# Why does the Sharpe ratio not change when the strategy is leveraged?

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?

• @Eric Your insight would be very helpful – labrynth Jan 7 '16 at 20:09
• should this be a comment to another question? – chollida Jan 7 '16 at 20:19
• Yes it is, but the editor asked me to phrase it as a new question. – labrynth Jan 7 '16 at 20:32
• @chollida I have re-framed my question. Hope you retract the negative score assigned to my question. – labrynth Jan 7 '16 at 21:19
• I haven't downvoted you. – chollida Jan 7 '16 at 22:02

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 $n\sigma$

Hence:

"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".