R中的多元概率密度函数

Question

I want to create a multivariate probability density function with R to plot it as a perspective plot. f(x,y) is x+y for [0,1]*[0,1], 0 else. I tried this:

x <- seq(-0.5,1.5,length=50)*seq(0.5,1.5,length=50)
y <- x

f <- function(x,y){
  if (0<=x<=1 && 0<=y<=1){
    x+y
  } else {
    0
  }
}
f(x,y)  

But the function is not working and I don't know how to fix it. Does anyone have an idea?

Answer 1

First of all your if statement is wrong. This would fix the if statement but would still give the error.

x <- seq(-0.5,1.5,length=50)*seq(0.5,1.5,length=50)
y <- x

f <- function(x,y){
  if (0<=x & x<=1 & 0<=y & y<=1){
    x+y
  } else {
    0
  }
}

f(x,y) 

f1 <- function(x,y){
  cond <- 0<=x &x<=1 & 0<=y&y<=1
  ifelse(cond, x+y, 0)
}

f1(x,y)
[1] 0
Warning message:
In if (0 <= x & x <= 1 & 0 <= y & y <= 1) { :
  the condition has length > 1 and only the first element will be used

You need to vectorize your function in order to give it vector arguments. One way is to use function ifelse which is vectorized version of if and else

f1 <- function(x,y){
  cond <- 0<=x & x<=1 & 0<=y & y<=1
  ifelse(cond, x+y, 0)
}

f1(x,y)
 [1] 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000
 [9] 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 0.04685548 0.11224490 0.18096626
[17] 0.25301958 0.32840483 0.40712203 0.48917118 0.57455227 0.66326531 0.75531029 0.85068721
[25] 0.94939608 1.05143690 1.15680966 1.26551437 1.37755102 1.49291962 1.61162016 1.73365264
[33] 1.85901708 1.98771345 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000
[41] 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000
[49] 0.00000000 0.00000000