# No-arbitrage and the sharpe ratio?

I'm reading a paper and it says that in a no-arbitrage market the sharpe ratio is the same for all bonds. I'm guessing that a difference in two bonds sharpe ratios would open the possibility of arbitrage, but why is that?

Regards

• Link to the paper? Does the author make additional assumptions? This isn't true in general. – Matthew Gunn Apr 25 at 13:41
• Maybe it is "assuming a one-factor model of interest rates the Sharpe Ratios are all the same"... or some kind of other simplifying assumption. – noob2 Apr 25 at 19:55
• Hi, it is indeed assuming a one-factor model of interest rates. I found the answer (I think) myself. Because bonds are subject to the same source of uncertainty in a one-factor model (the brownian motion in for example the Vasicek model) then if the market-price of risk is higher for one bond than the other it would be possible to short-sell the one bond and buy the other with the higher market price of risk. The bonds will depreciate and appreciate the same and it would thus be possible to deliver the short-selled bond at a price lower than the facev (z-coupon) you get from the first bond. – MC_nonmaster Apr 26 at 14:58

Consider a market with $$d$$ assets $$(S^1, \dots, S^d)$$ which is free ob arbitrage. Let $$B$$ denote the numeraire. According to the fundamental theorem of asset pricing there exists a martingale measure $$\Bbb Q$$. In particular: $$\Bbb E_{\Bbb Q} \Bigg[\frac{S_{1}^j - S_0^j} {S_0^j} \Biggr] = \frac{1}{S_0^j}\Bbb E_{\Bbb Q} \bigl[S_{1}^j\bigr] - 1 = \frac{B_1}{S_0^j}\frac{S_0^j}{B_0} - 1 = \frac{B_1 - B_0}{B_0}, \quad \ \text{for all} \ j \in \{1,\dots, d\}.$$
• Unless otherwise stated, a Sharpe ratio uses the expectation and standard deviation under the true measure $\mathbb{P}$, not the risk-neutral pricing measure $\mathbb{Q}$. Yes, under $\mathbb{Q}$ the expected return of every asset is the risk-free rate and so the Sharpe ratio would be zero for every asset, but I rather doubt that's what the OP (or the paper) is trying to refer to? – Matthew Gunn Apr 25 at 13:40
• Very true, I did not have this in mind when writing the answer. However under $\Bbb P$ this would not hold in general. – Cettt Apr 25 at 14:58