да при p=1 СБ. И что из этого следует что процесс стационарный? Я писал что нестационарный. FOXXХi просил дать определение СБ там оно есть. В чем проблема то? :)
I didn't ask you to give me a definition of SB, I asked you this:
But the cumulative sum of these SB increments will also give a normal distribution.
no. The normal will be the increment of this function over any time interval, since the sum of NRs is also NR. That is, F(tx)-F(ty) is normally distributed (which is a property of integrated processes), but not the SB itself as a function of time F(t) - its variance will increase with time. HP has a finite and fixed variance that is independent of time. The variance F(t)>F(t-1), unlike the variance of the first difference - it is constant and independent of time
J.L. DOUBE VERIOUS PROCESSES Translation from English Р. L. DOERUSHIN and A. M. YAGLOMA Edited by А. EDITED BY A. M. YAGLOMAH EDITOR'S NOTE STATE OF FOREIGN LITERATURE Moscow - (956
да при p=1 СБ. И что из этого следует что процесс стационарный? Я писал что нестационарный. FOXXХi просил дать определение СБ там оно есть. В чем проблема то? :)
I didn't ask you to give me a definition of SB, I asked you this:Still answer the question : "What is the distribution of the SB process? "Just leave the difference alone.
P.S. смысл есть все же говорить о приращениях, т.к. автор сформулировал задачу именно через приращения
But the cumulative sum of these SB increments will also give a normal distribution.
But the cumulative sum of these SB increments will also give a normal distribution.
Good for you, you wrote: "since the sum of NRs is also NRs." Hence:But the cumulative sum of these SB increments will also give a normal distribution.
VERIOUS
PROCESSES
Translation from English
Р. L. DOERUSHIN and A. M. YAGLOMA
Edited by
А. EDITED BY A. M. YAGLOMAH
EDITOR'S NOTE
STATE OF FOREIGN LITERATURE
Moscow - (956
А вот по усложненной задаче "с победителями" есть тоже закладочка.. ;)
Can I ask for a link to Wentzel's full text?