# Tag Info

3

Perhaps this works : http://en.wikipedia.org/wiki/Geometric_standard_deviation In particular, see under "Derivation"

1

For a random variable $\xi$, the variance is defined by $$mean\Big(\big(\xi -mean (\xi)\big)^2\Big).$$ Then the geometric variance should be defined by $$\prod_{i=1}^n\Bigg(1+ \bigg[x_i-\prod_{j=1}^n(1+x_j)^{1/n}\,\bigg]^2\, \Bigg)^{1/n}.$$ Addendum ---- The definition given in the link below is only a way of thinking. However, it does not provide a ...

4

Based on Ito's isometry, \begin{align*} E_t (r^2_{t+1}) &= E_t \bigg(\int_t^{t+1} \sigma_s dW_s \int_t^{t+1} \sigma_s dW_s\bigg)\\ &= E_t \bigg(\int_t^{t+1} \sigma_{\tau}^2 \,d\tau\bigg) \\ &= E_t\bigg(\int_0^1 \sigma_{\tau+t}^2 \,d\tau\bigg) \\ &=\int_0^1 E_t\big(\sigma_{\tau+t}^2\big) \,d\tau. \end{align*} The identity \begin{align*} E_t ...

0

The only correct way is using log returns. To keep everything consistent, take a arithmetic mean of log returns. Then calculate it net of the risk free (how do you subtract properly using geometric returns?). Then divide (how do you do this properly using geometric returns?) by the standard deviation (how would you calculate this properly with geometric ...

0

If you wanted to see the following (price $S_t$, log return $r$, simple return $R$) then $$r = \log(S_{t+1}) - \log(S_t) = \log(S_{t+1}/S_{t}),$$ and $$R = S_{t+1}/S_{t}-1,$$ thus $$R = \exp(r)-1$$ and $$r = \log(1+R).$$ Was this the question?

Top 50 recent answers are included