这是一份 Imperial帝国理工大学 MATH97234作业代写的成功案例
$$
S_{j}=\sum_{i=1}^{p} a_{j i} Z_{i}+\sigma_{j} \epsilon_{j}, \quad j=1, \ldots n .
$$
Here $a_{j i}$ describes the exposure of obligor $j$ to factor $i$, i.e. the so-called factor loading, and $\sigma_{j}$ is the volatility of the idiosyncratic risk contribution. In such a framework one can easily interfere default correlation from the correlation of the underlying drivers $S_{j}$. To do so, we define default indicators
$$
Y_{j}=1\left(S_{j} \leq D_{j}\right)
$$
where $D_{j}$ is the cut-off point for default of obligor $j$. The individual default probabilities are
$$
\pi_{j}=\mathrm{P}\left(Y_{j}=1\right)=\mathrm{P}\left(S_{j} \leq D_{j}\right),
$$
and the joint default probability is
$$
\pi_{i j}=\mathrm{P}\left(Y_{i}=1, Y_{j}=1\right)=\mathrm{P}\left(S_{i} \leq D_{i}, S_{j} \leq D_{j}\right) .
$$
If we denote by $\rho_{i j}=\operatorname{Corr}\left(S_{i}, S_{j}\right)$ the correlation of the underlying latent variables and by $\rho_{i j}^{D}=\operatorname{Corr}\left(Y_{i}, Y_{j}\right)$ the default correlation of obligors $i$ and $j$, then we obtain for the default correlation the simple formula
$$
\rho_{i j}^{D}=\frac{\pi_{i j}-\pi_{i} \pi_{j}}{\sqrt{\pi_{i} \pi_{j}\left(1-\pi_{i}\right)\left(1-\pi_{j}\right)}} .
$$
MATH97234 COURSE NOTES ：
The SPD is then the second derivative of the call option function with respect to the strike price:
$$
f^{*}(\cdot)=e^{r \tau} \frac{\partial^{2} C}{\partial K^{2}}=e^{r \tau} \tilde{S} \frac{\partial^{2} c}{\partial K^{2}} .
$$
The conversion is needed because $c(\cdot)$ is being estimated not $C(\cdot)$. The analytical expression of (8.13) depends on:
$$
\frac{\partial^{2} c}{\partial K^{2}}=\frac{\mathrm{d}^{2} c}{\mathrm{~d} M^{2}}\left(\frac{M}{K}\right)^{2}+2 \frac{\mathrm{d} c}{\mathrm{~d} M} \frac{M}{K^{2}}
$$
The functional form of $\frac{\mathrm{d} c}{\mathrm{~d} M}$ is:
$$
\frac{\mathrm{d} c}{\mathrm{~d} M}=\Phi^{\prime}\left(d_{1}\right) \frac{\mathrm{d} d_{1}}{\mathrm{~d} M}-e^{-r \tau} \frac{\Phi^{\prime}\left(d_{2}\right)}{M} \frac{\mathrm{d} d_{2}}{\mathrm{~d} M}+e^{-r \tau} \frac{\Phi\left(d_{2}\right)}{M^{2}}
$$