Correlations and Copulas - Ch 11 Hull

[Dr. Jayanthi Sankaran]

Hi David,

In the following Example 6.1

Suppose that lambda = 0.95 and that the estimate of the correlation between two variables X and Y on day n - 1 is 0.6. Suppose further that the estimates of the volatilities for X and Y on day n -1 are 1% and 2% respectively. From the relationship between correlation and covariance, the estimate of the covariance rate between X and Y on day n - 1 is

0.6 x 0.01 x 0.02 = 0.00012

Suppose that the percentage changes in X and Y on day n - 1 are 0.5% and 2.5% respectively. The variance rates and covariance rate for day n would be updated as follows:

(Sigma(x,n))^2 = 0.95 x 0.01 ^2 + 0.05 x 0.005^2 = 0.00009625

My question is that according to EWMA

Covn = lambda*Covn - 1 + (1 - lambda)xn - 1*Yn-1

Extending the above for variance
(Sigma(x,n))^2 = 0.95 x 0.01 ^2 + 0.05 x 0.005^2 = 0.00009625 - I don't understand how 0.005^2 comes into the picture. It is not a volatility but a percentage change.
I don't know if I have made myself clear. Can you please clarify this - I seem to be missing something?

Thanks
Jayanthi

 

[Dr. Jayanthi Sankaran]

Hi David,

In the 'Other Copulas' section of 'Page 76, Chapter 11 of Hull's Correlations and Copulas' - in the Student t-copula - it is said that the variables U1 and U2 are assumed to have a bivariate Student t-distribution instead of a bivariate normal distribution. To sample from a bivariate Student t-distribution with f degrees of freedom and correlation rho, the steps are:

1. Sample from the inverse chi-square distribution to get a value of chi (in Excel, the CHIINV function can be used. The first argument is RAND() and the second is f(.)
2. Sample from a bivariate normal distribution with correlation rho, as described earlier
3, Multiply the normally distributed samples by SQRT(f/chi-square value)

Can you please explain how we get the bivariate Student t-distribution from the f and chi-square values? How are they related?

Thanks
Jayanthi

 

[ShaktiRathore]

Hi
Jayanthi regarding your first post EWMA model is
sigma(x,n)^2=lambda*sigma(x,n-1)^2+(1-lambda)*u(x,n-1)^2
where variance on day n of x is lambda(persistance factor)times variance on day n-1 plus 1-lambda times percent change i.e. return on n-1day squared. We take squared return we have already accounted for volatility factor.
Thanks

 

[Dr. Jayanthi Sankaran]

Hi Shakti,

In the following Example 6.1:

Suppose that lambda = 0.95 and that the estimate of the correlation between two variables X and Y on day n - 1 is 0.6. Suppose further that the estimates of the volatilities for X and Y on day n -1 are 1% and 2% respectively. From the relationship between correlation and covariance, the estimate of the covariance rate between X and Y on day n - 1 is

0.6 x 0.01 x 0.02 = 0.00012

Suppose that the percentage changes in X and Y on day n - 1 are 0.5% and 2.5% respectively. The variance rates and covariance rate for day n would be updated as follows:

(Sigma(x,n))^2 = 0.95 x 0.01 ^2 + 0.05 x 0.005^2 = 0.00009625

Suppose that lambda = 0.95 and that the estimate of the correlation between two variables X and Y on day

(Sigma(y,n))^2 = 0.95 x 0.02 ^2 + 0.05 x 0.025^2 = 0.00041125

Covn = 0.95 x 0.00012 + 0.05 x 0.005 x 0.025 = 0.00012025

Thanks, Shakti for pointing out as to where I was going wrong in my thinking - I have illustrated it as above as in Hull!

Jayanthi