) = E\left( \sum \limits_{k=0}^p {p \choose k} (X - \mu)^k \mu^{p-k} \right)$$

$$= \sum \limits_{k=0}^p {p \choose k} \mu^{p-k} E\left((X - \mu)^k\right)$$

$$= \sum \limits_{k=0, \ k \text{ is even}}^p {p \choose k} \mu^{p-k} \sigma^k (k - 1)!!.$$

As a sanity check, in the moments section of the Wikipedia article on the normal distribution we can see that $E(X^6)$, the 6th noncentral moment of $X$, is supposed to be $\mu^6 + 15 \mu^4 \sigma^2 + 45 \mu^2 \sigma^4 + 15 \sigma^6$. We can confirm that the above formula produces this expression when $p = 6$.

I'm not convinced the title accurately describes your question. It appears to me that you want to find the variance of the product $X_1X_2 \cdots X_t,$ where the $X_i$ are independent.

If so, it is helpful here to use the solution from this question: Variance of product of multiple random variables :

\begin{align} \operatorname{var}(X_1 \cdots X_t) &= E[(X_1 \cdots X_t)^2]-\left(E[X_1\cdots X_t]\right)^2\\ &= E[X_1^2\cdots X_t^2]-\left(E[(X_1]\cdots E[X_t]\right)^2\\ &= E[X_1^2]\cdots E[X_t^2] - (E[X_1])^2\cdots (E[X_t])^2\\ &= \prod_{i=1}^t \left(\operatorname{var}(X_i)+(E[X_i])^2\right) - \prod_{i=1}^t \left(E[X_i]\right)^2 \end{align}

I don't think your scenarios will have identically distributed terms, but just in case, then this would simplify considerably to $$\operatorname{var}(X_1 \cdots X_t) = \left(\sigma^2+\mu^2 \right)^t - \mu^{2t}$$