### Options on portfolio loss

In this section, we derive formula for the price of a European call option on the portfolio loss. This will ease our transition into the CDOs world. Let K represent the strike price, T the maturity of the option, and r the risk free rate. The payoff the option at maturity is:

$$ \mbox{Payoff}=Max\left(l_{T}-K,0 \right) $$

And its value today is the expected value of the payoff under the risk neutral probability measure discounted to today:

$$ V_{0}=e^{-r T}\tilde{E}\left[ Max\left(l_{T}-K,0 \right) \right] $$ $$ V_{0}=e^{-r T}\tilde{E}\left[ l_{T} 1_{l_{T}>K} \right]-e^{-r T}\tilde{E}\left[ K 1_{l_{T}>K} \right] $$ $$ V_{0}=e^{-r T}\tilde{E}\left[ l_{T} 1_{l_{T}>K} \right]-K e^{-r T}\tilde{E}\left[ 1_{l_{T}>K} \right] $$ $$ V_{0}=e^{-r T}\tilde{E}\left[ l_{T} 1_{l_{T}>K} \right]-K e^{-r T}\tilde{P}\left[ l_{T}>K \right] $$ $$ V_{0}=e^{-r T}\tilde{E}\left[ \tilde{E}\left[ l_{T} \left( S \right) 1_{l_{T}\left( S \right)>K} \mid S \right] \right]-K e^{-r T}\tilde{P}\left[ l_{T}>K \right] $$

Now we know from the previous sections that:

$$ l_{T} \left( S \right)=P\left[ {D \mid S} \right] \; LGD = N\left[ \frac{Threshold-\sqrt{\rho} S}{\sqrt{1-\rho} } \right] \; LGD $$

From which it is easy to see that:

$$ l_{T} \left( S \right) > K $$ $$ \Rightarrow N\left[ \frac{Threshold-\sqrt{\rho} S}{\sqrt{1-\rho} } \right] \; LGD > K $$ $$ \Rightarrow S < \frac{Threshold-\sqrt{1-\rho} N^{-1} \left[ \frac{K}{LGD} \right]}{\sqrt{\rho}}$$ $$ \Rightarrow S < A_{K} $$

Thus:

$$ \tilde{E}\left[ \tilde{E}\left[ l_{T} \left( S \right) 1_{l_{T}\left( S \right)>K} \mid S \right] \right]=\tilde{E}\left[ N\left[ \frac{Threshold-\sqrt{\rho} S}{\sqrt{1-\rho} } \right] \; LGD \; 1_{S < A_{K}} \right] $$ $$ = LGD \int_{-\infty}^{A_{K}}{N\left[ \frac{Threshold-\sqrt{\rho} S}{\sqrt{1-\rho} } \right] n\left( S \right) } d S $$ $$ = LGD \; N_{2} \left[Threshold, A_{K}, \sqrt{\rho} \right] $$

For the second component, we have:

$$ \tilde{P}\left[ l_{T}>K \right] = 1-\tilde{P}\left[ l_{T} < K \right] $$ $$ \tilde{P}\left[ l_{T}>K \right] = 1-N\left[ \frac{\sqrt{1-\rho} N^{-1} \left[ \frac{K}{LGD} \right]-Threshold_{i}}{\sqrt{\rho}} \right] $$ $$ \tilde{P}\left[ l_{T}>K \right] = N\left[- \frac{\sqrt{1-\rho} N^{-1} \left[ \frac{K}{LGD} \right]-Threshold_{i}}{\sqrt{\rho}} \right] $$ $$ \tilde{P}\left[ l_{T}>K \right] = N\left[ A_{K} \right] $$

And hence:

$$ V_{0}=e^{-r T}\tilde{E}\left[ \tilde{E}\left[ l_{T} \left( S \right) 1_{l_{T}\left( S \right)>K} \mid S \right] \right]-K e^{-r T}\tilde{P}\left[ l_{T}>K \right] $$ $$ V_{0}=e^{-r T} LGD \; N_{2} \left[Threshold, A_{K}, \sqrt{\rho} \right] -K e^{-r T}N\left[ A_{K} \right] $$

Where

$$ A_{K}=\frac{Threshold-\sqrt{1-\rho} N^{-1} \left[ \frac{K}{LGD} \right]}{\sqrt{\rho}}$$ $$ Threshold= N^{-1} \left[ \widetilde{PD}_{T} \right] = N^{-1} \left[PD_{T} \right] +{\rho_{i,m} \lambda_{m} \sqrt{T}}$$

