# 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. Dec 9, 2016 at 12:26
• @ArtemKorol that is incorrect. The sharp ratio takes into account volatility which increases as well when you leverage.
– SRKX
Dec 9, 2016 at 12:44
• You need to read the discussion in this topic: quant.stackexchange.com/questions/3607/…
– AK88
Dec 9, 2016 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, 2016 at 18:32