Estimating Four-Year Loss Probability from a Sharpe Ratio of One

Summary

The note answers how likely a strategy with a zero risk-free rate and a Sharpe ratio of one is to lose money over four years. It assumes annual returns are independent and normally distributed, with positive mean and volatility equal to that mean, as implied by the stated Sharpe ratio. Under those assumptions, the sum of four annual returns is normal, with its mean and standard deviation scaled by the number of years. Standardizing a cumulative return of zero gives a Z-score of minus two, corresponding to an estimated loss probability of about 2.28%.

This is a conditional calculation, not a general prediction from the Sharpe ratio alone. The result depends on normally distributed returns, independence across years, a stable annual mean and volatility, and the interpretation of loss as a nonpositive cumulative sum of returns. Serial correlation, changing risk, non-normal tails, or a different compounding convention could change the probability. The note provides a model-based answer but no empirical strategy data or validation of these assumptions.

Key ideas

• A Sharpe ratio of one with a zero risk-free rate implies mean return equals return volatility under the stated annual convention.
• Assuming independent normal annual returns, four-year cumulative return is also normally distributed.
• The probability of a nonpositive cumulative return is estimated from the standardized threshold at zero.
• The numerical estimate depends on independence, normality, and stable return moments.

Tags

# Sharpe ratio 1 and probability to lose money


# Sharpe ratio 1 and probability to lose money












I came across the following interview problem and I am looking for a possible solution. We have a strategy with risk free return 0 and sharpe ratio 1. What is the probability to lose money over four years ?

## Answer by Charles0349 (score 4, accepted)

https://quant.stackexchange.com/a/78660

I would assume annual returns, because that's usually how the Sharpe ratio is stated. Let's assume returns are normally distributed and independent with mean $\mu > 0$. Then $R_i \sim{\mathcal{N}(\mu, \mu^2)}$ for $i = 1, 2, 3, 4$. Because of my independence assumption, $\sum_{i = 1}^4 R_i \sim{\mathcal{N}(4\mu, 4\mu^2)}$. The $Z$-score corresponding to 0 is $Z = \frac{0 - 4\mu}{2\mu} = -2$. Hence, $$ P\left(\sum_{i = 1}^4 R_i \leq 0\right) = P\left(Z \leq -2\right) \approx 2.28\%. $$

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