If A can either increase by 100% or decrease by 50% with equal probability, what will be the arithmetic mean return over n periods?

Question

This is more of a finance related question but deals with some discrete probability and or combinations. The question goes like this. If you buy stock A, and it has a 50% chance of going up 100% in a period, or 50% chance of going down 50%, what is the arithmetic mean return over n periods?

Some people are saying the answer is 25%. But, I don't see how that works out beyond 1 period. For illustration, consider all possible combination of returns over 1 and 2 periods. One has the following,

one period:

1 , 2

1 , 0.5

two periods:

1 , 2 , 1

1 , 0.5, 1

1 , 2 , 4

1 , 0.5 , 0.25

The total and arithmetic mean returns for each scenario are

one period:

100% 100%

-50% -50%

two periods:

0% 0%

0% 0%

300% 150%

-75% -37.5%

So, averaging the first periods mean arithmetic returns, one gets 25%. But, for two periods, averaging the mean arithmetic returns one gets a bit over 28%. Maybe I'm not understanding the definitions of arithmetic returns, but can anyone here tell me what I'm doing wrong? Hopefully someone has some finance knowledge and understands how to calculate arithmetic mean returns.

Answer 1

The return $X_i$ in a single-period is a binary random variable with probabilities

$$P(X_i = 1) = 1/2, \\ P(X_i = -1/2) = 1/2.$$

The expected return is

$$E(X_i) = \frac{1}{2}(1) + \frac{1}{2}\left(- \frac{1}{2}\right) = \frac{1}{4}.$$

The expected arithmetic mean return over $n$ periods is

$$E\left(\frac1{n}\sum_{i=1}^n X_i\right)= \frac1{n}\sum_{i=1}^n E(X_i)= \frac{1}{4}.$$

Just to be clear, the arithmetic mean considered here and the geometric mean are distinct conventions for calculating average return.

If we have a $100 \%$ return in period 1 and a $-50 \%$ return in period 2, then the arithmetic mean return is

$$R_A = (1 - 0.5)/2 = 0.25 = 25 \%,$$

but the geometric mean return is

$$R_G = [(1+1)(1- 0.5)]^{1/2}-1 = 0 \%.$$

Answer 2

For the first paragraph I use some statements which are related to the binomial distribution. After $n$ periods there are $n \choose x$ ways of loosing $x$ times and winning $n-x$ times. The probability of loosing $x$ times and winning $(n-x)$ times is $0.5^x\cdot 0.5^{n-x}=0.5^n$.

The result after loosing $x$ times and winning $(n-x)$ times in one specific way is $0.5^x\cdot 2^{n-x}$. In total the expected value is

$$E(x)=0.5^n\sum_{x=0}^n {n \choose x}\cdot 0.5^x\cdot 2^{n-x}$$

The binomial theorem says that

$ (a+b)^n=\sum_{x=0}^{n}{{n \choose x}\cdot a^{x}\cdot b^{n-x}} $

with $a=0.5$ an $b=2$ the sum is just $(2+0.5)^n=2.5^n$

$E(x)=0.5^n\cdot 2.5^n=(0.5\cdot 2.5)^n=1.25^n$

The average increase in n periods is $\large{\sqrt[n]{E(x)}-1=\sqrt[n]{1.25^n}}-1\normalsize{=1.25-1=0.25}$.