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I'm trying to understand example 2.6 taken from McNeil and Embrechts "Quantitative Risk Management". The example consists of obtaining the risk mapping of a portfolio of $d$ zero-coupon bonds.

  • The zero-coupon bond price at time $s$ and with maturity $T$ is $$p(s, T)= e^{-(T-s)y(s, T)}$$ where $y(s, T)$ is the yield curve

  • The loss $L_{t+1}$ over a time horizon $\Delta$ is given by the change in portfolio value given by: $$L_{t+1}=-\left( V_{t+1}-V_t \right)$$

  • The portfolio value is $V_t = \sum_{i=1}^d \lambda_i \ p(t\Delta, T_i)$, where $\lambda_i$ is the number of shares of the $i$th bond. Replacing for the bond price we have: $$V_t = \sum_{i=1}^d \lambda_i e^{-(T_i-t\Delta)y(t\Delta, T_i)}$$


My attempted solution:

To find the loss distribution, I take $V_t$ and write it in terms of the risk factor $Z_t=y(t\Delta,\ T_i)$, so basically: \begin{equation} L_{t+1}=-(V_{t+1}-V_t) = -\sum_{i=1}^d \lambda_i \left[e^{-(T_i-t\Delta) Z_t} -e^{-(T_i-(t+1)\Delta)(Z_t+X_{t+1})}\right] \ \ \ \ \ \ \ \ \ \ \text{(1)} \end{equation} where $X_{t+1}=Z_{t+1}-Z_t$ is the change of the risk factors.

Factoring $e^{-(T_i-t\Delta) Z_t} = p(t\Delta, T_i)$ from equation (1) gives me: \begin{equation} -\sum_{i=1}^d\lambda_i \ p(t\Delta, T_i) \left[ 1-e^{-(T_i-t\Delta) X_{t+1}} e^{\Delta (Z_t +X_{t+1})} \right] \ \ \ \ \ \ \ \ \ \ \text{(2)} \end{equation}


My question:

This is where I get stuck. The textbook says that taking derivatives of the loss function (what I suppose is equation (2) unless I've made a mistake) and using the first order approximation of the loss function: $$L_{t+1}^{\Delta} = \left[ f_t(t, Z_t) + \sum_{i=1}^d f_{z_i}(t, Z_t)X_{t+1, i} \right]$$

Where $f(t, Z_t)$ is the portfolio value function. They get the following answer:

$$ L_{t+1}^{\Delta} = -\sum_{i=1}^d \lambda_i \ p(t\Delta, T_i) \left[ y(t\Delta, T_i)\Delta - (T_i-t\Delta) X_{t+1, i} \right]$$

From the loss function that I have found, I haven't been able to get the answer. I'd be really grateful for any insight, explanation, or pointing to any mistake I've made. Thanks for reading this!

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