Re: [R] "rbinom" : Does randomness preclude precision?

From: Charles Annis, P.E. <Charles.Annis_at_statisticalengineering.com>
Date: Wed, 28 May 2008 10:26:50 -0400

What do you mean by "... *eventual* nature of the distribution?" If you simulated 100 samples, would you expect to see 1.5 successes? Or 1? Or 2? How many, in your thinking, is "eventual?"

Charles Annis, P.E.

Charles.Annis_at_StatisticalEngineering.com phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com  

-----Original Message-----
From: r-help-bounces_at_r-project.org [mailto:r-help-bounces_at_r-project.org] On Behalf Of Philip Twumasi-Ankrah
Sent: Wednesday, May 28, 2008 9:52 AM
To: ted.harding_at_manchester.ac.uk
Cc: r-help_at_r-project.org
Subject: Re: [R] "rbinom" : Does randomness preclude precision?

Teds reply is a bit comforting and as indicated in my post, I am resorting to using "sample" but as an academic issue, does randomness preclude precision?

Randomness should be in the sequence of zeros and ones and how they are simulated at each iteration of the process but not in the eventual nature of the distribution.

I mean if I simulated a Normal (0, 1) and got a Normal(1.5, 2) these would be very different distributions. It is the same with simulating a Binomial(1, p=0.15) and getting Binomial(1, 0.154)

Ted.Harding_at_manchester.ac.uk wrote: On 28-May-08 12:53:26, Philip Twumasi-Ankrah wrote:
> I am trying to simulate a series of ones and zeros (1 or 0) and I am
> using "rbinom" but realizing that the number of successes expected is
> not accurate. Any advice out there.
>
> This is the example:
>
> N<-500
> status<-rbinom(N, 1, prob = 0.15)
> count<-sum(status)
>
> 15 percent of 500 should be 75 but what I obtain from the "count"
> variable is 77 that gives the probability of success to be 0.154. Not
> very good.

The difference (77 - 75 =2) is well within the likely sampling variation when 500 values are sampled independently with P(1)=0.15:

The standard deviation of the resulting number of 1s is sqrt(500*0.15*0.85) = 7.98, so the difference of 2 is only 1/4 of a standard deviation, hence very likely to be equalled or exceeded.

Your chance of getting exactly 75 by this method is quite small:

  dbinom(75,500,0.15)
  [1] 0.04990852

and your chance of being 2 or more off your target is

  1 - sum(dbinom((74:76),500,0.15))
  [1] 0.8510483

> Is there another way beyond using "sample" and "rep" together?

It looks as though you are seeking to obtain exactly 75 1s, randomly situated, the rest being 0s, so in effect you do need to do something on the lines of "sample" and "rep". Hence, something like

  status <- rep(0,500)
  status[sample((1:500),75,replace=FALSE)] <- 1

Hoping this helps,
Ted.



E-Mail: (Ted Harding)
Fax-to-email: +44 (0)870 094 0861
Date: 28-May-08                                       Time: 14:19:24
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