Does leverage affect the Sharpe ratio? If my Sharpe is 2 at no leverage, does it change, fall by half say, at a different leverage?


4 Answers 4


Sharpe ratio is defined as $\frac{(x - r)}{\sigma}$ where $x$ is return, $r$ is the risk free rate and $\sigma$ is volatility. Now levering up $n$ times multiplies both the return and volatility by $n$. But shouldn't the ratio change since $r$ stays the same? Ah, but remember, leverage isn't free. You have to fund leverage, and that cuts out of your return. So if you can fund the leverage at the risk free rate, then you should subtract from your return $r(n-1)$. And thus, this is as if we were multiplying both the numerator and denominator by $n$. The Sharpe thus stays the same.

  • 1
    $\begingroup$ Wonderfully explained. +1 $\endgroup$
    – Jagra
    Commented Jun 17, 2015 at 23:07
  • $\begingroup$ Yup, I deleted my comment, you are correct. $\endgroup$ Commented Jun 17, 2015 at 23:37

The textbook academic answer is that Sharpe ratio is not impacted by leverage as explained by other answers.

However, reality tells a different tale entirely: Imagine you lever up your investments by such amount that your future performance will critically hinge on the following conditions:

  • That those who extended credit to you will not re-call their funding at any time, something which they are generally fully entitled to

  • That in case that your broker extended credit to you, it will not impose additional margin requirements or modity margin requirements that you will be subjected to.

  • That the assets you invest in will not exhibit staggering return volatility. If that is the case such as during the flash crash, during the financial crisis, or when the Swiss National Bank removed its Euro-Swissie peg, you will be fully exposed to forced liquidation in case of over-leverage.

There are couple other points but the above I deem the core issues in this question.

So, what will happen when your assets's return volatility rises and you are heavily leveraged? You will receive either a margin call and need to inject additional funding, or your funding supplier will pull his/her funding entirely or will modify leverage ratios. In some of those cases you will end up with forced liquidation at the worst possible time which in turn will cause a negative impact on your portfolio returns. As sharpe ratios measure your risk adjusted returns in fact your portfolio returns will be negatively impacted while your return variations will increase, causing a stark reduction in the sharpe ratio measure.

In summary, in practice I argue that nothing could be further from the truth that your leverage ratio both identically scales your returns and return volatility.

  • $\begingroup$ Using leverage, may require us to use tighter SL orders, correct? And this will change both expected return and standard deviation. So I think in practice, leveraged algorithm's sharpe ratio has no obvious relation with the unleveraged version. $\endgroup$
    – Dandelion
    Commented Dec 18, 2022 at 7:38

Generally no. Sharpe ratio should vary linearly.

Use leverage: the return increases, but so does volatility. De-lever" the return decreases but, so does volatility.

Sharpe ratio


No. Simple way of thinking of it from a statistical approach: Sharpe = (TR - r) / STDEV

TR = n(x bar), where x bar is the mean arithmetic return. E(cX) = cE(x) is a property of the expectation operator. Proof for the discrete case: express expectation in terms of summation, for which a constant can be pulled out.

Var(cX) = c^2Var(X) ==> STDEV(cX) = cSTDEV(X). Proof is slightly more complex so I won't include it, but it is basically done by relating variance to the first 2 moments of X.

Since the risk free rate is a minor factor, Sharpe ratio can be approximately expressed as: nE(X) / STDEV(X). Leveraging just multiplies each value of the random variable X by a constant, which can be pulled out and cancelled from the numerator and denominator. Thus Sharpe ratio is independent of leverage.


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