I was working with a project and I used the VaR() function from the PerformanceAnalytics package to calculate Value-at-risk. I wanted to find out the probability of a stock generating making a loss of 1% or more. I found a solution to the problem by plugging numbers in to the probability variable, and controlling to see if it was approaching -1%. However, I was curious if it was possible to flip the formula so that I can just plug in the output and then the function will produce what would have been the input.
Produced the loss at 97.5% probability:
VaR(DNOlog, p = 0.975)
Produced a loss of -1% by changing the probability until it fit:
VaR(DNOlog, p = 0.6512184)
Let's get a reproducible example to demonstrate how you would go about this:
library(PerformanceAnalytics)
set.seed(2)
returns <- rnorm(1000, sd = 0.01)
This gives us a sensible result from VaR
VaR(returns, p = 0.975)
#> [,1]
#> VaR -0.01893631
To reverse this, we can use uniroot. This is a function which uses an iterative approach to finding the input value that makes a function return 0:
inverse_VaR <- function(x, target) {
f <- function(p) VaR(x, p)[1, 1] - target
uniroot(f, c(0.6, 0.99999), tol = .Machine$double.eps)$root
}
In our example, if we want to find the p value that makes VaR give an output of -0.01 with our vector returns, we can do:
inverse_VaR(returns, -0.01)
#> [1] 0.848303
And to show this works, we can do:
VaR(returns, 0.848303)
#> [,1]
#> VaR -0.009999999
Created on 2022-04-16 by the reprex package (v2.0.1)