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

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.

• Much thanks, folks. I think I understand why my reasoning is flawed. May 12, 2016 at 21:25
• Did you also understand the reasoning behind the two answers ? May 12, 2016 at 23:04

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 $\large{\sqrt[n]{E(x)}-1=\sqrt[n]{1.25^n}}-1\normalsize{=1.25-1=0.25}$.

• A complete outsider to the field asks: What is the significance of taking the $n$-th root? Is this part of the definition of “arithmetic mean return”? May 12, 2016 at 21:01
• Not the arithmetic mean. It is the geometric mean. It is the average growth factor. If something have increased from 1 up to A in n periods then the average growth factor is $\sqrt[n]{A}$. And the average growth rate is $\sqrt[n]{A}-1$ May 12, 2016 at 21:07
• Thanks. I did the same calculation as you, up to that. But seems to me that since the situation and the setup are completely time-independent, it’s enough to make the calculation for just one period, which of course is $5/4$, as your final result comes out to anyway. May 13, 2016 at 3:22
• That´s right. But for me (and the OP) it is wasn´t not obvious from the beginning that the expected value is equal for each step. Hindsight is always easier than foresight. May 13, 2016 at 3:44

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

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