Deriving a Regression-Based Sharpe Ratio from Mean and Variance
Summary
The document asks how to substitute the mean and variance of a linearly modeled return into the Sharpe ratio. For a model with an intercept, a coefficient times a random factor, and an error term, the mean is the intercept plus the coefficient times the factor’s mean. Under the usual assumptions that the error has zero mean and is uncorrelated with the factor, the variance combines the scaled factor variance with the error variance. Substituting these moments gives the stated ratio.
The response describes the substitution as an application of expectation and variance rules, but its variance explanation is imprecise: a constant intercept contributes no variance, and the covariance between the factor and error must be zero for the displayed variance formula to hold. The document also does not discuss whether returns are excess returns, nor annualization or sampling frequency, all of which matter when interpreting a Sharpe ratio.
Key ideas
- A linear return model’s expected value follows from linearity of expectation.
- A constant intercept does not affect the variance of the modeled return.
- The displayed variance formula assumes the residual is uncorrelated with the predictor.
- A Sharpe ratio’s practical interpretation also depends on the return definition and measurement frequency.
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# step by step calculation of the sharpe ratio
# step by step calculation of the sharpe ratio
I am trying to calculate the Sharpe ratio. Suppose I have:
$$ x_t = \alpha + \beta y_{t} + \epsilon_{t}$$
$$E[x_{t}] = \alpha + \beta E[y_{t}]$$
$$var[x_{t}] = \beta^2var[y_t] + \sigma^2$$
The Sharpe ratio is:
$$\dfrac{E[x_{t}]}{\sqrt{var[x_{t}]}}$$
I am trying to go from the above Sharpe ratio to the following output (but step-by-step showing everything I am doing):
$$\dfrac{\alpha + \beta E[y_{t}]}{\sqrt{\beta^2 var[y_t] + \sigma^2}}$$
Does anybody have some step-by-step solution? Or if there is somewhere online where I can see step-by-step how its calculated.
## Answer by develarist (score 2)
https://quant.stackexchange.com/a/49744
The formulas given already explain the substitution you're wondering about, based on standard statistical laws surrounding the distribution of random variables, often focused on the first and second moments of that distribution: the mean $E(\cdot)$ and variance $Var(\cdot)$. You can find these rules being followed in the derivations of many economic models that are probabilistic.
Taking the expected value of random variable $x_t$ as $E(x_t)$ is simply due to the regression coefficient $\beta$ coming out as a scalar coefficient since it doesn't have an expected value being a scalar, whereas random variable $y_t$ is not deterministic and does have an expected value, therefore $E(y_t)$ is there. Intercept term $\alpha$ is intact when taking expectations, but not in the variance shown next.
The rules for taking the variance of a random variable consisting of a constant and random component are as follows, which has a longer derivation that can be found in the appendix of many mainstream econometrics textbooks:
- Variance of the scalar $\beta$ is the same scalar but squared, thus $Var(\beta)=\beta^2$ comes out in front of the random variable it is connected to
- Variance of a random variable is just as it is shown, $Var(y_t)$
- The $\sigma^2$ always comes out at the end as a residue of sorts from taking the variance of a random variable and intercept term $\alpha$ (having alot to do with the covariance of the model) again fully derived in the back of many econometrics textbooks.
Since the Sharpe ratio's numerator is the expected value of random variable $x_t$ and the denominator is the variance of the same random variable $x_t$ consisting of a deterministic and stochastic term, the $E(x_t)$ and $Var(x_t)$ are just put into place.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.