Question

我有以下脚本名为test.R，我在其中输入变量值，然后调用我在SABR.R中编写的函数来计算某些值。

然而，当我运行代码时，我不断收到诸如此类的错误错误：找不到函数SABR.BSIV 错误：找不到功能SABR.calibration

我在这里做错了什么？它还说＆＃34;找不到对象k＆＃34;即使我在我的test.R代码中非常清楚地声明它。

library(testthat)
source("SABR.R")
test_that("SABR Model Test", {
iv <- c(0.346, 0.280, 0.243, 0.208, 0.203, 0.192, 0.192, 0.201, 0.205, 0.223, 0.228, 0.247, 0.252, 0.271, 0.275, 0.293, 0.313) 
k <- c(12.0, 15.0, 17.0, 19.5, 20.0, 22.0, 22.5, 24.5, 25.0, 27.0, 27.5, 29.5, 30.0, 32.0, 32.5, 34.5, 37.0)
f <- 22.724
t <- 0.583
a <- 0.317
b <- 0.823
r <- 0.111
n <- 1.050

iv.model <- SABR.BSIV(t, f, k, a, b, r, n)
params <- SABR.calibration(t, f, k, iv)
iv.calibrated <- SABR.BSIV(t, f, k, params[1], params[2], params[3], params[4])

# Check whether initial model can produce market IV or not
for(i in length(k)){expect_equal(iv.model[i], iv[i], tolerance = 0.01*iv[i])}

# Check whether calibrated parameter can produce market IV or not
for(i in length(k)){expect_equal(iv.calibrated, iv[i], tolerance = 0.01*iv[i])}

})

这是SABR.R代码：

EPS <- 10^(-8)

# Sub function for SABR BS-IV (Black-Scholes IV?)
.x <- function(z, r){log((sqrt(1-2*r*z+z^2)+z-r)/(1-r))}
.z <- function(f, k, a, b, nu){nu/a*(f*k)^(0.5*(1-b))*log(f/k)}

# Variable transformation function
.t1  <- function(x){1/(1+exp(x))}
.t2  <- function(x){2/(1+exp(x)) -1}

# Black-Scholes IV apporoximation formula by Hagan
SABR.BSIV <- function(t, f, k, a, b, r, n)
{
  z <- .z(f, k, a, b, n)
  x <- .x(z, r)
  numerator <- 1 + ((1-b)^2/24*a^2/(f*k)^(1-b) + 0.25*r*b*n*a/(f*k)^(0.5*(1-b)) + (2-3*r^2)*n^2/24)*t
  denominator <- x*(f*k)^(0.5*(1-b))*(1 + (1-b)^2/24*(log(f/k))^2 + (1-b)^4/1920*(log(f/k))^4)
  ifelse(abs((f-k)/f) < EPS, a*numerator/f^(1-b), z*a*numerator/denominator)
}

# Parameter calibration function for SABR
SABR.calibration <- function(t, f, k, iv)
{
  # Objective function for optimization, variables are transformed because of satisfing the constraint conditions
  objective <- function(x){sum( (iv - SABR.BSIV(t, f, k, exp(x[1]), .t1(x[2]), .t2(x[3]), exp(x[4])))^2) }
  x <- nlm(objective, c(0.25, 0.5, 0.5, 0.5))
  # Return optimized parameters
  parameter <- x$estimate
  parameter <- c(exp(parameter[1]), .t1(parameter[2]), .t2(parameter[3]), exp(parameter[4]))
  names(parameter) <- c("Alpha", "Beta", "Rho", "Nu")
  parameter
}

即使我采购了

0 个答案: