# Reproduce CDS Index Default Probability via Tranche [0,100] Probability

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

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

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

• 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. – Magic is in the chain Nov 8 at 17:28