o****e
发帖数: 80 | 1 我明白 E[B(t)]=0
为什么 E[B(t)^3]=0?
还有怎么计算出下面的答案
E[ B(t)^ (2k+1) ] = 0
E[ B(t) ^ (2k) ] = (2k)! t^k / (2^k k!)
我是菜鸟，请大家多指教，祝大家早日拿到满意的offer. |
d*j
发帖数: 13780 | 2 by symmetry
use the above result, apply integration by parts |
o****e
发帖数: 80 | 3 thank daj!
B(t)~N(0,t)
int (-inf, inf) B(t)^2k+1=0 by symmetry
int(-inf,inf) B(t)^2k = (2k-1)!!t^k, not the one i copied from previous
post as above.
【在 o****e 的大作中提到】
: 我明白 E[B(t)]=0
: 为什么 E[B(t)^3]=0?
: 还有怎么计算出下面的答案
: E[ B(t)^ (2k+1) ] = 0
: E[ B(t) ^ (2k) ] = (2k)! t^k / (2^k k!)
: 我是菜鸟，请大家多指教，祝大家早日拿到满意的offer.
|
p*****k
发帖数: 318 | 4 ogtree, you could also use the moment generating function,
then take derivatives:
http://en.wikipedia.org/wiki/Normal_distribution#Characteristic_and_moment_generating_functions
also (2k-1)!! = (2k)!/(2^k k!) |
o****e
发帖数: 80 | 5 pcasnik
thank you.
but are u sure (2k-1)!! = (2k)!/(2^k k!), i tired k=4, they are not equal
【在 p*****k 的大作中提到】
: ogtree, you could also use the moment generating function,
: then take derivatives:
: http://en.wikipedia.org/wiki/Normal_distribution#Characteristic_and_moment_generating_functions
: also (2k-1)!! = (2k)!/(2^k k!)
|
f*****s
发帖数: 141 | |
l****u
发帖数: 84 | 7 Use martingale
【在 o****e 的大作中提到】
: 我明白 E[B(t)]=0
: 为什么 E[B(t)^3]=0?
: 还有怎么计算出下面的答案
: E[ B(t)^ (2k+1) ] = 0
: E[ B(t) ^ (2k) ] = (2k)! t^k / (2^k k!)
: 我是菜鸟，请大家多指教，祝大家早日拿到满意的offer.
|
