Why Positive Sharpe Can Coexist With Long-Run Portfolio Losses
Summary
The document explains why a positive Sharpe ratio calculated from arithmetic returns does not guarantee that compounded wealth will rise. In the example, the average daily return is positive, yet multiplying the daily growth factors produces a loss over the observed period. The accepted explanation distinguishes the arithmetic expected return from the expected logarithmic return: the former can be positive while the latter is negative.
For repeated returns, the average log return determines the long-run growth rate of wealth. The law of large numbers implies that, under the example’s assumptions, average log returns converge to their expectation; if that expectation is negative, wealth tends to decline over time. The example illustrates the distinction but does not establish that every positive-Sharpe strategy loses money. It also does not address estimation uncertainty, serial dependence, changing return distributions, or the effects of risk-free rates and trading costs.
Key ideas
- A positive arithmetic average return can coexist with a negative compounded outcome.
- The Sharpe ratio based on arithmetic returns does not directly measure long-run compounded growth.
- Expected log return determines the long-run growth rate under repeated returns.
- A negative expected log return implies declining wealth in the long run under the stated assumptions.
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# A positive Sharpe ratio when portfolio loses money, can that happen or bug in my code?
# A positive Sharpe ratio when portfolio loses money, can that happen or bug in my code?
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:
- Subtract $1.0$ to get percentage: $X=[0.15, 0.2, -0.3]$
- Calculate sample expectance: $\hat{E}=(0.15+0.2-0.3)/3=0.0166$
- Calculate sample standard deviation: $\hat{\sigma}=0.224$
- 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 by Matthew Gunn (score 3, accepted)
https://quant.stackexchange.com/a/43930
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: $$ \frac{1}{T} \sum_{t=1}^T \log (1 + R_t) \xrightarrow[]{\text{a.s.}} \operatorname{E}[\log (1 + R)]$$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.