It has been correctly stated that the Sharpe ratio of a strategy does not change when it is leveraged. I understand Eric's point that leveraging by $n$ multiplies both the return $x$ and volatility $\sigma$ by $n$. I also understand that we fund the leverage at risk free rate, and hence subtract $r(n-1)$ from the return. However, I cannot understand why this would not change the Sharpe ratio (sr) since $$sr = \frac{nx - (n-1)r}{n\sigma} = \frac{n(x-r)}{n\sigma} + \frac{r}{n\sigma} = \frac{(x-r)}{\sigma} + \frac{r}{n\sigma}$$ The sr is different, where am I going wrong?

  • $\begingroup$ @Eric Your insight would be very helpful $\endgroup$
    – labrynth
    Commented Jan 7, 2016 at 20:09
  • 1
    $\begingroup$ should this be a comment to another question? $\endgroup$
    – chollida
    Commented Jan 7, 2016 at 20:19
  • $\begingroup$ Yes it is, but the editor asked me to phrase it as a new question. $\endgroup$
    – labrynth
    Commented Jan 7, 2016 at 20:32
  • $\begingroup$ @chollida I have re-framed my question. Hope you retract the negative score assigned to my question. $\endgroup$
    – labrynth
    Commented Jan 7, 2016 at 21:19
  • $\begingroup$ I haven't downvoted you. $\endgroup$
    – chollida
    Commented Jan 7, 2016 at 22:02

1 Answer 1


Sharpe ratio = $\frac{r_p - r_f}{\sigma_p}$, where:

  • $r_p$ is the expected portfolio return
  • $\sigma_p$ is the portfolio's standard deviation
  • $r_f$ is the risk free rate.

When you leverage '$n$' times:

  1. The leveraged portfolio return is $n r_p - (n-1) r_f$ (subtracting the cost of borrowing the money)
  2. The standard deviation increases to $n\sigma$


"Leveraged Sharpe ratio" = $\frac{n r_p - (n-1) r_f - r_f}{n\sigma}=\frac{n(r_p-r_f)}{n\sigma}$ = $\frac{r_p - r_f}{\sigma_p}$ = Sharpe ratio.

Given any risky asset, one can generate an infinite expected return at the cost of added risk (by leveraging the investment). The Sharpe ratio mitigates "false advertising".


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