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I am trying to learn how to fit a probability distribution to a vector of data, using the program R, but there are a lot of potential probability distributions to use! So my question is, how do I find the best distribution for my data, and how do I prove that I have picked the right distribution? Can I acquire AIC values for a whole set of different distributions?

The data are observational count data of bees visiting flowers. Each species has a certain number of visits, hence the differing frequencies. The goal is to find the best distribution to describe the bee visitation, show that I have selected the right one, and then use that distribution to sample from randomly for a set of simulations.

Here is what the data looks like, it is a vector of count observations. It is zero inflated, with a long tailed distribution (maybe zero-inflated negative binomial?).

i.vec=c(0,63,1,4,1,44,2,2,1,0,1,0,0,0,0,1,0,0,3,0,0,2,0,0,0,0,0,2,0,0,0,0,
0,0,0,0,0,0,0,0,6,1,11,1,1,0,0,0,2)

And here are some basic parameters that I have calculated. I am using standard deviation for sigma, and phi is the proportion of zeroes in the data.

m=mean(i.vec)
#[1] 3.040816
sig=sd(i.vec)
#[1] 10.86078
tab<-table(i.vec)
zero.prop<-as.numeric(tab[1])/sum(as.numeric(tab))
#[1] 0.6122449

As you can see, the standard deviation is much greater than the mean, and I have a very high proportion of zeroes.

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    $\begingroup$ It would be helpful to have some context: where do the data come from? Why do you want to fit a distribution? To me, it's the 11, 44, and 63, rather than the 0's, that are unusual. $\endgroup$ – Karl Aug 26 '11 at 1:59
  • $\begingroup$ I will add in some information to my question. $\endgroup$ – Laura Aug 28 '11 at 22:45
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You can use Vuong test in pscl package to compare non-nested models. Here is an example

> m1 <- zeroinfl(i.vec ~ 1 | 1, dist = "negbin")
> summary(m1)

Call:
zeroinfl(formula = i.vec ~ 1 | 1, dist = "negbin")

Pearson residuals:
    Min      1Q  Median      3Q     Max 
-0.3730 -0.3730 -0.3730 -0.2503  7.3544 

Count model coefficients (negbin with log link):
            Estimate Std. Error z value Pr(>|z|)    
(Intercept)   1.1122     0.3831   2.903  0.00369 ** 
Log(theta)   -1.9256     0.2839  -6.784 1.17e-11 ***

Zero-inflation model coefficients (binomial with logit link):
            Estimate Std. Error z value Pr(>|z|)
(Intercept)   -9.815     96.462  -0.102    0.919
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

Theta = 0.1458 
Number of iterations in BFGS optimization: 579 
Log-likelihood: -80.51 on 3 Df


> m2 <- zeroinfl(i.vec ~ 1 | 1, dist = "poisson")
> summary(m2)

Call:
zeroinfl(formula = i.vec ~ 1 | 1, dist = "poisson")

Pearson residuals:
    Min      1Q  Median      3Q     Max 
-0.7242 -0.7242 -0.7242 -0.4860 14.2795 

Count model coefficients (poisson with log link):
            Estimate Std. Error z value Pr(>|z|)    
(Intercept)  2.05911    0.08205    25.1   <2e-16 ***

Zero-inflation model coefficients (binomial with logit link):
            Estimate Std. Error z value Pr(>|z|)
(Intercept)   0.4561     0.2933   1.555     0.12
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

Number of iterations in BFGS optimization: 11 
Log-likelihood: -233.7 on 2 Df


> vuong(m1, m2)
Vuong Non-Nested Hypothesis Test-Statistic: 1.946095 
(test-statistic is asymptotically distributed N(0,1) under the
 null that the models are indistinguishible)
in this case:
model1 > model2, with p-value 0.02582165

Vuong test also suggests that the zero-inflated negative binomial provides a better fit to your data compared to the ordinary negative binomial (not shown here, but you can fit both models and compare them).

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I don't think you necessarily need to inflate zeros.

Your data seem quite consistent with a negative binomial:

> library(MASS)
> table(rnegbin(49,mu=3.1,theta=0.075))

 0  1  2  3  5 18 20 21 31 61 
36  4  2  1  1  1  1  1  1  1 
> table(i.vec)
i.vec
 0  1  2  3  4  6 11 44 63 
30  8  5  1  1  1  1  1  1 

So lets get some estimates:

> mean(i.vec)
[1] 3.040816
> theta.ml(i.vec,3.041)
[1] 0.145777
attr(,"SE")
[1] 0.04136887

So let's just take a look at mu = 3.041 and theta at say 0.14 (there's lots of uncertainty in theta here):

Here's three random samples from that distribution:

> table(rnegbin(49,mu=3.041,theta=0.14))

 0  1  2  3  4  5  6  7  8 18 49 
30  5  1  3  2  2  1  2  1  1  1 
> table(rnegbin(49,mu=3.041,theta=0.14))

 0  1  2  3  4  7  9 15 29 31 47 56 
33  2  4  1  1  1  1  2  1  1  1  1 
> table(rnegbin(49,mu=3.041,theta=0.14))

 0  1  2  3  4  7  9 12 16 48 66 
36  4  1  1  1  1  1  1  1  1  1 

They look similar enough to your data. Negative binomial seems at least plausible.

You may like to play with the functions in MASS

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  • $\begingroup$ Thank you! It won't let me vote up, but I appreciate your help. $\endgroup$ – Laura Aug 28 '11 at 22:47
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Im not sure you can do much better than just plugin the empirical measure in that case, without further information on your data (especially since you have very few observations). And in that case the variance of your error should be of the order of the inverse of the number of your observations (via Efron-Stein).

Maybe you could use some convolution based estimator (like the density function in R, but with an integer supported kernel) to smooth things a little. Or see things as a mixture. But theres no reason to do so if you have no idea about where your data comes from.

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  • $\begingroup$ Thank you! It won't let me vote up, but I appreciate your help. $\endgroup$ – Laura Aug 28 '11 at 22:47

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