l*********o
发帖数: 88 | 1 A brownian motion,W(t), given W(t1) and W(t2) (t1
where t could be less t1, between t1 and t2, or greater than t2. |
G********d
发帖数: 10250 | 2 B(t1), B(t1)(t2-t)+B(t2)(t-t1)/t2-t1, B(t2)?
【在 l*********o 的大作中提到】
: A brownian motion,W(t), given W(t1) and W(t2) (t1: where t could be less t1, between t1 and t2, or greater than t2.
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l*********o
发帖数: 88 | 3 That looks right, how would you prove it?
【在 G********d 的大作中提到】
: B(t1), B(t1)(t2-t)+B(t2)(t-t1)/t2-t1, B(t2)?
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c**********e
发帖数: 2007 | 4 The 1st one should be B(t1)*t/t1.
【在 G********d 的大作中提到】
: B(t1), B(t1)(t2-t)+B(t2)(t-t1)/t2-t1, B(t2)?
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G********d
发帖数: 10250 | 5 you assume it starts at zero
【在 c**********e 的大作中提到】
: The 1st one should be B(t1)*t/t1.
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l*********o
发帖数: 88 | 6 do not assume it starts at zero.
【在 G********d 的大作中提到】
: you assume it starts at zero
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c**********e
发帖数: 2007 | 7 Then where does it start?
【在 l*********o 的大作中提到】
: do not assume it starts at zero.
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l*********o
发帖数: 88 | |
x******a
发帖数: 6336 | 9 The proof is based on the following observation: for t1< t
X=(w_t-w_{t1})-{t-t1}/{t2-t1} *(w_{t2}-w_{t1}) is independent of
Y=w_{t2}-w_{t1}.
then
0= E(X) =E(X|Y)= E( w_t-w_{t1}|Y) - {t-t1}/{t2-t1} (w_{t2}-w_{t1})
E(W_t|Y) =w_t1+ {t-t1}(w_{t2}- w{t1})/{t2-t1}
【在 l*********o 的大作中提到】
: That looks right, how would you prove it?
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o****e
发帖数: 80 | 10 没看懂，谁能解释一下这道题？
【在 x******a 的大作中提到】
: The proof is based on the following observation: for t1< t : X=(w_t-w_{t1})-{t-t1}/{t2-t1} *(w_{t2}-w_{t1}) is independent of
: Y=w_{t2}-w_{t1}.
: then
: 0= E(X) =E(X|Y)= E( w_t-w_{t1}|Y) - {t-t1}/{t2-t1} (w_{t2}-w_{t1})
: E(W_t|Y) =w_t1+ {t-t1}(w_{t2}- w{t1})/{t2-t1}
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T*****w
发帖数: 802 | 11 check Brownian Bridge
http://en.wikipedia.org/wiki/Brownian_bridge
【在 o****e 的大作中提到】
: 没看懂，谁能解释一下这道题？
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l******6
发帖数: 23 | 12 谁能解释一下, 当 t
given B1, E[B(t)] = B1?
【在 T*****w 的大作中提到】
: check Brownian Bridge
: http://en.wikipedia.org/wiki/Brownian_bridge
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c**********e
发帖数: 2007 | 13 For this part, the question is misleading. You could ignore this part.
Normally B(0) is specified, either 0 or fixed B_0.
Then it is the same as the 2nd part, as t is between 0 and t1,
and both B(0) and B(t1) are known.
【在 l******6 的大作中提到】
: 谁能解释一下, 当 t: given B1, E[B(t)] = B1?
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o****e
发帖数: 80 | 14 谢谢你，看明白了
谁能解释一下，t>t2的情况？
【在 T*****w 的大作中提到】
: check Brownian Bridge
: http://en.wikipedia.org/wiki/Brownian_bridge
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o****e
发帖数: 80 | 15 应该是t/t1*B1
E[w(t)-t/t1*w(t1)|w(t1)]=E[w(t)-t/t1*w(t1)](independent)=0
=E[w(t)|w(t1)]-t/t1*w(t1)==>E[w(t)|w(t1)]=t/t1*w(t1)
谁能解释一下， t>t2怎么做？ 谢谢
【在 l******6 的大作中提到】
: 谁能解释一下, 当 t: given B1, E[B(t)] = B1?
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t***s
发帖数: 4666 | 16 it's the easist part of this question.
【在 o****e 的大作中提到】
: 应该是t/t1*B1
: E[w(t)-t/t1*w(t1)|w(t1)]=E[w(t)-t/t1*w(t1)](independent)=0
: =E[w(t)|w(t1)]-t/t1*w(t1)==>E[w(t)|w(t1)]=t/t1*w(t1)
: 谁能解释一下， t>t2怎么做？ 谢谢
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o****e
发帖数: 80 | 17 明白了 martingale,谢谢
【在 t***s 的大作中提到】
: it's the easist part of this question.
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