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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.

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

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