楼主: schwereburg
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用Esscher measure推European call的问题 [推广有奖]

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楼主
schwereburg 发表于 2013-4-13 07:03:09 |AI写论文

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我现在到了C(s,0)=exp(-rt)*EQ{max(ST-k,0)},然后我把ST和K拆开,分别换成EQ(h*+1)和EQh*作为measurement来推,其中h*代表在h*的情况下,exp(-rt)*St是一个martingale。
换句话说,
我把
C(s,0)=exp(-rt)*EQ{max(ST-k,0)}拆开
我现在在
C=So*EQ(h*+1){ST1ST>k}-k*exp(-rT)*Qh*{1ST>k}
其中,1ST>k是特征方程,从exp(-rt)*EQ{ST1ST>k}变到So*EQ(h*+1){ST1ST>k}是用了Esscher的一个性质和h*的性质,这个不是关键。
我的目的是推出来C=So*N(d1)-k*exp(-rT)*N(d2)

然后我现在想做的是解ds/s=rdt+sigma*dzQ(h*+1),如果解出ST=So*exp(r+sigma^2/2)+sigma*Z>K就对了,EQ(h*+1){ST1ST>k}就正好是N(d1),
然后解ds/s=rdt+sigma*dz(Qh*),如果解出来是ST=So*exp(r-sigma^2/2)+sigma*Z>K就对了,Qh*{1ST>k}正好是N(d2)。

但是我卡在这一步了。按照刚才反推出来的话,如果:
ZQ(h*+1)=ZQ+sigma的积分
ZQh*=ZQ
就对了,ZQ是在risk neutral下的随机过程。
ZQh*=ZQ可以理解,但不知道怎么证出来,
ZQ(h*+1)为什么就多了一个sigma的积分呢?
我wiki了一下,
感觉已经很近了,但还是想不出来。
实在不好意思,表述不太容易懂。请高手指教。
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关键词:European Measure Esscher Europe call

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Chemist_MZ 发表于3楼  查看完整内容

You don't need to consider the problem in that complicated way. Anyway, your thinking is correct. What need to be shown is that: Under risk neutral measure where D=exp(rt) is used as a numeraire, the stock process should follow: dS=r*S*dt+sigma*S*dW (we have done it) while under the so-called forward risk neutral (FRN) measure where the stock St is used as a numeraire, the stock process ...

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沙发
schwereburg 发表于 2013-4-13 07:42:35
搞出来了,刚才又看到一条性质,h*=(r-mu)/sigma,代入最后的那个Eh(N(.....=N(mu+h....的那个公式,就解决了。

藤椅
Chemist_MZ 在职认证  发表于 2013-4-13 08:06:50
You don't need to consider the problem in that complicated way.

Anyway, your thinking is correct. What need to be shown is that:

Under risk neutral measure where D=exp(rt) is used as a numeraire, the stock process should follow:
dS=r*S*dt+sigma*S*dW (we have done it)

while under the so-called forward risk neutral (FRN) measure where the stock St is used as a numeraire, the stock process should follow:

dS=(r+sigma^2)S*dt+sigma*S*dW

So the problem is how can we show this?

Your problem now is C=exp(-rT)*EQ(ST*I{ST>K}) ( Q is a risk neutral measure), and we can construct a measure, say S, in which all the assets price C use S as a numeraire is a martingale.

C0/S0=ES(CT/ST);

In our case is C0/S0=ES(ST*I{ST>K}/ST)=ES(I{ST>K}), you know that there should be only one price for the option, so: C0=exp(-rT)*EQ(ST*I{ST>K})=S0*ES(I{ST>K}), which gives us

EQ(ST*exp(-rT)*I{ST>K})=ES(S0*I{ST>K}/ST) (the expectation is conditional at t=0 so S0 can be put into it)

compare left side to the right side, we know that EQ(ST*exp(-rT)/S0 *X)=ES(X), where X is any random variable in our case its the indicator function(not 特征函数)。

From change of measure theorem, we know that the Likelihood ratio, or the Radon–Nikodym derivative
dS/dQ=ST*exp(-rT)/S0, (of course, here ST is the process under Q measure)

