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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 $(2)$?

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  • $\begingroup$ 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. $\endgroup$ – Magic is in the chain Nov 8 at 17:28

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