R 中的概率 function

Question

I am trying to write the function s_i = P(X+X+Y=i) in R. I have the following distributions:

I write this code

a_i <- function(k){
  if (k == 0) return (3/4)
  if (k == 1) return (1/8)
  if (k == 2) return (1/8)
  else return(0)
}

b_i <- function(k){
  if (k == 0) return (1/10)
  if (k == 1) return (8/10)
  if (k == 2) return (1/10)
  else return(0)
}

s_i<-function(i){
  sas<-0
  for (x1 in 0:2){
    for(x2 in 0:2){ 
      for(y in 0:2){
        if(x1+x2+y==i){
        sum=a_i(x1)*a_i(x2)*b_i(y)
        sas=sas+sum
        }
      }
    }
  }
  return(sas)
}

I am not sure if I understand correctly this function. Any ideas?

Answer 1

First of all, your function is correct.

---

Your Method

Applying your functions, you can see the distribution in the format of data.frame like below

v <- 0:6
res1 <- data.frame(s = v, prob = sapply(0:6, s_i))

such that

> res1
  s      prob
1 0 0.0562500
2 1 0.4687500
3 2 0.2265625
4 3 0.1843750
5 4 0.0468750
6 5 0.0156250
7 6 0.0015625

---

Cross-Check

In what follows, I generate the whole sample space to verify your obtained distribution

space <- setNames(expand.grid(replicate(3,0:2,simplify = F)),c("X1","X2","Y"))
space$s <- rowSums(space)
space <- within(space, prob <- Vectorize(a_i)(X1)*Vectorize(a_i)(X2)*Vectorize(b_i)(Y))

where you can see the whole space in data frame is

> space
   X1 X2 Y s      prob
1   0  0 0 0 0.0562500
2   1  0 0 1 0.0093750
3   2  0 0 2 0.0093750
4   0  1 0 1 0.0093750
5   1  1 0 2 0.0015625
6   2  1 0 3 0.0015625
7   0  2 0 2 0.0093750
8   1  2 0 3 0.0015625
9   2  2 0 4 0.0015625
10  0  0 1 1 0.4500000
11  1  0 1 2 0.0750000
12  2  0 1 3 0.0750000
13  0  1 1 2 0.0750000
14  1  1 1 3 0.0125000
15  2  1 1 4 0.0125000
16  0  2 1 3 0.0750000
17  1  2 1 4 0.0125000
18  2  2 1 5 0.0125000
19  0  0 2 2 0.0562500
20  1  0 2 3 0.0093750
21  2  0 2 4 0.0093750
22  0  1 2 3 0.0093750
23  1  1 2 4 0.0015625
24  2  1 2 5 0.0015625
25  0  2 2 4 0.0093750
26  1  2 2 5 0.0015625
27  2  2 2 6 0.0015625

then the distribution can be summarized as

res2 <- aggregate(prob ~ s, space,sum)

such that

> res2
  s      prob
1 0 0.0562500
2 1 0.4687500
3 2 0.2265625
4 3 0.1843750
5 4 0.0468750
6 5 0.0156250
7 6 0.0015625

which is the same as res1