Quantifying Tail Risk

Implementation of Three Techniques to Model the Value-at-Risk of a Multi-Currency Portfolio over a 10-day time horizon.

- The Delta-Normal (Variance-Covariance) Method

- The Monte Carlo Simulation Method

- The Bootstrap Historical Simulation Method

Paul McAteer, MSc, MBA

pcm353@stern.nyu.edu

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Model Design¶

1) Value-at-Risk with the Delta-Normal (Variance-Covariance) Method.¶

The Delta-Normal method requires the construction of the covariance matrix of daily security returns $Σ$, which is a function of the correlation matrix and the sample variance of the constituent securities:

$$\mathit{\Sigma }\mathit{\ =\ DCD}$$

Where $C$ is the correlation matrix and $D$ is a diagonal matrix with the variance of securities across the diagonal and zeros elsewhere.

The 1-day variance of the portfolio $V_{\mathit{\Pi}_{1-day}}$ is determined by the weights, variances and covariances on the constituent assets:

$$V_{\mathit{\Pi}_{1-day}}=\ w'\mathit{\Sigma}w$$

Where we post-multiply $w'\mathit{\Sigma}$, the transpose of the weights vector and the covariance matrix, by the column vector of weightings $w$.

The volatility of the portfolio is simply the square root of variance, so:

$$\sigma_{\mathit{\Pi}_{1-day}}=\ \sqrt{w'\mathit{\Sigma}w}$$

Variance squares with time, $t$. However, volatility squares with the square root of time, so 10-day volatility is obtained in the following manner:

$$\sigma_{\mathit{\Pi}_{10-day}}=\ \sqrt{w'\mathit{\Sigma}w}\ \times \ \sqrt{10}$$

To find the Delta-Normal Value-at-Risk at a particular confidence level, multiply the portfolio volatility by the appropriate standard normal deviate, denoted by $\alpha$:

$${VaR}_{\mathit{\Pi},10-day}=\ \alpha \sqrt{w'\mathit{\Sigma}w}\ \times \ \sqrt{10}$$

So, for example, to calculate the VaR at a 95% confidence level, alpha would be set to $\mathrm{\sim}$1.645.

2) Value-at-Risk with the Monte Carlo Simulation Method.¶

The Monte Carlo Method involves simulating the security returns $r$ at each time step $t$ in a portfolio composed of correlated assets. The multivariate normal distribution of returns is specified by by a mean vector of security returns and the covariance matrix of returns:

$$r_t\sim MVN({\mu },\Sigma)$$
Now, to simulate a vector of security returns, it is necessary to capture the structure of the covariance matrix. This can be achieved by decomposing the matrix $\mathrm{\Sigma }$ into the product of a lower triangular matrix $L$ and upper triangular matrix $L'$.

We use Cholesky Factorization¹ to find $L$ such that:

$$\mathit{\Sigma}=LL'$$

The simulation of the correlated returns of the portfolio constituents at each time step , can be generated by:

$${\mathrm{r}}_t = \mu\ + \ {L}{\mathrm{z}}_t$$

Where $u=\mathrm{(}{\mathrm{u}}_1\mathrm{\dots }{\mathrm{u}}_n\mathrm{)}\mathrm{\ }$ is the column vector of expected security returns and $L$ is the lower traiangular matrix. $\mathrm{z}$ is a column vector of independent standard normal random variables $\mathrm{(}{\mathrm{z}}_1\mathrm{\dots }{\mathrm{z}}_n\mathrm{)}\mathrm{\ }$ , i.e.:

$$z\sim N({0}, 1)$$

$Lz$ produces a random sample of deviations corresponding to our correlated and nonstandard returns. Importantly, we can demonstrate that the covariance matrix of the elements of $Lz$ , denoted by $V(Lz)$ has covariance matrix $\mathit{\Sigma}$, that is:

$$V(Lz)= Σ$$

Given that values of $({\mathrm{z}}_1\mathrm{\dots }{\mathrm{z}}_n\mathrm{)}\mathrm{\ }$ are independently distributed (i.e. have zero covariance) and have a variance of 1, the covariance matrix of the elements of z is an identity matrix, $I$:

$$\ V\left(z\right)=V\left[ \begin{array}{c} {\mathrm{z}}_1 \\ {\mathrm{z}}_2 \\ {\mathrm{z}}_3 \\ \vdots \\ {\mathrm{z}}_n \end{array} \right]=I_n\left[ \begin{array}{ccccc} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{array} \right]$$

So the covariance matrix of the elements of $Lz$, $V(Lz)$ , will be:

$$V(Lz)= L I L'$$

Which simplifies to:

$$V(Lz)= L L'$$

We know that:

L