You now have the likelihood ratio ( or R-N derivative) which can be simplified as:
LR(likelihood ratio) =exp(-0.5sigma^2*T+sigma*WT)

=exp(-∫(0 to T)0.5 (-sigma)^2 dt-∫(0 to T) -sigma*dW )

set -sigma=theta

From Girsanov theorem, if a W(t) is a brownian motion under A measure, then W(t)+theta*t is also a browian motion under B measure, where the change in the two measures are linked with the likelihood ratio
=exp(-∫(0 to T) 0.5(theta)^2 dt-∫(0 to T) theta*dW )

In our case W(t) is a brownian motion under Q( risk neutral measure), then, W(t)-sigma*t is a brownian motion under FRN measure where St is used as a numeraire.

Then every thing is simple. dW(t)_S=dW(t)_Q-sigma*dt=>dW_Q=dW(t)_S+sigma*dt

under S measure: dS=r*S*dt+sigma*S*(dW(t)_S+sigma*dt)

so dS=(r+sigma^2)*S*dt+sigma*S*dW(t)_S

done!

the following links are for reference

1. Change of numeraire:
http://en.wikipedia.org/wiki/Numéraire

2. Radon-Nikodym theorem
http://en.wikipedia.org/wiki/Radon–Nikodym_theorem

3. Girsanov theorem
http://en.wikipedia.org/wiki/Girsanov_theorem









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板凳
schwereburg 发表于 2013-4-14 03:03:33
Chemist_MZ 发表于 2013-4-13 08:06
You don't need to consider the problem in that complicated way.

Anyway, your thinking is correct. ...
谢谢!我看懂了。只是这个measurement不是称作Qs吗?如果它是FRN的话,那他和W-forward neutral world和T-forward neutral world是并列得关系了?
如果记W-forward neutral world为Qw,这个Qw/Q=exp(-(1/2)sigma^2*T-sigma*WT)
那Qt/Q是什么?
正好在看这一块,还有1个问题:
Definition of forward-measures and associated numéraire. Martingale property of relative prices.
我认为题目说的就是W和T,但是他们Martingale性质是怎么定的?我知道Q的martingale性质。

报纸
Chemist_MZ 在职认证  发表于 2013-4-14 04:30:15
schwereburg 发表于 2013-4-14 03:03
谢谢!我看懂了。只是这个measurement不是称作Qs吗?如果它是FRN的话,那他和W-forward neutral world和 ...
哈哈,看来我没白写,比较欣慰。

天下只有两个世界,一个real world,另一个叫forward risk neutral world。

在真实世界理论上你也能搞pricing只是很困难而已,因为你很难构造一个martingale(但是没有说不可以)

在FRN的世界,只要两个asset,f,g 是由相同的risk 驱动的

你就一定能找到一个measure,使得f/g是个martingale( g is the so-called numeraire),即f0/g0=EQi(fT/gT), 至于这个measure的记号是什么,你不用死板,自己定义就行,没有人说过risk neutral一定要用Q

不同的FRN世界有不同的叫法,几个特殊情况:

1. g是money market account,Qi叫risk neutral measure

2. g是zero coupon bond,Qi一般写作T,叫做forward measure

3. g是annuity,Qi一般写作A,叫做forward swap measure

4.其他,比如这里的以S为numeraire,那就只能叫做FRN respect to S measure了

very simple,Hull的8ed书上27章有讨论,可以参考。
理论上定价可以在任何世界进行,只是risk neutral大家比较好理解,所以用得最多。forward measure也非常常用,像Black(1976)以及现在的SABR(2002) model都是在forward measure下。
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地板
schwereburg 发表于 2013-4-14 19:07:46
Chemist_MZ 发表于 2013-4-14 04:30
哈哈,看来我没白写,比较欣慰。

天下只有两个世界,一个real world,另一个叫forward risk neutral w ...
我基本上理解透彻了。以前好像模模糊糊有一个概念,现在清楚了,太感谢了!

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