CODEWITHSUNDEEP
rnormal-distributionprobability-densityprobability-distribution
Eric
Eric

Reputation: 501

Why doesn't "dnorm" sum up to one as a probability?

This may be some basic/fundamental question on 'dnorm' function in R. Let's say I create some z scores through z transformation and try to get the sum out of 'dnorm'.

 data=c(232323,4444,22,2220929,22323,13)
 z=(data-mean(data))/sd(data)
 result=dnorm(z,0,1)
 sum(result)
 [1] 1.879131

As above, the sum of 'dnorm' is not 1 nor 0.

Then let's say I use zero mean and one standard deviation even in my z transformation.

 data=c(232323,4444,22,2220929,22323,13)
 z=(data-0)/1
 result=dnorm(z,0,1)
 sum(result)
 [1] 7.998828e-38

I still do not get either 0 or 1 in sum.

If my purpose is to get sum of the probability equal to one as I will need for my further usage, what method do you recommend using 'dnorm' or even using other PDF functions?

Upvotes: 3

Views: 2693

Answers (1)

Anders Ellern Bilgrau
Anders Ellern Bilgrau

Reputation: 10223

dnorm returns the values evaluated in the normal probability density function. It does not return probabilities. What is your reasoning that the sum of your transformed data evaluated in the density function should equate to one or zero? You're creating a random variable, there is no reason it should ever equate exactly zero or one.

Integrating dnorm yields a probability. Integrating dnorm over the entire support of the random variable yields a probability of one:

integrate(dnorm, -Inf, Inf)
#1 with absolute error < 9.4e-05

In fact, integrate(dnorm, -Inf, x) conceptually equals pnorm(x) for all x.

Edit: In light of your comment.

The same applies to other continuous probability distributions (PDFs):

integrate(dexp, 0, Inf, rate = 57)
1 with absolute error < 1.3e-05

Note that the ... argument(s) from ?integrate is passed to the integrand.

Recall also that the Poisson distribution, say, is a discrete probability distribution and hence integrating it (in the conventional sense) makes no sense. A discrete probability distribution have a probability mass function (PMF) and not a PDF which actually return probabilities. In that case, it should sum to one.

Consider:

dpois(0.5, lambda = 2)
#[1] 0
#Warning message:
#In dpois(0.5, lambda = 2) : non-integer x = 0.500000

Summing from 0 to a 'very' large number (i.e. over the support of the Poisson distribution):

sum(dpois(0:1000000, lambda = 2)) 
#[1] 1

Upvotes: 6

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