Tag Info

New answers tagged


It is really simple and probably not a question for this forum. You just need: $\alpha * .08 + (1-\alpha) * .02 = .12$. Solve for alpha and then check the standard deviation that should be .25.


This will depend on the definition of "return on the long run". If we define the annualized return on the long run by $\frac{1}{T}\ln \frac{S_T}{S_0}$ for a certain time $T$ in the future, then \begin{align*} E\left( \frac{1}{T}\ln \frac{S_T}{S_0} \right) = \mu-\frac{1}{2}\sigma^2, \end{align*} as claimed. Note that $\mu$ is the instant, or instantaneous, ...


Trying to shed some light here: What we also see using this here, is that if returns are log-normally distributed, ie. $$ 1 + r = \exp(\mu + \sigma Z), $$ with $Z$ standard-normal, then $$ E[1+r] = \exp(\mu + \frac 12 \sigma^2) $$ holds. But the geometric mean $GM$ is given by $\exp(\mu)$ and we have $$ \log(GM) = \mu = \log(E[1+r]) - \sigma^2 /2 $$ and ...

Top 50 recent answers are included