3
$\begingroup$

I came across the following question and am trying to understand it better. I was hoping you could share your intuitions.

For a given stock, you are certain that for the next 100 days, it will move either 10% up or 10% down each day with equal probability. You can invest now, but if you chose to do so, you have to hold it for the entire 100 days. Would you do so?

In general, it can easily be shown that the expected value at any point in this process is equal to the initial value of the stock. However, it can also be shown that the expected geometric growth rate is negative.

Ignoring the time value of money etc., the expected value of the stock suggests one should be indifferent to making the trade. However, the negative geometric growth rate suggests one would lose money in the long term.

What is the correct way to interpret these two values? I feel like I am misunderstanding what they are telling us and there is a simple resolution to the 'paradox'. Would you make the trade and how would your answer change if the stock moved up by 11% instead of 10%?

$\endgroup$
1
  • 1
    $\begingroup$ Expected arithmetic return is zero so and expected stock price constant. Expected log return / growth rate is negative. A risk averse agent would not take the bet nor would a growth rate / Kelly optimizer. $\endgroup$
    – fes
    Commented Jul 5, 2021 at 16:46

1 Answer 1

1
$\begingroup$

The number of up moves of the stock $S$ after 100 days follows binomial distribution. To calculate expected value of the stock we have to weight values by probability mass function. After 100 days we have $k$ up moves of $1+10\%$ and $100-k$ down moves of $1-10\%$ i.e. the value of the stock is $S_0*(1+10\%)^k*(1-10\%)^{(100-k)}$ with probability ${100}\choose{k}$$*p^k*p^{(100-k)}$ where $p=\frac{1}{2}$. The expected value at T=100 is: $$E[S_{100}]= \sum_{k=0}^{100} S_0*(1+10\%)^k*(1-10\%)^{(n-k)}{{100}\choose{k}}*p^k*p^{(n-k)}$$ which is equal to $S_0$ but the expected geometric return is: $$E[Return]= \sum_{k=0}^{100} (1+10\%)^k*(1-10\%)^{(n-k)}{{100}\choose{k}}*p^k*p^{(n-k)} - 1$$ which when calculated is equal to 0. Therefore mean price of the stock is equal to initial value and the mean geometric return is 0.

EDIT: If we consider mean geometric return i.e. return of the form $\sqrt[n]{x_1*x_2*...*x_n} -1$ then we have: $$E[Mean GeomReturn]= \sum_{k=0}^{100} [\sqrt[n]{(1+10\%)^k*(1-10\%)^{(n-k)}} -1]{{100}\choose{k}}*p^k*p^{(n-k)}$$ Then because mean geometric return is concave function we have: $$E[Mean GeomReturn] \leq E[Return]$$ It is the same for logarithmic returns (concave) i.e. $$E[MeanLogReturn]= \sum_{k=0}^{100} Log{(1+10\%)^k*(1-10\%)^{(n-k)}} ]{{100}\choose{k}}*p^k*p^{(n-k)}$$ we have: $$E[MeanLogReturn] \leq E[Return]$$

In fact for the stock process you described, the expected log return is negative (after 100 periods equals to -50%). You should be really careful with that, as you mentioned expected stock price is equal to its initial value - and that's true, but both expected log return and mean geometric return are negative. This is due to the fact that logarithm flattens high returns and inflates negative returns. For example if the stock is worth 100, next period 199 (geom return of 99%) or 1 (geom return of -99%), then what are the log returns? +68% and -460% respectively. What is expected log return? -390%. It doesn't make much sense to say that on average you earn -390% log return. But if you would calculate average stock prices next period $(0.5*199 + 0.5*1)$ and calculate log return on average prices you would get a number that is easily interpretable. $$log((0.5*199 + 0.5*1)/100)=0%$$ Therefore the order of calculations is crucial here. For ease of intepretation I would advise to firstly calculate expected Stock price $E[S_{100}]$ and then calculate logarithm return on that. You should get 0%.

EDIT 2: There is nice interpretation of expected log returns (rather than log of expected return) (thanks @fesman) - if they expected log returns are negative that means in the long run there is higher probability of losing money than earning money. In our example we have one period return of $(1+X)$ where $X$ can take value of +10% and -10% with equal probability. After N periods we have: $$(1+x_1)*(1+x_2)*...*(1+x_n)$$ the log return is $$ln((1+x_1)*...*(1+x_N))=ln(1+x_1)+ln(1+x_2)+...(1+x_N)$$ If $N$ is sufficiently large, then this sum can be approximated by normal distribution (due to CLT). The mean of $ln(1+X)$ is $$\hat x=E[ln(1+X)]$$ the variance is: $$\sigma^2=E[(ln(1+X))^2]-E[(ln(1+X))]^2$$ Therefore the sum is approximately normally distributed with mean $N*\hat x$ and variance $N*\sigma^2$. This can be represented as: $$N\hat x + \sqrt N*\sigma*Z$$ where $Z$ is standard normal rv. Let's calculate the probability that we actually lose money on that investment: $$P(N\hat x + \sqrt N*\sigma*Z < 0)=P(Z<\frac{-N*\hat x}{{\sqrt N}*\sigma})=\Phi(\frac{-N*\hat x}{{\sqrt N}*\sigma})$$ where $\Phi$ is CDF of standard normal distribution. After substituting the numbers into equation we get that after $N=100$ periods the probability of losing money is ~69% (high risk of losing money) and increases when N increases. Therefore risk averse investor wouldn't take this deal. In order to get probability of 50% it is easy to check that the expected log return has to be zero, not negative.

$\endgroup$
8
  • $\begingroup$ Thanks very much for the clear explanation. I believe your calculation of the geometric return uses the ensemble average. When I said it was negative in the question, I had computed a time average (ignoring that the stock is held for 100 days), which gives 0.5 log(0.9) + 0.5 log(1.1). Would that affect the conclusion at all? $\endgroup$
    – FoxCharles
    Commented Jul 5, 2021 at 16:28
  • $\begingroup$ @FoxCharles: I have edited my post to address your comment. Hope that helps! $\endgroup$
    – emot
    Commented Jul 5, 2021 at 19:57
  • $\begingroup$ Thanks again - answer accepted. I had a feeling this would be explained by either a concave utility function or the effect of applying a time average. I’m still struggling to use the results to decide if the trade is worthwhile or not - especially in the 11% / -10% example where you have positive EV but a negative mean geometric return. I feel like the correct answer is to decline the trade and we are seeing the effect of ‘volatility tax’. Happy to hear your thoughts! $\endgroup$
    – FoxCharles
    Commented Jul 5, 2021 at 20:07
  • $\begingroup$ It is standard to calculate mean log-returns, you should not average before taking the log. Expected log-return has the interpretation of long run growth rate, which here is negative. $\endgroup$
    – fes
    Commented Jul 6, 2021 at 6:26
  • 3
    $\begingroup$ The process in your example is a martingale that tends to zero almost surely i.e. negative growth rate. Understanding this difference is important to learning how stochastic processes work. $\endgroup$
    – fes
    Commented Jul 6, 2021 at 8:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.