CODEWITHSUNDEEP
rfunctionnan
MyName
MyName

Reputation: 109

NaNs in a function

I am trying to run the following function but I get NaNs I don't understand my mistake... any ideas?

preg<-seq(0.15,0.42,length=100) 
sigma<-seq(0.01,0.1, length=100) 

y <- c(0.1851852,0.4210526,0.3243243)
lf <- function(preg, sigma) prod(dunif(y, preg, sigma))
lf(0.3, 0.02)

Thaks!

Upvotes: 1

Views: 67

Answers (1)

akrun
akrun

Reputation: 887951

With dunif, the arguments are x, min, max. According to ?dunif

dunif(x, min = 0, max = 1, log = FALSE)

Here, 'preg' is starting as a higher value (for min) when compared to sigma

dunif(y, preg[1], sigma[1])
#[1] NaN NaN NaN

dunif(y, 0.02, 0.3)
#[1] 3.571429 0.000000 0.000000

Upvotes: 1

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