The tranche survival probability up to time $t$ between attachment $K_1$ and detachment $K_2$ is defined as $$Q(t,K_1,K_2) \quad=\quad 1 - \mathbb{E}[L(t,K_1,K_2)]$$ with tranche loss function $$L(t,K_1,K_2) \quad=\quad \frac{\min(L(t),K_2) - \min(L(t),K_1)}{K_2 - K_1}$$ and index loss function $$L(t) \quad=\quad \frac{1}{N} \cdot \sum_{i=1}^N (1-R_i)\cdot 1_{\{\tau_i<t\}}$$

Now, if we set $K_1=0\%$ and $K_2=100\%$ we get $$Q(t,0,1) \quad =\quad 1 - \frac{\mathbb{E}[\min(L(t),1)] - 0}{1-0} \quad =\quad 1 - (1-R)\cdot \mathbb{P}(\tau<t)\tag{1}$$ (assuming $R_i\equiv R$ and $\mathbb{P}(\tau_i<t)\equiv \mathbb{P}(\tau<t)$ for simplicity)

However, if $K_1=0\%$ and $K_2=100\%$, should we not recover the pure index probabilities? That is $$Q(t,0,1) \quad=\quad 1 -\mathbb{P}(\tau<t)\tag{2}$$ It seems $\color{red}{(1-R)}$ is somehow incorrectly showing up in formula $(1)$.
How does one reconcile formula $(1)$ and $(2)$?

  • $\begingroup$ If every name defaults, default rate will be 100% but total loss will be PD * (1-R)=1-R, so maximum K2 of 100% does not make sense, should be 1-R. I assumed homogeneous portfolio but same carries over to the non homogeneous case. $\endgroup$ Nov 8, 2019 at 17:28
  • $\begingroup$ @Magicisinthechain The formula from (1) does not change if we set $K_2=1-R$ (as the max). So we still have (1) and (2) mismatching. Does this mean the 0-100 Tranche does NOT replicate the Index? $\endgroup$
    – Phil-ZXX
    Jan 24, 2020 at 18:59

1 Answer 1


It is actually that you forgot your $1 - R$ in formula (2) :) The index survival curve is defined similarly to the tranche's : $Q\left(t\right) = 1 - \mathbb{E} \left[L\left(t\right)\right] = 1 - \left(1 - R\right)\mathbb{P}\left(\tau < t\right)$. Hence, your formula for the 0-100 tranche survival curve does coincide with the index'.

That history of loss consistency between index and tranche is handled the following ways :

  • On the index, the protection seller pays the loss and will receive coupons on $1-w_i$ of their notional, $w_i$ being the weight of the defaulted entity. $1 - \sum_{i = 1}^N{w_i1_{\tau_i < t}}$ is often called the index factor.
  • On the tranche, the equity protection seller pays the loss and receives coupons on $1 - L \left(t, 0, K\right)$ of their notional. The super senior tranche holder, though they do not take any loss, receive coupons on $1 - \sum_{i = 1}^N{w_iR_i1_{\tau_i < t}}$, i.e. on a diminished notional. Practitionners say they are attacked by the top.

This is to ensure that when all names have defaulted, a protection buyer would not pay remaining coupons on $R$ of the notional, which would not make any sense. A good reference on the subject is O'Kane's textbook on credit derivatives (2008).

  • $\begingroup$ Thanks. Just to clarify one thing in your second bullet point: Looking at O'Kane's book (section 12.5.2, page 233) it says coupons are paid on $1-L(t,K_1,K_2)$. So for a super senior tranche, we have $1-L(t,K,1) = 1$ if the loss amount $L(T)$ is still less than $K$ (= no losses taken)? And if $K>L(T)$ (= losses are taken), then it would in fact be $1-L(t,K,1) =\frac{1 - \sum_{i = 1}^N{w_i\color{red}{(1-R_i)}1_{\tau_i < t}}}{1-K}$? Note the $1-R_i$. $\endgroup$
    – Phil-ZXX
    Jan 27, 2020 at 12:56
  • $\begingroup$ No, when it is not affected by the index loss (i.e. $L\left(T\right) < K$, with $K$, the lower strike), the super senior tranche is reduced by the amount of the recoveries of the defaulted credits. WHEN the SS starts to be affected by defaults, i.e. all subordinated tranches have been wiped out, it is attacked by the top (recoveries) and the bottom (losses). $\endgroup$
    – siou0107
    Jan 27, 2020 at 13:09
  • $\begingroup$ I see (just read section 12.5.4 "The Senior Tranche" from O'Kane's book). So in short, the NPV of the 0-100 Tranche and Regular Index are different, because of the fee leg dynamics upon default: the 0-100 Tranche reduces Notional by $w_i R_i$ whereas the Index Notional reduces by $w_i$? $\endgroup$
    – Phil-ZXX
    Jan 27, 2020 at 14:31
  • $\begingroup$ You're almost there: since a 0-100 tranche has no subordinated tranche, it is immediately attacked both by the top (recoveries on the defaulted credits) and by the bottom (losses on the defaulted credits), which sums up to $w_i$ at each default. Thus, 0-100 tranche and regular index are the same :) $\endgroup$
    – siou0107
    Jan 27, 2020 at 14:34
  • $\begingroup$ Ah right, top and bottom. Thanks. This was very useful. $\endgroup$
    – Phil-ZXX
    Jan 27, 2020 at 14:38

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