# Relationship between risk and return for GBM and riskless bond

Suppose we have $$S$$, a stock following geometric Brownian motion ($$dS_t = S_t (\mu dt + \sigma dZ_t)$$ for $$Z =$$ Brownian motion) and $$B$$, a zero coupon bond with rate $$r$$, i.e. $$dB_t = rB_t dt$$.

In trying to explain/derive the Sharpe ratio using these two assets ($$= (\mu - r)/\sigma$$), a set of lecture notes that I'm reading states that if we invest some proportion $$w \in [0,1]$$ in $$S$$, then the expected return is $$w\mu + (1-w) r$$ and the volatility is $$w \sigma$$ and hence any security with this volatility should give the same expected return. i.e. Any asset with volatility $$w \sigma$$ must give return excess of $$r$$ of $$w(\mu - r)$$ and thus

$$\frac{\text{Excess return}}{\text{Volatility}} = \frac{w(\mu-r)}{w \sigma} = \frac{\mu -r}{\sigma}$$

This confuses me, because the expected return of the stock is actually $$\exp(\mu t)$$ and of the bond is $$\exp(rt)$$. What is the rationale here? I've attached the slide I'm referring to in particular. Is the argument supposed to be purely heuristic over a short period? I've attached the slide I'm interested in below.

• This is based on the instantaneous return. $exp(\mu t)$ is the expected gross return between $0$ and $t$. – fesman Oct 31 '20 at 16:05