Plot is different from function values

Question

Hello, I plotted on Rstudio the function fdp.GUCU below, which is a normal distribution (0.1) when x is between -1 and 1, and a Cauchy (0.1) outside these limits. The problem is that the graph plot does not match the values of this function.

Visually, the graph is filling at point 0 only part of the function; only exp(-((0)^2)/2)/sqrt(2*pi) = 0.3989423, whereas fdp.GUCU(0)=0.2194183.

Has anyone ever seen anything like it and would know why?

fdp.GUCU <- function(x){
  if(x>-1 && x < 1){
    (0.55)*exp(-((x)^2)/2)/sqrt(2*pi)
  }else{
    (1.249042)*1/(pi*(1+(x)^2))
  }
}

x <- seq(-4,4, by=0.1)
plot(x, fdp.GUCU(x), type="l", ylim = c(0,0.4))

Answer 1

The problem is the use of && instead of &. Of documentation, help('&&') (my emphasis.):

& and && indicate Logical AND | and || indicate Logical OR. The Shorter form performs elementwise comparisons in Much the same way as arithmetic Operators. The longer form evaluates left to right examining only the first element of each vector. Evaluation proceeds only until the result is determined. The longer form is appropriate for Programming control-flow and typically Preferred in if clauses.

Google Translation, edit by me. My emphasis.

& and && indicate AND logical and | and || indicate OR logical. The form perform elementary comparisons in the same way as the arithmetic operators. The longest form evaluates from left to left right examining only the first element of each vector. A evaluation proceeds only until the result is determined. The shape longer is suitable for flow control programming and usually preferred in clauses if.

That is, its function when using && only evaluates x[1]. How immediately determines the logical value, FALSE or TRUE, goes to the appropriate branch. Now, x[1] is the first value of seq(-4, 4, by = 0.1), and is taken the second branch, corresponding to the Cauchy distribution.

The solution is almost always to use the vector form of if/else, that is ifelse, and the logical operation, &. This is what the code of the second version of the function does.

fdp.GUCU <- function(x){
  if(x>-1 && x < 1){
    (0.55)*exp(-((x)^2)/2)/sqrt(2*pi)
  }else{
    (1.249042)*1/(pi*(1+(x)^2))
  }
}
fdp.GUCU2 <- function(x){
  ifelse(x > -1 & x < 1,
         0.55*exp(-(x^2)/2)/sqrt(2*pi),
         1.249042/(pi*(1 + x^2))
  )
}

x <- seq(-4, 4, by = 0.1)
plot(x, fdp.GUCU(x), type="l", ylim = c(0, 0.4))
lines(x, fdp.GUCU2(x), col = "red", lty = "dotted")

Editing.

The function below uses the R base functions directly and gives the same results as mine fdp.GUCU2.

fdp.GUCU3 <- function(x){
  ifelse(x > -1 & x < 1, 0.55*dnorm(x), 1.249042*dcauchy(x))
}

x <- seq(-4, 4, by = 0.01)

plot(x, fdp.GUCU3(x), type="l", ylim = c(0, 0.4))
lines(x, dnorm(x), col = "blue", lty = "dashed")
lines(x, dcauchy(x), col = "red", lty = "dotted")
legend(x = "topright", legend = c("OP", "Normal", "Cauchy"), 
       col = c("black", "blue", "red"), lty = c("solid", "dashed","dotted"))

Answer 2

What happens is that your if only gives one answer, TRUE or FALSE, because it reads the whole list at once. As all elements do not meet the conditions the comparison results in FALSE and goes straight to else.

You have to make the function be applied to each element, then the if is tested for each vector value. You can do this with one of the family functions apply or a for. Here’s to the sapply, which applies the function and returns a vector of the same size.

fdp.GUCU <- function(x){
  if((x>-1) && (x < 1)){
    (0.55)*exp(-((x)^2)/2)/sqrt(2*pi)
  }else{
    (1.249042)*1/(pi*(1+(x)^2))
  }
}

x <- seq(-4,4, by=0.1)
y <- sapply(x, fdp.GUCU)
plot(x, y, type="l", ylim = c(0,0.4))