### Simplified single-tranche CDO

We can use the above European call option on portfolio loss to derive the price of a 'Simplified' single tranche CDO, with attachment and detachment points \( K_{1} \)and \( K_{2} \), respectively. The protection cover the portfolio losses in excess of \( K_{1} \) but only up-to \( K_{2} \). Whilst the protection buyer of a single tranche CDO normally pays premium each quarter on the remaining tranche notional,which reduces by the incurred losses, and the protection seller pays amount equal to the losses when defaults occur, we assume a simplified product that, like the more familiar stock option, pays premium at the start of the contract, and receives the protection amount, if losses have exceeded the attachment point, at the maturity of the contract. The payoff at maturity is:

$$ \mbox{Payoff}=\frac{Max\left(l_{T}-K_{1},0 \right)-Max\left(l_{T}-K_{2},0 \right)}{K_{2}-K_{1} } $$

We can then use the formula from the previous section to compute the price of the tranche:

$$ V_{0}=\frac{e^{-r T}}{K_{2}-K_{1}} \left( LGD \; N_{2} \left[N^{-1}\left[ \widetilde{PD}_{T} \right], A_{K_{1}}, \sqrt{\rho} \right] -K_{1} e^{-r T}N\left[ A_{K_{1}} \right] - LGD \; N_{2} \left[N^{-1}\left[ \widetilde{PD}_{T} \right], A_{K_{2}}, \sqrt{\rho} \right] + K_{2} e^{-r T}N\left[ A_{K_{2}} \right] \right) $$

### Single-tranche CDO

We now remove the simplifying assumption that the full premium for the duration of the tranche CDO contract is paid at the start of the contract, and protection payment is made at the maturity of the contract. We instead assume that the premium is paid on a generic set of dates \( T_{1}, T_{2},, T_{N} \). We will also assume that the protection amount is also paid at one of these dates on any incremental losses from the previous premium date, as long as the total loss is within the range covered by the tranche attachment and detachment points. Again we let \( K_{1} \)and \( K_{2} \) represent the attachment and detachment points, respectively.

The key to pricing this structure is the risk-neutral survival curve. Once we have the survival rate at each of the premium date, we can compute the expected amount of premium at each date as we know that the premium is paid on the remaining or 'surviving' notional of the tranche. Knowing the surviving balances at the then current and previous cash flow dates then also tell us what has been lost in the period, which is exactly the quantity we need to compute the protection amount. Discounting the risk neutral premium and protection cash flows at the risk free rate then should give us the price of the tranche.

We have already done most of the work needed to derive the risk-neutral survival curve. Let \( l_{T_i} \) represent the portfolio loss rate at time \( T_i \). We know from the previous section that this correspond to a tranche cumulative loss rate of (we called it payoff there):

$$ \mbox{Tranche loss rate at time } T_i=\frac{Max\left(l_{T_i}-K_{1},0 \right)-Max\left(l_{T_i}-K_{2},0 \right)}{K_{2}-K_{1} } $$

And the survival rate is then:

$$ Q_{T_i}= 1- \frac{Max\left(l_{T_i}-K_{1},0 \right)-Max\left(l_{T_i}-K_{2},0 \right)}{K_{2}-K_{1} } $$

And its expected value under the risk neutral measure is:

$$ \tilde{E}\left[ Q_{T_i} \right]= 1- \frac{ \tilde{E} \left[ Max\left(l_{T_i}-K_{1},0 \right) \right]- \tilde{E} \left[ Max\left(l_{T_i}-K_{2},0 \right) \right] }{K_{2}-K_{1} } $$ $$ \tilde{E}\left[ Q_{T_i} \right]= 1- \frac{ LGD \; N_{2} \left[N^{-1}\left[ \widetilde{PD}_{T_i} \right], A_{K_{1}}, \sqrt{\rho} \right] -K_{1} e^{-r T}N\left[ A_{K_{1}} \right] - LGD \; N_{2} \left[N^{-1}\left[ \widetilde{PD}_{T_i} \right], A_{K_{2}}, \sqrt{\rho} \right] + K_{2} e^{-r T}N\left[ A_{K_{2}} \right] }{K_{2}-K_{1} } $$