R:如何得到两个分布的总和？

我有一个简单的问题.我想总结两个非参数分布.

这是一个例子.有两个城市有10个房子.我们知道每个房子的能源消耗.(编辑)我想得到从每个城市中选择的随机房屋总和的概率分布.

A1 <- c(1,2,3,3,3,4,4,5,6,7) #10 houses' energy consumption for city A
B1 <- c(11,13,15,17,17,18,18,19,20,22) #10 houses' energy consumption for city B

我有A1和B1的概率分布,我怎样才能得到A1 + B1的概率分布？如果我只是A1+B1在R中使用,它会给出12 15 18 20 20 22 22 24 26 29.但是,我不认为这是对的.因为房子里没有秩序.

当我改变房屋的顺序时,它会产生另一个结果.

# Original
A1 <- c(1,2,3,3,3,4,4,5,6,7)
B1 <- c(11,13,15,17,17,18,18,19,20,22)
#change order 1
A2 <- c(7,6,5,4,4,3,3,3,2,1) 
B2 <- c(22,20,19,18,18,17,17,15,13,11)
#change order 2
A3 <- c(3,3,3,4,4,5,6,7,1,2) 
B3 <- c(17,17,18,18,19,13,20,11,22,15)
sum1 <- A1+B1; sum1
sum2 <- A1+B2; sum2
sum3 <- A3+B3; sum3

红线是sum1,sum2和sum3.我不知道如何分配两个发行版的总和.请给我任何想法.谢谢!

(如果这些分布是正态分布或均匀分布,我可以很容易地得到分布的总和,但这些不是正常的,没有顺序)

理论上,两个随机变量的和分布是它们的PDF卷积,细节如下:

PDF(Z)= PDF(Y)*PDF(X)

所以,我认为这种情况可以通过计算convolution.

# your data
A1 <- c(1,2,3,3,3,4,4,5,6,7) #10 houses' energy consumption for city A
B1 <- c(11,13,15,17,17,18,18,19,20,22) #10 houses' energy consumption for city B

# compute PDF/CDF
PDF_A1 <- table(A1)/length(A1)
CDF_A1 <- cumsum(PDF_A1)

PDF_B1 <- table(B1)/length(B1)
CDF_B1 <- cumsum(PDF_B1)

# compute the sum distribution 
PDF_C1 <- convolve(PDF_B1, PDF_A1, type = "open")

# plotting
plot(PDF_C1, type="l", axe=F, main="PDF of A1+B1")
box()
axis(2)
# FIXME: is my understand for X correct?
axis(1, at=seq(1:14), labels=(c(names(PDF_A1)[-1],names(PDF_B1))))

注意:

CDF:累积分布函数

PDF:概率密度函数

## To make the x-values correspond to actually sums, consider
## compute PDF
## pad zeros in probability vectors to convolve
r <- range(c(A1, B1))
pdfA <- pdfB <- vector('numeric', diff(r)+1L)
PDF_A1 <- table(A1)/length(A1)                        # same as what you have done
PDF_B1 <- table(B1)/length(B1)
pdfA[as.numeric(names(PDF_A1))] <- as.vector(PDF_A1)  # fill the values
pdfB[as.numeric(names(PDF_B1))] <- as.vector(PDF_B1)

## compute the convolution and plot
res <- convolve(pdfA, rev(pdfB), type = "open")
plot(res, type="h", xlab='Sum', ylab='')

## In this simple case (with discrete distribution) you can compare
## to previous solution
tst <- rowSums(expand.grid(A1, B1))
plot(table(tst) / sum(as.vector(table(tst))), type='h')