b******k
发帖数: 58 | 1 A) take a data set of any time-series variable Xt. Xt could be
normally
distributed or not (just make sure all the elements in your set are
positive) - poisson, power law, pick a distribution.
(1) estimate the annualized standard deviation of Xt (call it Sx).
This
is our estimate of the "normal vol."
(2) take the logs (Xi+1/Xi), yielding a "new" data set Zt [with one
less
element], and estimate the annualized standard deviation of Zt (Sz
by
the same convention) - the "lognormal vol."
- what is | c**********e
发帖数: 2007 | 2 why Z_t is one less element than X_t. As Z_t=log(X_t+1/X_t), aren't
the number of elements the same?
Does "the annualized standard deviation of Xt" mean "the annualized standard
deviation of return of Xt"? Otherwise, how is it annualized?
Thanks. | p*******i
发帖数: 309 | 3 if time interval is a year, it is vol_annulized,
if a day, there is sqrt(days a year)*Vol_daily. Am I right??
niu niu shows me some light!!
And How about B, confused！！！！！！！！！！！！！ | t********n
发帖数: 10 | 4 answer to B (not sure):
if f(x) is concave, {a1 * f(x1) + a2 * f(x2) + ... + an * f(xn)} < f(sum(ai
* xi)), where sum(ai) = 1
f(x) = log(x + 1/x) is concave on x > 1 (looks in this question, x > 1)
so, Zt < log(Xt + 1/Xt) | l********n
发帖数: 5 | 5 For A, I suppose the question should be:
(1) ...
(2) take the log[X(i+1)/Xi], hence yielding one less element inside the new
data set.
Any one got the solution for this question? | b******e
发帖数: 118 | 6 What's position? Is it entry-level or experienced?
【在 b******k 的大作中提到】
: A) take a data set of any time-series variable Xt. Xt could be
: normally
: distributed or not (just make sure all the elements in your set are
: positive) - poisson, power law, pick a distribution.
: (1) estimate the annualized standard deviation of Xt (call it Sx).
: This
: is our estimate of the "normal vol."
: (2) take the logs (Xi+1/Xi), yielding a "new" data set Zt [with one
: less
: element], and estimate the annualized standard deviation of Zt (Sz
| z****i
发帖数: 406 | 7 A) normal_vol is approximately lognormal_vol*mean(Xt) ?
B) If the spot price is given, we can find the implied vol for each K. If
implied-vol is roughly flat across K, then market-implied distribution of
the underlying is lognormal. If implied_vol*K is roughly flat across K, then
the market-implied distribution of the underlying is normal. | a****y
发帖数: 4 | 8 I guess:
(1) calculate covariance matrix of Xt(assume column vector), R=E{Xt*Xt'}, if
the eigenvalues of R are the same or very close (depends on the length of
Xt), then we may say r.v. in Xt are normally distributed; std = sqrt(
eigenvalue) |
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