Basic example, assume that the expected daily return of the S&P 500 stock market index is 0, i.e., the return on any given day of the stock market is 0.0%. But, we also (generally) expect that when compounded over long periods of time--say 20 years--that the expected value of that compounded return (e.g., compounded daily returns each with expectation to be 0) is greater than 0. Is it possible for both of these assumptions to be true, i.e., a 0 expectation for each individual instance but an expected value of the compounded value to be greater than 0?

Assuming ${X_i}$~$\mathcal{\pi}(mean = 0,...)$ for all $i\in\mathbb{Z}^+$ ($\pi(\cdot)$ being some arbitrary distribution with mean = 0). Assume $Cov[X_{i},X_{j}]=0$ for every $i\ne{j}$ (i.e., $X$ is independent and identically distributed). I believe the compounded (net) return over $n>1$ periods is $$R_n=\prod_{i=1}^n{(1+X_i)}-1$$ Therefore, the question is: Is it possible for $$E[R_{n}]>0$$ when $E[X_{i}]=0$ for all $i$?

It would seem to me this would not be possible if you assumed the random variable came from a symmetric distribution, e.g., a normal distribution. However, if you assumed the random variable was generated by a distribution with positive skewness then this might be possible. Is this correct? Are there other features of a distribution that would make both of these assumptions hold true.

  • $\begingroup$ Is it possible for a random VARIABLE with 0 mean to have a positive expectation after compounding many observations? $\endgroup$ – Antoine Aug 10 '15 at 22:03
  • $\begingroup$ It actually is possible--in fact, inevitable--even for Normal random variates, if by "compounded" you mean the variable represents the instantaneous return and "compounding" amounts to accumulating the return over longer periods. Is this what you mean? $\endgroup$ – whuber Aug 11 '15 at 0:43
  • $\begingroup$ A minor correction (doesn't change the question): $\text{cov}(X_i, X_j)$ is not equivalent to "independent and identically distributed". If $X_i$ are normally distributed, then the condition is equivalent to independence (but not being equally distributed). You may require simply that $X_i$ be iid with mean zero, without introducing much notation. $\endgroup$ – gappy Aug 11 '15 at 21:48

If $X_i$ are iid with $EX_i=0$, then $ER_n=0$, as you showed in the comments. However, the situation changes if your asset has $E\log(X_i) =0$; this is like stating that the expected rate of return of an asset is zero. In this case, counterintuitively, you can have positive expected returns. Check out the "volatility pumping" example in Luenberger's "Investment Science" (ex 15.2, page 422), or this paper on Parrondo Games.


I understand the "compounding" of a random variable $X$ over $n\ge 1$ periods to be the variable

$$X_n = (1 + X)^n - 1.$$

When $X$ is the interest on an asset after one period, $X_n$ is the interest after $n$ periods (assuming $X$ describes the return for every period).

Since $n$ is finite there is no problem with expanding the right hand side and employing the linearity of expectation to compute

$$\eqalign{\mathbb{E}(X_n) &= \mathbb{E}((1+X)^n - 1) \\ &= \mathbb{E}\left(\sum_{i=0}^n \binom{n}{i} X^i - 1\right)\\ &= \sum_{i=1}^n \binom{n}{i}\mathbb{E}(X^i). }$$

When $X$ has a symmetric distribution and $j$ is any positive integer,

$$\mathbb{E}(X^{2j-1}) = \mathbb{E}((-X)^{2j-1}) = \mathbb{E}(-X^{2j-1}) = -\mathbb{E}(X^{2j-1})$$

implies $\mathbb{E}(X^{2j-1})=0$: all odd moments are zero. Unless $X$ is almost surely zero, all even moments $\mathbb{E}(X^{2j})$ are strictly positive. Therefore, when $n\ge 2$,

$$\mathbb{E}(X_n) = \sum_{j=1}^{\lfloor n/2\rfloor} \binom{n}{2j}\mathbb{E}(X^{2j}) \gt 0.$$

Thus: symmetry of the distribution of $X$ (and non-degeneracy of $X$) guarantee that the return compounded for more than one period will be strictly positive.

Regardless of symmetry, when $n=2$ the return is $\mathbb{E}(X^2)$ which is strictly positive.

In general, for non-symmetric distributions and arbitrary $n$, you would have to evaluate the full sum to determine the sign of the return--and that could vary with $n$. As an example let $X$ take on the value $-5$ with probability $1/6$ and $1$ with probability $5/6$. $X$ has zero expectation, but for $n=3$

$$\mathbb{E}((1+X)^3 - 1) = ((-5+1)^3-1) \times \frac{1}{6} + ((1+1)^3-1) \times \frac{5}{6} = -5 \lt 0.$$

This shows that a negative compounded return for more than $n=2$ periods is possible.

  • $\begingroup$ Assuming ${X_i}$~$\mathcal{N}(0,sigma)$, I believe the compounded return is $$X_n=\prod_{i=1}^n{(1+X_i)}-1$$ Is this equivalent to your representation? I don't believe so, but not certain. $\endgroup$ – Ursus Frost Aug 11 '15 at 15:31
  • $\begingroup$ No, it's not the same. When the $X_i$ are independent, so are the $1+X_i$, whence $$\mathbb{E}(X_n) = \prod_{i=1}^n\mathbb{E}(1+X_i)-1=\left(\prod_{i=1}^n 1\right)-1=0.$$ This is why it was important to specify precisely what you mean by "compounding" in the question. $\endgroup$ – whuber Aug 11 '15 at 16:38
  • $\begingroup$ Understood. My apologies. I added clarifying notation. $\endgroup$ – Ursus Frost Aug 11 '15 at 20:35

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