Bounding Annual Loss Probability from a Strategy’s Sharpe Ratio

Summary

The document examines what an annualized Sharpe ratio can imply about the chance of a losing year. It standardizes annual PnL by its mean and standard deviation, then uses a symmetry assumption and Chebyshev’s inequality to derive an upper bound. For the example Sharpe ratio of 2, the questioner obtains a bound below 12.5% under the stated assumptions.

A reply gives a much smaller probability, about 2.275%, when returns are assumed independent and normally distributed and the risk-free rate is zero. That figure is a model-based estimate, not a distribution-free guarantee. The exchange illustrates how loss probabilities depend on assumptions about return shape and independence; a Sharpe ratio alone does not determine the probability of a negative year. The initial argument also relies on symmetry, while the normal estimate requires a stronger distributional model.

Key ideas

• A Sharpe ratio relates expected excess return to return variability but does not uniquely determine loss probability.
• The example uses symmetry and Chebyshev’s inequality to derive an upper bound below 12.5%.
• Under independent normal returns and a zero risk-free rate, the reply estimates a loss probability of about 2.275%.
• The normal estimate is conditional on its assumptions and should not be treated as a universal bound.

Tags

# What can we say about the probability a strategy losing money in a year if it has an annualized Sharpe of say 2?


# What can we say about the probability a strategy losing money in a year if it has an annualized Sharpe of say 2?












If we imposed the restriction that the strategy is not skewed, then using Chebyshev's Inequality I can show that the probability of it losing money in a year is less than 12.5%.

Let $X$ be the yearly PnL of a strategy. Let $\mathbf{E}(X) = \mu$ and $\mathbf{Var}(X)=\sigma$. We know that $\frac{\mu}{\sigma} = 2$.

Then, $\mathbf{P}(X < 0) = \mathbf{P}(\frac{X - \mu}{\sigma} < -2) = \mathbf{P}(Z < -2) = 0.5\mathbf{P}(|Z| > 2)$. The last equality makes use of the fact that $X$ is not skewed. Then, by Chebyshev's Inequality, $\mathbf{P}(|Z| > 2) < \frac{1}{2^2}$. Therefore, $\mathbf{P}(X < 0) < 0.125$.

Is there a tighter bound to this problem?

## Answer by Newquant (score -1)

https://quant.stackexchange.com/a/73956

Sure. With 0 rfr and assuming independent and normally distributed returns that’s equivalent to taking N(-2) which is ≈ 2.275%.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.