A replacement for qnorm?-CodePudding

I have this problem and I know that it can be carried out in several ways.

Assume that the returns of a security X are distributed according to a normal law with mean m=0 and standard deviation s=5.

What is the value at risk at 1 percent (i.e., that minimum value below which it will go in 1 percent of the cases)?

I solved it this way but I would like to know if there are other ways

qnorm(0.01,mean=0,sd=5)
pnorm(-11.63174,mean=0,sd=5)

CodePudding user response：

Taking into account Extreme Value Theory and practice the returns of a securities are usually not normally distributed because of long tails. Then it is assumed (with heuristic) e.g. a t distribution with a very small df (around 6), no matter how many observations is there.

(qt(0.01, 6)   0) * 5
# [1] -15.71334
qnorm(0.01, mean = 0, sd = 5)
# [1] -11.63174

From the technical perspective we mostly work with a sample, so empirical data. Then the data is sorted and value extracted or simply quantile taken.

set.seed(1234)
# some random sample
x <- rnorm(10000, mean = 0, sd = 5)
quantile(x, 0.01)
#        1% 
# -11.52401

CodePudding user response：

A. Here is the most obvious other way, if you can entertain the notion:

qnorm(0.01)*5 0

B. The other way is to solve the cdf function 1/2[1 erf((x-mu)/(sigma sqrt(2))] = 0.01 for x. The inverse erf can be approximated by a Maclaurin series erf^(-1)(z) = sum_(k=0)^infty ck/(2k 1)(sqrt(pi)/2 z)^(2k 1) where c0 = 1, c1 =1, and ck = sum_(m=0)^(k-1) cm ck-1-m / (m 1)(2m 1). Using k=1000 for a very good approximation, we can solve the cdf function for x to get the quantile.

c=c(1, 1)
K=1000
for (k in 2:K) {
  c[k 1]=0
  for (m in 0:(k-1)) {
    c[k 1]=c[k 1] c[m 1]*c[k-1-m 1]/(m 1)/(2*m 1)
  }
}
c

z = 0.01*2-1
  e = 0
  for (k in 0:K) {
    e = e c[k 1]/(2*k 1) * (sqrt(pi) / 2*z)^(2*k 1)
  }
e



e*5*sqrt(2)   0

qnorm(0.01, 0, 5)