全部版块 我的主页
论坛 金融投资论坛 六区 金融学(理论版) 金融工程(数量金融)与金融衍生品
5621 8
2009-05-21
<p>The spread for a FTD is 395bp, while the spreads for its constituents, IBM, MS and APPLE is 120,100,200 respectively. 395<420, why is that? that is, why the FTD's constituents are riskier than itself? Can someone give me an intuitive explanation?</p>
二维码

扫码加我 拉你入群

请注明:姓名-公司-职位

以便审核进群资格,未注明则拒绝

全部回复
2009-5-22 00:04:00

FTD insures the first default name,the spread should be smaller than one covering all four stocks. Think about it in this way, if there are two banks, one issuing FTD underlying three names,another bank issues three CDS with the same underlyings. When the first default occurs, two banks has to pay the same amount of money. When the second occurs bank one is free of any obligation because of the feature of FTD, but bank two has its obiligation because the CDS is written individually on each name. So FTD gives less "credit insurance" to policy holder, it only cover the first default. That's why FTD spread is less than the sum of all underlying CDS.

In the extreme case when all underlyings of the FTD have perfect correlations with each other (one defaults then all default) then buying one CDS is equivalent to buying FTD, so the price or the spread of FTD should be the lowest no matter how many names it covers (it costs less becasue it insures less randomness). On the other side of the spectrum where there is no correlations among all underlyings, in this case FTD will costs you a lot (it insures more randomness). But the spread of FTD will never exceed the sum of all CDS spread as I explained in the beginning. In your case the theoretical spread of FTD will be in the range of (200,420), how expensive FTD is depends on correlations, less correlation less cost. In today's financial crisis when correlations go up, CDS spead rockets while FTD climb up as well but relatively slower than amount of  aggregation. You will see the spread for the most risky name is much more approaching the spread for its FTD.

[此贴子已经被作者于2009-5-22 2:13:06编辑过]

二维码

扫码加我 拉你入群

请注明:姓名-公司-职位

以便审核进群资格,未注明则拒绝

2009-5-22 00:54:00

I say the FTD cannot be lower than 200.

[此贴子已经被作者于2009-5-23 1:04:19编辑过]

二维码

扫码加我 拉你入群

请注明:姓名-公司-职位

以便审核进群资格,未注明则拒绝

2009-5-22 00:56:00
以下是引用bmlf_001在2009-5-21 20:18:00的发言:

The spread for a FTD is 395bp, while the spreads for its constituents, IBM, MS and APPLE is 120,100,200 respectively. 395<420, why is that? that is, why the FTD's constituents are riskier than itself? Can someone give me an intuitive explanation?

大于420或者小于200都是arbitrage

二维码

扫码加我 拉你入群

请注明:姓名-公司-职位

以便审核进群资格,未注明则拒绝

2009-5-22 01:53:00

恩,用概率的交并补就好

假设 IBM, MS and APPLE的违约概率为P(A),P(B),P(C),对应的spread为120,100,200

max(P(A),P(B),P(c))<=P(FTD)=P(A or B or C)<=P(A)+P(B)+P(C),而spread和违约概率是单调对应的,所以就得到irvingy的那个答案 

二维码

扫码加我 拉你入群

请注明:姓名-公司-职位

以便审核进群资格,未注明则拒绝

2009-5-22 04:08:00
以下是引用垃圾树在2009-5-22 1:53:00的发言:

恩,用概率的交并补就好

假设 IBM, MS and APPLE的违约概率为P(A),P(B),P(C),对应的spread为120,100,200

max(P(A),P(B),P(c))<=P(FTD)=P(A or B or C)<=P(A)+P(B)+P(C),而spread和违约概率是单调对应的,所以就得到irvingy的那个答案 

我完全同意你和irvingy的分析,我一开始是这么想的,FTD只保护其中一个credit,但如果我单独买single name的,他们可能一起default,所以当然更risky。但有一个问题,考虑FTD的settle方式,比如FTD的notional是20m,再假设recovery rate都为40%,无论其中哪一个credit default了,protection seller都需要支付12m;

如果分别买这个三个公司的single name CDS,say IBM 10m,MS 5m,APPLE 5m,只有当他们一起default的时候seller才支付12m 与FTD相同,其余任何时候损失都会较少,不知道你们怎么看?

[此贴子已经被作者于2009-5-22 4:13:52编辑过]

二维码

扫码加我 拉你入群

请注明:姓名-公司-职位

以便审核进群资格,未注明则拒绝

点击查看更多内容…
相关推荐
栏目导航
热门文章
推荐文章

说点什么

分享

扫码加好友,拉您进群
各岗位、行业、专业交流群