Should the Sharpe ratio of a portfolio change when it is leveraged?

I am trying to understand why the Sharpe ratio changes (increases) when I simulate leveraging my portfolio by multiplying all the time series of daily returns by a leverage factor (e.g. 5).

I understand that the Sharpe ratio should not change when a portfolio is leveraged (other things being equal).

However I find that the annualized Sharpe ratio (calculated geometrically with formula: return = (product of 1+ daily returns ^ (262/number of returns))-1, stdev = stdev(returns)*(sqrt(262)) deos increase (e.g. from 3.1 to 4.3).

However, the daily Sharpe ratio (calculated as the arithmetic average of returns divided by standard deviation) remains identical (mathematically identical).

I am assuming a risk free rate of zero, so the Sharpe is simply return divided by stdev.

I'm sure it's something obvious, but can anyone explain why?

• It depends on how you calculate the returns. If you calculate as a % of the total portfolio value, then when you leverage the trade you are getting larger returns on the same amount of capital thus Sharpe ratio is increasing. – Artem Korol Dec 9 '16 at 12:26
• @ArtemKorol that is incorrect. The sharp ratio takes into account volatility which increases as well when you leverage. – SRKX Dec 9 '16 at 12:44
• You need to read the discussion in this topic: quant.stackexchange.com/questions/3607/… – AK88 Dec 9 '16 at 14:05

Let us assume:

• a constant risk-free rate $r$
• a risky asset with returns $X$
• with expected value $\mathbb{E}(X)=\mu_X$
• and variance $\text{Var}(X)=\sigma_X^2$
• a portfolio investing $w$ in the risky asset and $(1-w)$ in the risk-free asset

Then you can compute the expected value of the portfolio:

$$\mu_P = \mathbb{E}(P) = w \mu_X + (1-w)r$$

and variance

$$\sigma_P^2 = \text{Var}\left[wX + (1-w)r\right] = w^2\sigma_X^2$$

If you use the definition of the Sharpe ratio, you have:

$$\text{Sharpe}(P) = \frac{\mu_P - r}{\sigma_P} = \frac{w (\mu_X - r)}{w \sigma_X} = \frac{\mu_X - r}{\sigma_X}$$

Clearly, the weight $w$ gets simplified and disappears in the Sharpe's computation which means that the Sharpe ratio stays the same $\forall w$.

However, this assumes that the mean is estimated as:

$$\hat{\mu}(X) = \frac{1}{n}\sum_{i=1}^n r_{X,i}$$

and in particular:

$$\hat{\mu}(wX) = \frac{1}{n}\sum_{i=1}^n wr_{X,i} = w \frac{1}{n}\sum_{i=1}^n r_{X,i} = w \hat{\mu}_X$$

which is fine.

However, what you are doing is

$$\hat{\mu}(X) = \left(\prod_{i=1}^n 1 + r_{X,i}\right)^{\frac{1}{N}}$$

and in particular

$$\hat{\mu}(wX) = \left(\prod_{i=1}^n 1 + wr_{X,i}\right)^{\frac{1}{N}} \neq w\hat{\mu}(X)$$

Hence, you are not computing the expected value strictly-speaking, so you're not really computing a Sharpe ratio. Furthermore, your version loses the property of being independent of $w$.

A lot of people use the same approach in the industry, but it's fair to say that this is not the exact definition of the Sharpe.

• Comprehensive, thanks. I (think I) get it. You can leverage up an arithmetic mean (and stdev) by multiplying by a factor and the sharpe stays constant because the weight (leverage factor in this case) cancels out. However, for a geometric mean, this does not work - geometric mean x leverage factor is not the same as geometric mean calculated by leveraging all the returns first. Thanks. – Will Dec 9 '16 at 18:32
• @Will if your question is answered, please mark the answer as accepted – SRKX Dec 10 '16 at 5:08