# The Sharpe ratio and loss probability

Assume we have a portfolio with Sharpe ratio of $S_r = 1$. What is the probability of the portfolio losing value over a 4 year time horizon?

We have to make some assumptions before we can find an answer. A common, albeit unrealistic assumption is that the portfolio returns are normally distributed. Denote the returns of the portfolio over the years by $X_1, X_2, X_3, X_4$ and assume that returns are independent.

To answer the question let’s first reduce the problem to one period. The Sharpe ratio is defined as the risk adjusted return over one year

$\displaystyle S_r = \frac{E[X_1]}{\sqrt{\mathrm{Var}(X_1)}}.$

Assuming that $X_1$ is Gaussian with mean $\mu = E[X_1]$ and $\sigma = \sqrt{\mathrm{Var}(X_1)}$ we are interested in the probability of a loss which can be rewritten as

$\displaystyle \mathbb{P}(X_1 < 0) = \mathbb{P}\left(\frac{X_1-\mu}{\sigma} < -\frac{\mu}{\sigma} \right).$

This is helpful! The random variable $Z_1 = (X_1-\mu)/\sigma$ is a standard normal distribution, i.e. $Z_1 \sim \mathcal{N}(0, 1)$ and its cumulative distribution function can be easily computed using the statistical software of our choice. We now write the above problem

$\displaystyle \mathbb{P}(X_1 < 0) = \mathbb{P}\left(Z_1 < -\frac{\mu}{\sigma} \right) = \mathbb{P}\left(Z_1 < -1 \right).$

Using, for example, the statistical software R this can be calculated as pnorm(-1)=0.159 (up to three digits).

Now for the answer for 4 year period. The new random variable of interest is given by the portfolio $P = X_1 + X_2 + X_3 + X_4$. We need to calculate the expectation and variance of $P$, subsequently we can proceed as before

$\displaystyle \begin{array}{ll} \mu_P & = E[X_1] + E[X_2] + E[X_3] + E[X_4] = 4 \mu, \\ \sigma_P^2 & = Var[X_1] + Var[X_2] + Var[X_3] + Var[X_4] = 4 \sigma^2 \end{array}$

that is $\sigma_P = \sqrt{4 \sigma^2} = 2 \sigma$. The 4 year loss probability can be calculated as $\displaystyle \mathbb{P}\left(\frac{P - \mu_P}{\sigma_P} < - \frac{4 \mu}{2\sigma} \right) = \mathbb{P}\left(\frac{P - \mu_P}{\sigma_P} < - 2\frac{\mu}{\sigma} \right) = \mathbb{P}\left(\frac{P - \mu_P}{\sigma_P} < - 2 \right)$

because we know $\mu/\sigma = 1$! The final answer is thus pnorm(-2) = 0.0228.