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$$ \begin{split} B(0,T) & = \color{fuchsia}{\text{EV[non-default scenario]}} + \color{blue}{\text{EV[default scenario]}}\\ & = E\left[\color{fuchsia}{\exp\left(-\int_0^T\!\!\! r_t dt\right)·\mathbf{1}_{\{T<\tau\}}} + \color{blue}{\int_0^T\!\!\! RR\cdot \exp\left(-\int_0^t\!\!\! r_s ds\right) · P(t ≤ \tau < t+dt )} \right]\\ & = E\left[\color{fuchsia}{\exp\left(-\int_0^T\!\!\! r_t dt\right)\cdot \exp\left(-\int_0^T\!\!\! h_t dt\right)}\right. \\ &\qquad\qquad\quad\qquad + \left.\color{blue}{\int_0^T RR\cdot \exp\left(-\int_0^t\!\!\! r_s ds\right) \cdot h_t\exp\left(-\int_0^t\!\!\! h_s ds\right)dt} \right]\\ & = E\left[\color{fuchsia}{\exp\left(-\int_0^T\!\! (r_t+h_t) dt\right)} + \color{blue}{\int_0^T RR \cdot h_t\cdot \exp\left(-\int_0^\color{red}{t} \!\!(r_s+h_s) ds\right) \color{red}{dt}} \right] \end{split} $$ In this expression, $r_t$ and $h_t$ is just some function with respect to time. RR (recovery ratio) is the some constant value. In the blue part of the expression, inner integral is from 0 to t, and the outer integral is from 0 to T (T is the full time span). I don't know how to simplify further but the textbook says it can be simplified to:

enter image description here

hint given is to use the approximation: $e^x = 1+x$.

Can anyone show me the steps in between? much appreciated

Daniele Tampieri
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Jeremy
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1 Answers1

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We expand in small $T$. Note that $e^x = 1+x+O(x^2)$. We have \begin{align*} \cdots &= \exp\left(-\int_0^T (r_t+h_t) dt\right) + \int_0^T RR \cdot h_t \exp\left(-\int_0^t(r_s+h_s) ds\right) dt \\ &= \left(1 - \int_0^T (r_t+h_t) dt + O(T^2)\right) + \int_0^T RR\cdot h_t \left( 1 - \underbrace{\int_0^t(r_s+h_s)ds}_{O(t)} + O(t^2) \right)dt \\ &= 1 - \int_0^T(r_t+\underbrace{(1-RR)h_t)}_{s_t}dt + O(T^2) \\ &= \exp\left( - \int_0^T(r_t+s_t)dt \right) + O(T^2) \end{align*} How well this approximation works depends on the specific form for $r_t$ and $h_t$. We assume that $r_t$ and $h_t$ are analytic on some open interval centered on $t=0$ and containing $[0,T]$. Generally corrections are $O(T^2)$. If $h_0=0$, corrections are $O(T^3)$. If $r_0=h_0=0$, corrections are $O(T^4)$.

user26872
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