Calculating One-Day VaR with Student-t GARCH Forecasts
Summary
The document explains how to convert a one-day GARCH(1,1) forecast with Student-t innovations into a 99% value-at-risk estimate for a one-million-dollar position. Its central correction is that the fitted Student-t quantile must be rescaled because the model’s shape parameter describes a t distribution whose variance is not one. The adjustment uses the square root of the degrees-of-freedom ratio, shape divided by shape minus two, before combining the quantile with the forecasted mean and standard deviation.
The example reports a fitted shape parameter of 5.483, a forecast mean of -0.001711227, and a forecast standard deviation of 0.02180995, and gives a VaR of $54,687 as the reference answer. The note presents a calculation matching that result. It does not discuss alternative VaR conventions, tail-loss sign conventions, model validation, or whether the GARCH and Student-t assumptions adequately capture actual risk.
Key ideas
- Student-t quantiles need variance standardization when used with a GARCH forecast standard deviation.
- The standardization factor depends on the fitted degrees of freedom.
- The example scales the one-day return VaR by a one-million-dollar position.
- The calculation relies on the fitted GARCH model and does not evaluate its broader risk assumptions.
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Full text
# Calculating the VaR from a GARCH(1,1) with Student-t innovations
# Calculating the VaR from a GARCH(1,1) with Student-t innovations
I'm self-studying several questions on Ruey S. Tsay's teaching page. I'm experiencing some difficulty getting the correct answer for final exam 2013 Problem B Question 3.
Given a Student-t GARCH (1,1) model, I believe that the correct way to calculate 1-Day $VaR$ would be to take the 1-Day predicted mean ($\mu_t$) and standard deviation ($\sigma_t$) and apply the formula: $VaR_{0.99} = \mu_t+t_{0.99}\cdot \sigma_t$. To get the $VaR$ in dollar terms we multiply this by the position size, $1 million.
In applying the results below, I took $\mu_t=-0.001711227, \sigma_t=0.02180995$ and the t-distribution degrees of freedom from the shape parameter in the results, $5.483$. However, this gives the wrong answer. The correct answer is $VaR = $$54,687, which can be found in the solutions manual
The results are here:
```
> summary(m3)
Title: GARCH Modelling
Call: garchFit(formula = ~garch(1,1), data=xt, cond.dist="std", trace = F)
Mean and Variance Equation:
data ~ garch(1, 1) [data = xt]
Conditional Distribution: std
Error Analysis:
Estimate Std. Error t value Pr(>|t|)
mu -1.711e-03 3.698e-04 -4.627 3.71e-06 ***
omega 6.235e-06 2.499e-06 2.495 0.0126 *
alpha1 4.833e-02 1.027e-02 4.707 2.51e-06 ***
beta1 9.421e-01 1.227e-02 76.812 < 2e-16 ***
shape 5.483e+00 5.379e-01 10.192 < 2e-16 ***
---
Standardised Residuals Tests:
Statistic p-Value
Ljung-Box Test R Q(10) 14.9856 0.1325876
Ljung-Box Test R^2 Q(10) 5.575123 0.849608
Information Criterion Statistics:
AIC BIC SIC HQIC
-4.780660 -4.770210 -4.780667 -4.776892
> predict(m3,1)
meanForecast meanError standardDeviation
1 -0.001711227 0.02180995 0.02180995
```
## Answer by José Fuentes (score 1)
https://quant.stackexchange.com/a/34760
I know this post is quite old, but someone else just might face a question like this one. I got the, I think, correct answer as follows in R:
`VaR <- (0.001711227-(qt(p = 0.99, df = 5.483)/sqrt(5.483/3.483))*0.02180995)*1000000`
You can see 0.001711227 is the forecasted mean, 0.02180995 the forecasted sd and 5.483 your fitted shape parameter.
About the method, that's pretty much how it appears in a lecture file I found here. You have to standardize your t-student quantile before, and use `sqrt(shape/(shape-2))`.
Hope this helps.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.