Question

我有一个表示两个变量X和Y之间的依存关系的模数。我想计算以下公式：E（X |Y≤1％）。这是X的期望值，条件是Y低于1％。我看到有人问了一个there类似的问题，但是提供的R代码没有给出我想要的值。以下是有关系动词和边缘分布的一些详细信息。

library(VineCopula)
   library(copula)
#I estimate my Copula and assumes normal distribution for the two marginals
copula_dist <- mvdc(copula=claytonCopula(param=1.0), margins=c("norm","norm"),
                    paramMargins=list(list(mean=0, sd=5),list(mean=0, sd=5)))

#I take a sample of 500 events
sim <- rMvdc(500,copula_dist)
# Compute the density
pdf_mvd <- dMvdc(sim, my_dist)
# Compute the CDF
cdf_mvd <- pMvdc(sim, my_dist)

Answer 1

您必须评估这个双重积分：integral of x*pdf(x,y), -oo < x < +oo, -oo < y < 1%，然后将其除以Pr(Y < 1%)。这是在下面完成的。我还通过模拟执行近似以进行检查。

library(copula)

# the distribution
copula_dist <- mvdc(copula=claytonCopula(param=1.0), margins=c("norm","norm"),
                    paramMargins=list(list(mean=0, sd=5),list(mean=0, sd=5)))

### we will calculate E[X | Y < y0]
y0 <- 1/100

### approximation of E[X | Y < y0] using simulations
sim <- rMvdc(100000, copula_dist)
mean(sim[sim[,2]<y0,1])
# [1] -1.967642

### approximation of E[X | Y < y0] using numerical integration
### this is E[X * 1_{Y<y0}] / P(Y < y0)
library(cubature)
# PDF of the distribution 
pdf <- function(xy) dMvdc(xy, copula_dist)
# P(Y < y0)
denominator <- pnorm(y0, mean=0, sd=5)
# integrand
f <- function(xy) xy[1] * pdf(xy)
# integral
integral <- hcubature(f, lowerLimit = c(-Inf, -Inf), upperLimit = c(Inf, y0))
integral$integral / denominator
# [1] -1.942691

R中的copula的期望值

1 个答案: