A positive Sharpe ratio when portfolio loses money, can that happen or bug in my code?

Question

I'm having a trouble calculating (annualized from daily performance) Sharpe ratio, even though I've read some related posts here. Say I have a daily performance, for example: $$[1.15, 1.2, 0.7]$$ means that after 3 trading days my cumulative (for, say 1 dollar investment) wealth is: cum_wealth=$1*1.15*1.2*0.7=0.966$ which states that I actually lost money. When I calculate the Sharpe ratio I follow these steps:

1. Subtract $1.0$ to get percentage: $X=[0.15, 0.2, -0.3]$
2. Calculate sample expectance: $\hat{E}=(0.15+0.2-0.3)/3=0.0166$
3. Calculate sample standard deviation: $\hat{\sigma}=0.224$
4. Assuming the risk-free rate is 0%. Divide and annualize: Sharpe=$E/\sigma * \sqrt{252}=1.176$

I've constructed the example above specifically to show that I get positive Sharpe ratio when actually the portfolio loses money, which is counter-logic (if the math is correct)... What am I missing? Regards

Answer 1

I'll expand on @AlexC's excellent comment. Let $R$ denote a random variable that takes the values .15, .2, and -.3 with equal probability.

We can see that both:

$$\operatorname{E}[R] > 0 \quad \quad \operatorname{E}[\log(1+R)] < 0$$

While the expected return $\operatorname{E}[R]$ is indeed positive, the expected log return $\operatorname{E}[\log(1+R)]$ is negative which leads the value process $v_T = \prod_{t=1}^T (1 + R_t$) to decline almost surely in the long run.

By the law of large numbers, the logarithm of the geometric mean return converges (as $T \rightarrow \infty$) to the expectation of the log return. The geometric mean return is:

$$ \left( \prod_{t=1}^T (1 + R_t) \right)^{\frac{1}{T}}$$

Take the log to obtain:

$$ \frac{1}{T} \sum_{t=1}^T \log (1 + R_t)$$

By the law of large numbers: $$ \frac{1}{T} \sum_{t=1}^T \log (1 + R_t) \xrightarrow[]{\text{a.s.}} \operatorname{E}[\log (1 + R)]$$